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10.3: Re-examining the Quadratic Formula - Mathematics


10.3: Re-examining the Quadratic Formula - Mathematics

Mathematics Part I (solutions) for Class 10 Math Chapter 2 - Quadratic Equations

Mathematics Part I (solutions) Solutions for Class 10 Math Chapter 2 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 10 students for Math Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I (solutions) Book of Class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I (solutions) Solutions. All Mathematics Part I (solutions) Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 34:

Question 1:

Write any two quadratic equations.

Answer:

Two quadratic equations are
x 2 + 10 x - 200 = 0 and x 2 + 5 x - 6 = 0 .

Page No 34:

Question 2:

Decide which of the following are quadratic equations.
(1) x 2 + 5x &ndash 2 = 0
(2) y 2 = 5y &ndash 10
(3) y 2 + 1 y = 2
(4) x + 1 x = - 2
(5) (m + 2) (m &ndash 5) = 0
(6) m 3 + 3m 2 &ndash 2 = 3m 3

Answer:

(1) x 2 + 5x &ndash 2 = 0
Only one variable x.
Maximum index = 2
So, it is a quadratic equation.

(2) y 2 = 5y &ndash 10
Only one variable y.
Maximum index = 2
So, it is a quadratic equation.

(3) y 2 + 1 y = 2
⇒ y 3 + 1 = 2 y
Only one variable y.
Maximum index = 3
So, it is not a quadratic equation.

(4) x + 1 x = - 2
⇒ x 2 + 1 = - 2 x
Only one variable x.
Maximum index = 2
So, it is a quadratic equation.

(5) (m + 2) (m &ndash 5) = 0
⇒ m 2 - 3 m - 10 = 0
Only one variable m.
Maximum index = 2
So, it is a quadratic equation.

(6) m 3 + 3m 2 &ndash 2 = 3m 3
Only one variable m.
Maximum index = 3
So, it is not a quadratic equation.

Page No 34:

Question 3:

Answer:

(1) 2y = 10 &ndash y 2
⇒ y 2 + 2 y = 10 ⇒ y 2 + 2 y - 10 = 0
So, it is of the form ax 2 + bx + c = 0 where a = 1, b = 2 and c = &minus10.

(2) (x &ndash 1) 2 = 2x + 3
⇒ x 2 - 2 x + 1 = 2 x + 3 ⇒ x 2 - 4 x - 2 = 0
So, it is of the form ax 2 + bx + c = 0 where a = 1, b = &minus4 and c = &minus2.

(3) x 2 + 5x = &ndash(3 &ndash x)
⇒ x 2 + 5 x + 3 - x = 0 ⇒ x 2 + 4 x + 3 = 0
So, it is of the form ax 2 + bx + c = 0 where a = 1, b = 4 and c = 3.

(4) 3m 2 = 2m 2 &ndash 9
⇒ 3 m 2 - 2 m 2 + 9 = 0 ⇒ m 2 + 0 m + 9 = 0
So, it is of the form ax 2 + bx + c = 0 where a = 1, b = 0 and c = 9.

(5) p(3+6p) = &ndash5
⇒ 3 p + 6 p 2 + 5 = 0 ⇒ 6 p 2 + 3 p + 5 = 0
So, it is of the form ax 2 + bx + c = 0 where a = 6, b = 3 and c = 5.

(6) x 2 &ndash 9 = 13
⇒ x 2 - 9 - 13 = 0 ⇒ x 2 - 22 = 0 ⇒ x 2 + 0 x - 22 = 0
So, it is of the form ax 2 + bx + c = 0 where a = 1, b = 0 and c = &minus22.

Page No 34:

Question 4:

Determine whether the values given against each of the quadratic equation are the roots of the equation.
(1) x 2 + 4x &ndash 5 = 0 , x = 1, &ndash1
(2) 2m 2 &ndash 5m = 0, m = 2 ,   5 2

Answer:

(1) x 2 + 4x &ndash 5 = 0 , x = 1, &ndash1
F o r   x = 1 1 2 + 4 1 - 5 = 0 ⇒ 1 + 4 - 5 = 0 ⇒ 5 - 5 = 0
So, x = 1 is a solution of the given equation.
For x = &ndash1
- 1 2 + 4 - 1 - 5 = 0 ⇒ 1 - 4 - 5 = 0 ⇒ - 3 - 5 = 0 ⇒ - 8 ≠ 0
So, x = &ndash1 is not a solution of the given equation.
Thus, only x = 1 is root of the given equation.

(2) 2m 2 &ndash 5m = 0, m = 2 ,   5 2
W h e n   m = 2 2 2 2 - 5 2 = 0 ⇒ 8 - 10 = 0 ⇒ - 2 ≠ 0
So, m = 2 is not a solution of the given equation.
When   m = 5 2 2 5 2 2 - 5 5 2 = 0 ⇒ 25 2 - 25 2 = 0
So, m = 5 2 is a solution of the given quadratic equation.
Thus, only m = 5 2 is a root of the given quadratic equation.

Page No 34:

Question 5:

Answer:

Given x = 3 is a root of equation kx 2 &ndash 10 x + 3 = 0
So, x = 3 must satisfy the given quadration equation.
k 3 2 - 10 3 + 3 = 0 ⇒ 9 k - 30 + 3 = 0 ⇒ 9 k - 27 = 0 ⇒ 9 k = 27 ⇒ k = 27 9 = 3
Thus, k = 3.

Page No 34:

Question 6:

One of the roots of equation 5m 2 + 2m + k = 0 is - 7 5 . Complete the following activity to find the value of 'k'.

Answer:

- 7 5 is a root of the quadratic equation 5m 2 + 2m + k = 0.
&there4 Put m = - 7 5 in the equation.
5 × - 7 5 2 + 2 × - 7 5 + k = 0 49 5 + - 14 5 + k = 0 7 + k = 0 k = - 7

Page No 36:

Question 1:

Solve the following quadratic equations by factorisation.
(1) x 2 &ndash 15x + 54 = 0
(2) x 2 + x &ndash 20 = 0
(3) 2y 2 + 27y + 13 = 0
(4) 5m 2 = 22m + 15
(5) 2x 2 &ndash 2x + 1 2 = 0
(6) 6 x - 2 x = 1
(7) 2 x 2 + 7 x + 5 2 = 0
to solve this quadratic equation by factorisation, complete the following activity.
(8) 3 x 2 - 2 6 x + 2 = 0
(9) 2 m m - 24 = 50
(10) 25 m 2 = 9
(11) 7 m 2 = 21 m
(12) m 2 - 11 = 0

Answer:

(1) x 2 &ndash 15x + 54 = 0
⇒ x 2 - 9 x - 6 x + 54 = 0 ⇒ x x - 9 - 6 x - 9 = 0 ⇒ x - 6 x - 9 = 0 ⇒ x - 6 = 0   or   x - 9 = 0 ⇒ x = 6   or   x = 9  
So, 6 and 9 are the roots of the given quadratic equation.

(2) x 2 + x &ndash 20 = 0
⇒ x 2 + 5 x - 4 x - 20 = 0 ⇒ x x + 5 - 4 x + 5 = 0 ⇒ x - 4 x + 5 = 0 ⇒ x - 4 = 0   or   x + 5 = 0 ⇒ x = 4   or   x = - 5  
So, 4 and - 5 are the roots of the given quadratic equation.

(3) 2y 2 + 27y + 13 = 0
2 y 2 + 26 y + y + 13 = 0 ⇒ 2 y y + 13 + y + 13 = 0 ⇒ 2 y + 1 y + 13 = 0 ⇒ 2 y + 1 = 0   or   y + 13 = 0 ⇒ y = - 1 2   or   y = - 13
So, - 1 2   and   - 13 are the roots of the given quadratic equation.

(4) 5m 2 = 22m + 15
⇒ 5 m 2 - 22 m - 15 = 0 ⇒ 5 m 2 - 25 m + 3 m - 15 = 0 ⇒ 5 m m - 5 + 3 m - 5 = 0 ⇒ 5 m + 3 m - 5 = 0 ⇒ 5 m + 3 = 0   or   m - 5 = 0 ⇒ m = - 3 5   or   m = 5
So, - 3 5   and   5 are the roots of the given quadratic equation.

(5) 2x 2 &ndash 2x + 1 2 = 0
⇒ 4 x 2 - 4 x + 1 = 0 ⇒ 2 x 2 - 2 × 1 × 2 x + 1 = 0 U sin g   t h e   i d e n t i t y ,   a - b 2 = a 2 + b 2 - 2 a b 2 x 2 - 2 × 1 × 2 x + 1 = 0 ⇒ 2 x - 1 2 = 0 ⇒ x = 1 2 , 1 2
So, 1 2   and   1 2 are the roots of the given quadratic equation.

(6) 6 x - 2 x = 1
⇒ 6 x 2 - 2 = 1 x ⇒ 6 x 2 - 1 x - 2 = 0 ⇒ 6 x 2 - 4 x + 3 x - 2 = 0 ⇒ 2 x 3 x - 2 + 1 3 x - 2 = 0 ⇒ 3 x - 2 2 x + 1 = 0 ⇒ 3 x - 2 = 0   or   2 x + 1 = 0 ⇒ x = 2 3   or   x = - 1 2  
2 3   and   - 1 2 are the roots of the given quadratic equation.

(7) 2 x 2 + 7 x + 5 2 = 0
2 x 2 + 5 x + 2 x + 5 2 = 0 x 2 x + 5 + 2 2 x + 5 = 0 2 x + 5 x + 2 = 0 2 x + 5 = 0   or   x + 2 = 0 ∴ x = - 5 2   or   x = - 2 ∴ - 5 2     and   - 2   are   the   roots   of   the   equation .

(8) 3 x 2 - 2 6 x + 2 = 0
⇒ 3 x 2 - 6 x - 6 x + 2 = 0 ⇒ 3 x 3 x - 2 - 2 3 x - 2 = 0 ⇒ 3 x - 2 3 x - 2 = 0 ⇒ x = 2 3 , 2 3

(9) 2 m m - 24 = 50
⇒ m m - 24 = 25 ⇒ m 2 - 24 m = 25 ⇒ m 2 - 24 m - 25 = 0 ⇒ m 2 - 25 m + m - 25 = 0 ⇒ m m - 25 + 1 m - 25 = 0 ⇒ m + 1 m - 25 = 0 ⇒ m = - 1 , 25

(10) 25 m 2 = 9
⇒ 25 m 2 - 9 = 0 ⇒ 5 m 2 - 3 2 = 0 ⇒ 5 m - 3 5 m + 3 = 0                                       ∵ x - a x + a = x 2 - a 2 ⇒ 5 m - 3 = 0   o r   5 m + 3 = 0 ⇒ m = 3 5 , - 3 5  

(11) 7 m 2 = 21 m
⇒ 7 m 2 - 21 m = 0 ⇒ 7 m m - 3 = 0 ⇒ 7 m = 0   or   m - 3 = 0 ⇒ m = 0   or   m = 3

(12) m 2 - 11 = 0
⇒ m 2 - 11 2 = 0 ⇒ m - 11 m + 11 = 0 ⇒ m = 11 ,   - 11

Page No 39:

Question 1:

Solve the following quadratic equations by completing the square method.
(1) x 2 + x &ndash 20 = 0
(2) x 2 + 2x &ndash 5 = 0
(3) m 2 &ndash 5m = &ndash3
(4) 9y 2 &ndash 12y + 2 = 0
(5) 2y 2 + 9y +10 = 0
(6) 5x 2 = 4x + 7

Answer:

(1) x 2 + x &ndash 20 = 0
x 2 + x + 1 2 2 - 1 2 2 - 20 = 0 ⇒ x + 1 2 2 = 1 2 2 + 20 ⇒ x + 1 2 2 = 1 4 + 20 ⇒ x + 1 2 2 = 81 4 ⇒ x + 1 2 = 9 2   or   x + 1 2 = - 9 2   ⇒ x = 4   or   x = - 5
(2) x 2 + 2x &ndash 5 = 0
⇒ x 2 + 2 x + 2 2 2 - 2 2 2 - 5 = 0 ⇒ x 2 + 2 x + 1 - 1 - 5 = 0 ⇒ x + 1 2 - 6 = 0 ⇒ x + 1 2 = 6 ⇒ x + 1 2 = 6 2 ⇒ x + 1 = 6   or   x + 1 = - 6 ⇒ x = 6 - 1   or   x = - 6 - 1

(3) m 2 &ndash 5m = &ndash3
⇒ m 2 - 5 m + - 5 2 2 - - 5 2 2 = - 3 ⇒ m 2 - 5 m + 25 4 - 25 4 = - 3 ⇒ m - 5 2 2 = - 3 + 25 4 ⇒ m - 5 2 2 = 13 4 ⇒ m - 5 2 2 = 13 2 2 ⇒ m - 5 2 = 13 2   or   m - 5 2 = - 13 2 ⇒ m = 13 2 + 5 2   or   m = - 13 2 + 5 2 ⇒ m = 13 + 5 2   or   m = 5 - 13 2

(4) 9y 2 &ndash 12y + 2 = 0
⇒ y 2 - 12 9 y + 2 9 = 0 ⇒ y 2 - 4 3 y + 2 9 = 0 ⇒ y 2 - 4 3 y + 4 3 2 2 - 4 3 2 2 + 2 9 = 0 ⇒ y 2 - 4 3 y + 2 3 2 - 2 3 2 + 2 9 = 0
⇒ y 2 - 4 3 y + 4 9 - 4 9 + 2 9 = 0 ⇒ y - 2 3 2 - 2 9 = 0 ⇒ y - 2 3 2 = 2 9 ⇒ y - 2 3 2 = 2 3 2
⇒ y - 2 3 = 2 3   or   y - 2 3 = - 2 3 ⇒ y = 2 3 + 2 3   or   y = - 2 3 + 2 3 ⇒ y = 2 + 2 3   or   y = - 2 + 2 3

(5) 2y 2 + 9y +10 = 0
⇒ y 2 + 9 2 y + 5 = 0 ⇒ y 2 + 9 2 y + 9 2 2 2 - 9 2 2 2 + 5 = 0 ⇒ y 2 + 9 2 y + 9 4 2 - 9 4 2 + 5 = 0 ⇒ y + 9 4 2 = 81 16 - 5 ⇒ y + 9 4 2 = 1 16
⇒ y + 9 4 2 = 1 4 2 ⇒ y + 9 4 = 1 4   or   y + 9 4 = - 1 4 ⇒ y = 1 4 - 9 4 = - 8 4 = - 2   or   y = - 1 4 - 9 4 = - 10 4 = - 5 2 ⇒ y = - 2 , - 5 2
(6) 5x 2 = 4x + 7
⇒ 5 x 2 - 4 x - 7 = 0 ⇒ x 2 - 4 5 x - 7 5 = 0 ⇒ x 2 - 4 5 x + 4 5 2 2 - 4 5 2 2 - 7 5 = 0 ⇒ x 2 - 4 5 x + 2 5 2 - 2 5 2 - 7 5 = 0 ⇒ x - 2 5 2 = 7 5 + 4 25 = 35 + 4 25 = 39 25 ⇒ x - 2 5 2 = 39 25
⇒ x - 2 5 2 = 39 5 2 ⇒ x - 2 5 = 39 5   or   x - 2 5 = - 39 5 ⇒ x = 2 + 39 5   or   x = 2 - 39 5

Page No 43:

Question 1:

Compare the given quadratic equations to the general form and write values of a,b, c.
(1) x 2 &ndash 7x + 5 = 0
(2) 2m 2 = 5m &ndash 5
(3) y 2 = 7y

Answer:

(1) x 2 &ndash 7x + 5 = 0
General form of the quadratic equation is a x 2 + b x + c = 0 .
Comparing x 2 &ndash 7x + 5 = 0 with the general form we have a = 1, b = - 7 and c = 5.

(2) 2m 2 = 5m &ndash 5
⇒ 2 m 2 - 5 m + 5 = 0
General form of the quadratic equation is a x 2 + b x + c = 0 .
Comparing 2m 2 = 5m &ndash 5 with the general form we have a = 2, b = - 5 and c = 5.

(3) y 2 = 7y
⇒ y 2 - 7 y + 0 = 0

General form of the quadratic equation is a x 2 + b x + c = 0 .
Comparing y 2 - 7 y + 0 = 0 with the general form we have a = 1, b = - 7 and c = 0.

Page No 43:

Question 2:

Answer:

(1) x 2 + 6x + 5 = 0
On comparing with the equation a x 2 + b x + c = 0 ,
a = 1, b = 6 and c = 5
Now b 2 - 4 a c = 6 2 - 4 × 1 × 5 = 36 - 20 = 16
x = - b ± b 2 - 4 a c 2 a
x = - 6 ± 16 2 × 1 = - 6 ± 4 2 ⇒ x = - 6 + 4 2   or   x = - 6 - 4 2 ⇒ x = - 1   or   x = - 5

(2) x 2 &ndash 3x &ndash 2 = 0
On comparing with the equation a x 2 + b x + c = 0 ,
a = 1, b = - 3 and c = - 2
Now b 2 - 4 a c = - 3 2 - 4 × 1 × - 2 = 9 + 8 = 17
x = - b ± b 2 - 4 a c 2 a
x = 3 ± 17 2 × 1 = 3 ± 17 2 ⇒ x = 3 + 17 2   or   x = 3 - 17 2

(3) 3m 2 + 2m &ndash 7 = 0
On comparing with the equation a x 2 + b x + c = 0 ,
a = 3, b = 2 and c = - 7
Now b 2 - 4 a c = 2 2 - 4 × 3 × - 7 = 4 + 84 = 88
x = - b ± b 2 - 4 a c 2 a
x = - 2 ± 88 2 × 3 = - 2 ± 88 6 ⇒ x = - 2 + 88 6   or   x = - 2 - 88 6 ⇒ x = - 1 + 22 3   or   - 1 - 22 3

(4) 5m 2 &ndash 4m &ndash 2 = 0
On comparing with the equation a x 2 + b x + c = 0 ,
a = 5, b = &ndash 4and c = - 2
Now b 2 - 4 a c = - 4 2 - 4 × 5 × - 2 = 16 + 40 = 56
x = - b ± b 2 - 4 a c 2 a
x = - 4 ± 56 2 × 5 = - 4 ± 56 10 ⇒ x = - 4 + 56 10   or   x = - 4 - 56 10 ⇒ x = - 2 + 14 5   or   - 2 - 14 5

(5) y 2 + 1 3 y = 2
Multiplying the equation by 3
3 y 2 + y = 6 ⇒ 3 y 2 + y - 6 = 0
On comparing with the equation a x 2 + b x + c = 0 ,
a = 3, b = 1 and c = - 6
Now b 2 - 4 a c = 1 2 - 4 × 3 × - 6 = 1 + 72 = 73
x = - b ± b 2 - 4 a c 2 a
x = - 1 ± 73 2 × 3 = - 1 ± 73 6 ⇒ x = - 1 + 73 6   or   x = - 1 - 73 6

(6) 5x 2 + 13x + 8 = 0
On comparing with the equation a x 2 + b x + c = 0 ,
a = 5, b = 13 and c = 8
Now b 2 - 4 a c = 13 2 - 4 × 5 × 8 = 169 - 160 = 9
x = - b ± b 2 - 4 a c 2 a
x = - 13 ± 9 2 × 5 = - 13 ± 3 10 ⇒ x = - 13 + 3 10   or   x = - 13 - 3 10 ⇒ x = - 10 10   or   x = - 16 10 ⇒ x = - 1   or   x = - 8 5

Page No 44:

Question 3:

With the help of the flow chart given below solve the equation x 2 + 2 3 x + 3 = 0 using the formula.

Answer:

Compare   equations x 2 + 2 3 x + 3 = 0   and a x 2 + b x + c = 0   find   the   values   of   a , b , c .   → Find   value   of b 2 - 4 a c → Write   formula to   solve quadratic equation . → Substitute   values   of a , b , c   and   find   roots

comparing x 2 + 2 3 x + 3 = 0 with a x 2 + b x + c = 0 we get
a = 1 , b = 2 3   and   c = 3
b 2 - 4 a c = 2 3 2 - 4 × 1 × 3 = 12 - 12 = 0
Formula to solve a quadratic equation will be
x = - b ± b 2 - 4 a c 2 a ⇒ x = - 2 3 ± 0 2 × 1 = - 2 3 2 = - 3
Thus, x = - 3 , - 3 .

Page No 49:

Question 1:

Fill in the gaps and complete.


(3)If &alpha, &beta are roots of quadratic equation,

Answer:


If α   and   β are the roots of the quadratic equation then the quadratic equation can be written as
x 2 - α + β x + α β = 0
So, the quadratic equation with sum of roots as α + β = - 7 and product of roots as α β = 5 will be
x 2 + 7 x + 5 = 0

Sum of roots = - b a = - - 4 2 = 2
Product of roots = c a = - 3 2

Page No 49:

Question 2:

Find the value of discriminant.
(1) x 2 + 7x &ndash 1 = 0
(2) 2y 2 &ndash 5y + 10 = 0
(3) 2 x 2 + 4 x + 2 2 = 0

Answer:

(1) x 2 + 7x &ndash 1 = 0
Comparing the given equation with a x 2 + b x + c = 0 ,
a = 1 , b = 7   and   c = - 1
So, the discriminant b 2 - 4 a c = 7 2 - 4 × 1 × - 1 = 49 + 4 = 53

(2) 2y 2 &ndash 5y + 10 = 0
Comparing the given equation with a x 2 + b x + c = 0 ,
a = 2 , b = - 5   and   c = 10
So, the discriminant b 2 - 4 a c = - 5 2 - 4 × 2 × 10 = 25 - 80 = - 55

(3) 2 x 2 + 4 x + 2 2 = 0
Comparing the given equation with a x 2 + b x + c = 0 ,
a = 2 , b = 4   and   c = 2 2
So, the discriminant b 2 - 4 a c = 4 2 - 4 × 2 × 2 2 = 16 - 16 = 0

Page No 49:

Question 3:

Determine the nature of roots of the following quadratic equations.
(1) x 2 &ndash 4x + 4 = 0
(2) 2y 2 &ndash 7y +2 = 0
(3) m 2 + 2m + 9 = 0

Answer:

(1) x 2 &ndash 4x + 4 = 0
Comparing the given equation with the quadratic equation a x 2 + b x + c = 0 ,
a = 1 , b = - 4 , c = 4
Discriminant, △ = b 2 - 4 a c = - 4 2 - 4 × 1 × 4 = 16 - 16 = 0
Since the discriminant = 0 so, the roots of the given quadratic equation are real and equal.

(2) 2y 2 &ndash 7y +2 = 0
Comparing the given equation with the quadratic equation a x 2 + b x + c = 0 ,
a = 2 , b = - 7 , c = 2
Discriminant, △ = b 2 - 4 a c = - 7 2 - 4 × 2 × 2 = 49 - 16 = 33
Since the discriminant > 0 so, the roots of the given quadratic equation are real and unequal.

(3) m 2 + 2m + 9 = 0
Comparing the given equation with the quadratic equation a x 2 + b x + c = 0 ,
a = 1 , b = 2 , c = 9
Discriminant, △ = b 2 - 4 a c = 2 2 - 4 × 1 × 9 = 4 - 36 = - 32
Since the discriminant < 0 so, the roots of the given quadratic equation are not real.

Page No 50:

Question 4:

Form the quadratic equation from the roots given below.
(1) 0 and 4
(2) 3 and &ndash10
(3) 1 2 , - 1 2
(4) 2 - 5 , 2 + 5

Answer:

(1) 0 and 4
Sum of roots = 0 + 4 = 4
Product of roots = 0 × 4 = 0
The general form of the quadratic equation is x 2 - Sum   of   roots x + Product   of   roots = 0
So, the quadratic equation obtained is x 2 - 4 x + 0 = 0
⇒ x 2 - 4 x = 0 ⇒ x x - 4 = 0

(2) 3 and &ndash10
Sum of roots = 3 + (&ndash10) = &minus7
Product of roots = 3 × &ndash10 = &ndash30
The general form of the quadratic equation is x 2 - Sum   of   roots x + Product   of   roots = 0
So, the quadratic equation obtained is x 2 - - 7 x + - 30 = 0
x 2 + 7 x - 30 = 0

(3) 1 2 , - 1 2
Sum of roots = 1 2 + - 1 2 = 0
Product of roots = 1 2 × - 1 2 = - 1 4
The general form of the quadratic equation is x 2 - Sum   of   roots x + Product   of   roots = 0
So, the quadratic equation obtained is x 2 - 0 x + - 1 4 = 0
⇒ x 2 - 1 4 = 0

(4) 2 - 5 , 2 + 5
Sum of roots = 2 - 5 + 2 + 5 = 4
Product of roots = 2 - 5 2 + 5 = 4 - 5 = - 1
The general form of the quadratic equation is x 2 - Sum   of   roots x + Product   of   roots = 0
So, the quadratic equation obtained is x 2 - 4 x + - 1 = 0
⇒ x 2 - 4 x - 1 = 0

Page No 50:

Question 5:

Answer:

Let the roots of the quadratic equation be α   and   β .
Sum of roots = α + β
Product of roots = α β
Given that sum of roots = double the product
⇒ α + β = 2 α β . (I)
Given: x 2 &ndash 4k x + k +3 = 0 . (II)
General form of the quadratic equation is x 2 - Sum   of   roots x + Product   of   roots = 0
⇒ x 2 - α + β x + α β = 0 ⇒ x 2 - 2 α β x + α β = 0                           . . . . . III
On comparing (II) and (III) we have
4 k = 2 α β   and   k + 3 = α β
Solving these equations we have
4 k = 2 k + 3 ⇒ 2 k = k + 3 ⇒ 2 k - k = 3 ⇒ k = 3

Page No 50:

Question 6:

Answer:

&alpha, &beta are roots of y 2 &ndash 2y &ndash7 = 0
a = 1 , b = - 2 , c = - 7 a = 1 , b = - 2 , c = - 7
(1)
α + β = - b a = - - 2 1 = 2 α β = c a = - 7 1 = - 7
α 2 + β 2 = α + β 2 - 2 α β = 2 2 - 2 - 7 = 4 + 14 = 18

Page No 50:

Question 7:

The roots of each of the following quadratic equations are real and equal, find k.
(1) 3y 2 + ky +12 = 0
(2) kx (x &ndash 2) + 6 = 0

Answer:

(1) 3y 2 + ky +12 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
b 2 - 4 a c = 0 ⇒ k 2 - 4 × 3 × 12 = 0 ⇒ k 2 - 144 = 0 ⇒ k 2 = 144 ⇒ k = ± 12

(2) kx (x &ndash 2) + 6 = 0
⇒ k x 2 - 2 k x + 6 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
b 2 - 4 a c = 0 ⇒ - 2 k 2 - 4 × k × 6 = 0 ⇒ 4 k 2 - 24 k = 0 ⇒ 4 k k - 6 = 0 ⇒ 4 k = 0   or   k - 6 = 0 ⇒ k = 0   or   k = 6
But k cannot be equal to 0 since then there will not be any quadratic equation.
So, k = 6

Page No 52:

Question 1:

Product of Pragati&rsquos age 2 years ago and 3 years hence is 84. Find her present age.

Answer:

Let Pragati's present age be x years.
Her age 2 years ago = x - 2
Her age 3 years hence = x + 3
Product of Pragati&rsquos age 2 years ago and 3 years hence is 84.
x - 2 x + 3 = 84 ⇒ x 2 + x - 6 = 84 ⇒ x 2 + x - 90 = 0 x = - b ± b 2 - 4 a c 2 a ⇒ x = - 1 ± 1 2 - 4 × 1 × - 90 2 = - 1 ± 361 2 x = - 1 - 19 2   or   - 1 + 19 2 ⇒ x = - 10   or   9
But age cannot be negative so, x = 9.
Thus, Pragati's present age is 9 years.

Page No 52:

Question 2:

The sum of squares of two consecutive natural numbers is 244 find the numbers.

Answer:

Disclaimer: There is an error in the given question. Here it should be two consecutive even natural numbers.

Let the two consecutive natural numbers be x   and   x + 2 .
Sum of squares of these two consecutive natural numbers is 244
x 2 + x + 2 2 = 244 ⇒ x 2 + x 2 + 4 + 4 x = 244 ⇒ 2 x 2 + 4 x - 240 = 0 ⇒ x 2 + 2 x - 120 = 0 ⇒ x = - b ± b 2 - 4 a c 2 a ⇒ x = - 2 ± 2 2 - 4 × 1 × - 120 2 ⇒ x = - 2 ± 4 + 480 2
⇒ x = - 2 ± 484 2 ⇒ x = - 2 ± 22 2 ⇒ x = - 2 + 22 2 , - 2 - 22 2 ⇒ x = 20 2 , - 24 2 ⇒ x = 10 , - 12
But the natural number cannot be negative so,
The two numbers are 10 and 10 + 2 = 12.

Page No 52:

Question 3:

In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.

Answer:

Let the number of trees in each column be x.
Number of trees in a row = x + 5
Total trees = Number of rows × Number of columns = x x + 5 = 150
⇒ x x + 5 = 150 ⇒ x 2 + 5 x - 150 = 0 ⇒ x 2 + 15 x - 10 x - 150 = 0 ⇒ x x + 15 - 10 x + 15 = 0 ⇒ x + 15 x - 10 = 0 ⇒ x = 10 , - 15
But number of columns cannot be negative so, number of columns = 10.
Number of rows = 10 + 5 = 15

Page No 52:

Question 4:

Vivek is older than Kishor by 5 years. The sum of the reciprocals of their ages is 1 6 . Find their present ages.

Answer:

Let the present age of Kishor be x years.
Vivek is older than Kishor by 5 years.
So, Vivek's age will be x + 5.
The sum of the reciprocals of their ages is 1 6 .
1 x + 1 x + 5 = 1 6 ⇒ x + 5 + x x x + 5 = 1 6 ⇒ 5 + 2 x x 2 + 5 x = 1 6 ⇒ 30 + 12 x = x 2 + 5 x ⇒ x 2 - 7 x - 30 = 0 ⇒ x 2 - 10 x + 3 x - 30 = 0 ⇒ x x - 10 + 3 x - 10 = 0 ⇒ x + 3 x - 10 = 0 ⇒ x = - 3 , 10
But age cannot be negative so, age of Kishore is 10 years and that of Vivek is 10 + 5 = 15 years.

Page No 52:

Question 5:

Answer:

Let the score in the first test be x.
Suyash's score in second test is 10 more than that in the first.
So, his score in the second test = x + 10
5 times the score of the second test is the same as square of the score in the first test.
5 x + 10 = x 2 ⇒ 5 x + 50 = x 2 ⇒ x 2 - 5 x - 50 = 0 ⇒ x 2 - 10 x + 5 x - 50 = 0 ⇒ x x - 10 + 5 x - 10 = 0 ⇒ x + 5 x - 10 = 0 ⇒ x = 10 , - 5
The score is not negative so, Suyash scored 10 marks in first test.

Page No 52:

Question 6:

Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is Rs 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is Rs 600, find production cost of one pot and number of pots he makes per day.

Answer:

Let the number of pots Mr. Kasam makes in one day be x.
Production cost of each pot is Rs 40 more than 10 times total number of pots, he makes in one day = 10x + 40
Production cost of all pots per day is Rs 600
(10x + 40)x = 600
10 x 2 + 40 x = 600 ⇒ x 2 + 4 x - 60 = 0 ⇒ x 2 + 10 x - 6 x - 60 = 0 ⇒ x x + 10 - 6 x + 10 = 0 ⇒ x - 6 x + 10 = 0 ⇒ x = 6 , - 10
But the number of pots cannot be negative so,
x ≠ - 10 ⇒ x = 6
Production cost of 1 pot = 10 × 6 + 40 = 60 + 40 = Rs   100

Page No 52:

Question 7:

Pratik takes 8 hours to travel 36 km down stream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.

Answer:

Let the speed of water current = y km/h
Speed of boat in still water = 12 km/h
Speed upstream = 12 - y
Speed downstream = 12 + y
Distance travelled = 36 km downstream and 36 km upstream
Speed = Distance Time ⇒ Time = Distance Speed
36 12 - y + 36 12 + y = 8 ⇒ 36 1 12 - y + 1 12 + y = 8 ⇒ 12 + y + 12 - y 144 - y 2 = 1 4 ⇒ 24 144 - y 2 = 1 4 ⇒ 144 - y 2 = 24 × 4 ⇒ y 2 = 36 ⇒ y 2 - 36 = 0 ⇒ y = ± 6
Speed cannot be negative so, the speed of water current = 6 km/h.

Page No 52:

Question 8:

Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.

Answer:

Let the number of days taken by Nishu be x.
So, number of days taken by Pintu will be x + 6
Nishu's 1 day work = 1 x
Pintu's one day work = 1 x + 6
Work done together in 4 days = 1 4
1 x + 1 x + 6 = 1 4 ⇒ x + 6 + x x x + 6 = 1 4 ⇒ 6 + 2 x x 2 + 6 x = 1 4 ⇒ 24 + 8 x = x 2 + 6 x ⇒ x 2 - 2 x - 24 = 0 ⇒ x 2 - 6 x + 4 x - 24 = 0 ⇒ x - 6 x + 4 = 0 ⇒ x = 6 , - 4
Number of days cannot be negative so, x = 6.
Thus, number of days taken by Nishu = 6 and that by Pintu = 6 + 6 = 12.

Page No 52:

Question 9:

If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and diviser.

Answer:

Let the divisor be x.
Dividend = 460
Quotient = 5x + 6
Remainder = 1
We know
Divident = Divisor × Quotient + Remainder
460 = x 5 x + 6 + 1 ⇒ 460 = 5 x 2 + 6 x + 1 ⇒ 5 x 2 + 6 x - 459 = 0 ⇒ x = - b ± b 2 - 4 a c 2 a ⇒ x = - 6 ± 6 2 - 4 × 5 × - 459 2 × 5 ⇒ x = - 6 ± 36 + 9180 10 ⇒ x = - 6 ± 9216 10
⇒ x = - 6 ± 96 10 ⇒ x = - 6 + 96 10 , - 6 - 96 10 ⇒ x = 9 , - 10 . 2
Thus, divisor = 9, quotient = 5x + 6 = 5 × 9 + 6 = 45 + 6 = 51

Page No 52:

Question 10:

In the adjoining fig. □ ABCD is a trapezium AB || CD and its area is 33 cm 2 . From the information given in the figure find the lengths of all sides of the □ ABCD. Fill in the empty boxes to get the solution.

Answer:

□ ABCD is a trapezium.
AB || CD
A( □ ABCD) = 1 2 AB + CD × AM
33 = 1 2 x + 2 x + 1 × x - 4 ∴ 33 × 2 = 3 x + 1 × x - 4 ∴ 3 x 2 + 11 x - 70 = 0 ∴ 3 x x - 7 + 10 x - 7 = 0 ∴ 3 x + 10 x - 7 = 0 ∴ 3 x + 10 = 0   or   x - 7 = 0 ∴ x = - 10 3   or   x = 7
But length is never negative.
∴ x ≠ - 10 3       ∴ x = 7
AB = 7, CD = 15, BC = 5

Page No 53:

Question 1:

Choose the correct answers for the following questions.
(1) Which one is the quadratic equation ?

Answer:

(1) The general form of a quadratic equation is a x 2 + b x + c = 0 .
For (A), 5 x - 3 = x 2
5 - 3 x = x 3
Thus, (A) is not quadratic equation.
For (B) x(x + 5) = 2
x 2 + 5 x - 2 = 0
Thus, (B) is a qudratic equation.
(C) is also not a quadratic equation.
(D) 1 x 2 x + 2 = x
x + 2 = x 3 which is also not quadratic.
Hence, the correct answer is option (B).

(2) (A) x 2 + 4x = 11 + x 2
⇒ 4 x = 11
Thus, (A) is not a quadratic equation.
(B) and (C) can be written as x 2 - 4 x + 0 = 0 and 5 x 2 - 90 + 0 = 0 respectively.
So, B and C are quadratic equations.
(D) 2x &ndash x 2 = x 2 + 5 can be written as 2 x 2 - 2 x + 5 = 0 .
So, it also forms a quadratic equation.
Hence, the correct answer is option (A).

(3) Given quadratic equation is x 2 + kx + k = 0
For real and equal roots, D = 0
b 2 - 4 a c = 0 ⇒ k 2 - 4 × 1 × k = 0 ⇒ k 2 - 4 k = 0 ⇒ k k - 4 = 0 ⇒ k = 0 , k = 4
Hence, the correct answer is option C.

(4) 2 x 2 - 5 x + 2 = 0
a = 2 , b = - 5 , c = 2 D = b 2 - 4 a c = - 5 2 - 4 × 2 × 2 = 25 - 8 = 17
Hence, the correct answer is option B.

(5) The roots of the quadratic equation x 2 &ndash 8x + 15 = 0 are
x 2 - 5 x - 3 x + 15 = 0 ⇒ x x - 5 - 3 x - 5 = 0 ⇒ x - 3 x - 5 = 0 ⇒ x = 3 , 5
Hence, the correct answer is option B.

(6) For the equation 3x 2 + 15x + 3 = 0
Sum   of   roots = - b a = - 15 3 = - 5
Hence, the correct answer is option D.

(7) For the given quadratic equation 5 m 2 - 5 m + 5 = 0
D = b 2 - 4 a c = - 5 2 - 4 × 5 × 5 = 5 - 20 = - 15
Since D < 0 so, the roots are not real.
Hence, the correct answer is option C.

(8) For the equation, x 2 + mx &ndash 5 = 0
one of the roots is 2. So, 2 should satisfy the given equation.
2 2 + m × 2 - 5 = 0 ⇒ 4 + 2 m - 5 = 0 ⇒ 2 m - 1 = 0 ⇒ m = 1 2
Hence, the correct answer is option C.

Page No 54:

Question 2:

Which of the following equations is quadratic ?
(1) x 2 + 2 x + 11 = 0
(2) x 2 - 2 x + 5 = x 2
(3) x + 2 2 = 2 x 2

Answer:

General form of a quadratic equation is
a x 2 + b x + c = 0
(1) x 2 + 2 x + 11 = 0 is in the form of the general quadratic equation.
(2) x 2 - 2 x + 5 = x 2
⇒ - 2 x + 5 = 0
It is not in the form of quadratic equation.
(3) x + 2 2 = 2 x 2
⇒ x 2 + 4 + 4 x = 2 x 2 ⇒ x 2 - 4 x - 4 = 0
It is also in the form of a quadratic equation.
Hence, (1) and (3) are quadratic equations.

Page No 54:

Question 3:

Find the value of discriminant for each of the following equations.
(1) 2 y 2 - y + 2 = 0
(2) 5 m 2 - m = 0
(3) 5 x 2 - x - 5 = 0

Answer:

(1) 2 y 2 - y + 2 = 0
For the quadratic equation, a x 2 + b x + c = 0
D = b 2 - 4 a c
Here,
a = 2 , b = - 1 , c = 2 D = - 1 2 - 4 × 2 × 2 = 1 - 16 = - 15

(2) 5 m 2 - m = 0
For the quadratic equation, a x 2 + b x + c = 0
D = b 2 - 4 a c
Here,
m = 5 , b = - 1 , c = 0 D = - 1 2 - 4 × 5 × 0 = 1

(3) 5 x 2 - x - 5 = 0
For the quadratic equation, a x 2 + b x + c = 0
D = b 2 - 4 a c
Here,
a = 5 , b = - 1 , c = - 5 D = - 1 2 - 4 × 5 × - 5 = 1 + 20 = 21

Page No 54:

Question 4:

One of the roots of quadratic equation 2 x 2 + k x - 2 = 0 is &ndash2. find k.

Answer:

2 x 2 + k x - 2 = 0
One of the roots is &ndash2 so, it will satisfy the given equation.
2 - 2 2 + k - 2 - 2 = 0 ⇒ 8 - 2 k - 2 = 0 ⇒ 6 - 2 k = 0 ⇒ 6 = 2 k ⇒ k = 3

Page No 54:

Question 5:

Two roots of quadratic equations are given frame the equation.
(1) 10 and &ndash10
(2) 1 - 3 5   and   1 + 3 5
(3) 0 and 7

Answer:

(1) 10 and &ndash10
Sum of roots = 10 + (&ndash10) = 0
Product of roots = 10 × - 10 = - 100
The general form of the quadratic equation is
x 2 - Sum   of   roots x + product   of   roots = 0
So, the quadratic equation will be
x 2 - 0 x - 100 = 0 ⇒ x 2 - 100 = 0

(2) 1 - 3 5   and   1 + 3 5
Sum of roots = 1 - 3 5 + 1 + 3 5 = 2
Product of roots = 1 - 3 5 1 + 3 5 = 1 - 45 = - 44
The general form of the quadratic equation is
x 2 - Sum   of   roots x + product   of   roots = 0
So, the quadratic equation will be
x 2 - 2 x - 44 = 0

(3) 0 and 7
Sum of roots = 0 + 7 = 7
Product of roots = 0 × 7 = 0
The general form of the quadratic equation is
x 2 - Sum   of   roots x + product   of   roots = 0
So, the quadratic equation will be
x 2 - 7 x + 0 = 0 ⇒ x 2 - 7 x = 0

Page No 54:

Question 6:

Determine the nature of roots for each of the quadratic equation.
(1) 3 x 2 - 5 x + 7 = 0
(2) 3 x 2 + 2 x - 2 3 = 0
(3) m 2 - 4 x - 3 = 0

Answer:

(1) 3 x 2 - 5 x + 7 = 0
In order to find the nature of the roots we need to find the discriminant.
D = b 2 - 4 a c a = 3 , b = - 5 , c = 7 D = - 5 2 - 4 × 3 × 7 = 25 - 84 = - 59
Since, the D < 0 so, there is no real root.

(2) 3 x 2 + 2 x - 2 3 = 0
In order to find the nature of the roots we need to find the discriminant.
D = b 2 - 4 a c a = 3 , b = 2 , c = - 2 3 D = 2 2 - 4 × 3 × - 2 3 = 2 + 24 = 26 > 0
Since, the D > 0 so, real and unequal roots.

(3) m 2 - 4 x - 3 = 0
In order to find the nature of the roots we need to find the discriminant.
D = b 2 - 4 a c a = 1 , b = - 4 , c = - 3 D = - 4 2 - 4 × 1 × - 3 = 16 + 12 = 28
Since, the D > 0 so, real and unequal roots.

Page No 54:

Question 7:

Solve the following quadratic equations.
(1) 1 x + 5 = 1 x 2
(2) x 2 - 3 x 10 - 1 10 = 0
(3) 2 x + 3 2 = 25
(4) m 2 + 5 m + 5 = 0
(5) 5 m 2 + 2 m + 1 = 0
(6) x 2 - 4 x - 3 = 0

Answer:

(1) 1 x + 5 = 1 x 2
⇒ x 2 = x + 5 ⇒ x 2 - x - 5 = 0 ⇒ x = - - 1 ± - 1 2 - 4 × 1 × - 5 2 × 1 ⇒ x = 1 ± 1 + 20 2 = 1 ± 21 2
(2) x 2 - 3 x 10 - 1 10 = 0
⇒ 10 x 2 - 3 x - 1 = 0 ⇒ x = - - 3 ± - 3 2 - 4 × 10 × - 1 2 × 10 ⇒ x = 3 ± 9 + 40 20 = 3 ± 49 20 ⇒ x = 3 ± 7 20 ⇒ x = 3 + 7 20 , 3 - 7 20 ⇒ x = 10 20 , - 4 20 ⇒ x = 1 2 , - 1 5

(3) 2 x + 3 2 = 25
⇒ 4 x 2 + 9 + 12 x = 25 ⇒ 4 x 2 + 12 x - 16 = 0 ⇒ x 2 + 3 x - 4 = 0 ⇒ x 2 + 4 x - x - 4 = 0 ⇒ x x + 4 - 1 x + 4 = 0 ⇒ x - 1 x + 4 = 0 ⇒ x = 1 , - 4

(4) m 2 + 5 m + 5 = 0
⇒ m = - 5 ± 5 2 - 4 × 1 × 5 2 ⇒ m = - 5 ± 25 - 20 2 ⇒ m = - 5 ± 5 2

(5) 5 m 2 + 2 m + 1 = 0
⇒ m = - 2 ± 2 2 - 4 × 1 × 5 2 ⇒ m = - 2 ± 4 - 20 2 ⇒ m = - 2 ± - 16 2
So, the roots are not real as the discriminant is negative.

(6) x 2 - 4 x - 3 = 0
⇒ x = - - 4 ± - 4 2 - 4 × 1 × - 3 2 × 1 ⇒ x = 4 ± 16 + 12 2 ⇒ x = 4 ± 28 2 ⇒ x = 4 ± 2 7 2 ⇒ x = 2 ± 7

Page No 54:

Question 8:

Find m if (m &ndash 12) x 2 + 2(m &ndash12) x + 2 = 0 has real and equal roots.

Answer:

For real and equal roots, the Discriminant should be 0.
(m &ndash 12) x 2 + 2(m &ndash12) x + 2 = 0
∆ = b 2 - 4 a c = 0 a = m - 12 , b = 2 m - 12 , c = 2
2 m - 12 2 - 4 m - 12 × 2 = 0 ⇒ 4 m 2 + 576 - 96 m - 8 m + 96 = 0 ⇒ 4 m 2 - 104 m + 672 = 0 ⇒ m 2 - 26 m + 168 = 0 ⇒ m 2 - 14 m - 12 m + 168 = 0 ⇒ m m - 14 - 12 m - 14 = 0 ⇒ m - 14 m - 12 = 0 m = 12 , 14
But if m = 12, the quadratic equation will be formed hence, the vaalue of m is 14.

Page No 54:

Question 9:

The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.

Answer:

Let the roots be α   and   β .
Sum of roots = 5
α + β = 5
Also,
α 3 + β 3 = 35 ⇒ α 3 + β 3 = α + β 3 - 3 α β α + β ⇒ 35 = 5 3 - 3 α β 5 ⇒ 15 α β = 125 - 35 ⇒ α β = 6
The general quadratic equation is
x 2 - α + β x + α β = 0
So, the required quadratic equation will be
x 2 - 5 x + 6 = 0 ⇒ x 2 - 5 x + 6 = 0

Page No 54:

Question 10:

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation 2 x 2 + 2 p + q x + p 2 + q 2 = 0

Answer:

2 x 2 + 2 p + q x + p 2 + q 2 = 0
Let the roots of the given quadratic equation be α and β .
Sum of roots, α + β = - 2 p + q 2 = - p + q
⇒ α + β 2 = p + q 2 . (A)
α β = p 2 + q 2 2
α - β 2 = α + β 2 - 4 α β = p + q 2 - 4 p 2 + q 2 2 = p + q 2 - 2 p 2 + q 2 = - p - q 2                                                               . . . . . B
A + B = p + q 2 - p - q 2 = p + q - p + q p + q + p - q                                       ∵ x 2 - y 2 = x - y x + y = 4 p q
A B = p + q 2 - p - q 2 = - p 2 + q 2 + 2 p q p 2 + q 2 - 2 p q = - p 2 2 + q 2 2 - 2 p 2 q 2 = - p 2 - q 2 2
General form of quadratic equation is
x 2 - A + B x + A B = 0
Putting the value of A and B we get
x 2 - 4 p q - p 2 - q 2 2 = 0 .

Page No 54:

Question 11:

Answer:

Let amount with Sagar be Rs x.
Amount with Mukund = Rs x + 50
The product of the amount they have is 15,000.
x x + 50 = 15000 ⇒ x 2 + 50 x = 15000 ⇒ x 2 + 50 x - 15000 = 0 ⇒ x 2 + 150 x - 100 x - 15000 = 0 ⇒ x x + 150 - 100 x + 150 = 0 ⇒ x - 100 x + 150 = 0 ⇒ x = 100 , - 150
But amount cannot be negative so,
Amount with Sagar = Rs 100 and that with Mukund is Rs 150.

Page No 54:

Question 12:

The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.

Answer:

Let the smaller number be x.
Larger number be y.
The square of smaller number is twice the greater number.
x 2 = 2 y
The difference between squares of two numbers is 120.
y 2 - x 2 = 120 ⇒ y 2 - 2 y - 120 = 0 ⇒ y 2 - 12 y + 10 y - 120 = 0 ⇒ y y - 12 + 10 y - 12 = 0 ⇒ y + 10 y - 12 = 0 ⇒ y = 12 , - 10
So, the numbers are 12 as the larger number and the smaller number is
x 2 = 2 × 12 = 24 ⇒ x = ± 24

Page No 54:

Question 13:

Answer:

Total oranges = 540
Let the total number of students be x.
Orange each student gets will be 540 x
If 30 more students are there so, the total number of students will be x + 30
Number of oranges per person will be 540 x + 30 .
If 30 students were more each would get 3 oranges less = 540 x - 3
540 x + 30 = 540 x - 3
⇒ 3 = 540 1 x - 1 x + 30 ⇒ 1 = 180 x + 30 - x x x + 30 ⇒ 1 = 180 30 x 2 + 30 x ⇒ x 2 + 30 x = 5400 ⇒ x 2 + 30 x - 5400 = 0 ⇒ x 2 + 90 x - 60 x - 5400 = 0 ⇒ x x + 90 - 60 x + 90 = 0 ⇒ x - 60 x + 90 = 0 ⇒ x = 60 , - 90
But number of students cannot be negative so, the number of students is 60.

Page No 54:

Question 14:

Mr. Dinesh owns an agricultural farm at village Talvel. The length of the farm is 10 meter more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is 1 3 of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and of the pond.

Answer:

Breadth of farm = x
Length of farm = 2 x + 10
Area of farm = l b = x 2 x + 10 = 2 x 2 + 10 x
Side of pond = x 3
Area of pond = s 2 = x 3 2
Area of farm = 2(Area of pond)
⇒ 2 x 2 + 10 x = 20 × x 2 9 ⇒ 18 x 2 + 90 x = 20 x 2 ⇒ 9 x 2 + 45 x = 10 x 2 ⇒ x 2 - 45 x = 0 ⇒ x x - 45 = 0 ⇒ x = 0 , 45
Length of farm = 2 × 45 + 10 = 100   units
Breadth of farm = 45 units
Side of pond = 45 3 = 15 units


COURSE OBJECTIVES:

As the result of instructional activities, students will be able to:

  1. Determine the x- and y- intercepts of a linear equation algebraically and graphically
  2. Calculate the slope of a line
  3. Write an equation of a line given the y-intercept and the slope
  4. Write an equation of a line given one point and the slope
  5. Write an equation of a line given two points
  6. Write an equation of a vertical line
  7. Write an equation of a horizontal line
  8. Use a graphing calculator to draw a scatterplot
  9. Use a graphing calculator to find a linear regression model, where appropriate
  10. Use a linear regression equation to make predictions
  11. Solve a linear inequality
  12. Determine if a given relation is a function
  13. Identify the family to which a function belongs
  14. Identify the domain of a function algebraically and graphically
  15. Identify the range of a function graphically
  16. Evaluate a function
  17. Graph a function
  18. Apply transformations to basic functions
  19. Evaluate a piecewise defined function
  20. Create cost, revenue and profit functions
  21. Find a break-even point
  22. Find the equilibrium quantity and price given supply and demand functions
  23. Determine whether a parabola opens upward or downward
  24. Determine the vertex of a parabola graphically and algebraically using –b/2a
  25. Determine the axis of symmetry of a parabola
  26. Determine the maximum or minimum value of a quadratic function
  27. Find the x-intercepts of a quadratic function algebraically by factoring and using the quadratic formula
  28. Find the x-intercepts of a quadratic function using a graphing calculator
  29. Find the y-intercept of a quadratic function algebraically and by using a graphing calculator
  30. Use a graphing calculator to find a quadratic regression model, where appropriate
  31. Use a quadratic regression equation to make predictions
  32. Use the simple interest formula
  33. Use the compound interest formula
  34. Determine whether a system of equations is consistent and independent, dependent, or inconsistent
  35. Solve a system of linear equations graphically
  36. Solve a system of linear equations algebraically using substitution
  37. Solve a system of linear equations algebraically using elimination
  38. Solve a system of linear equations using a matrix
  39. Graph a linear inequality in two variables
  40. Graph a system of linear inequalities and identify the feasible region
  41. Formulate a linear programming model
  42. Solve a linear programming model graphically
  43. Define set, subset, empty set, universal set
  44. List the elements of a set
  45. Identify the number of elements in a set
  46. Use the set operations of union, intersection, and complementation
  47. Draw Venn Diagrams to illustrate relationships between sets
  48. Determine the sample space of an experiment
  49. Determine if two events are disjoint (mutually exclusive)
  50. Calculate a factorial
  51. Distinguish between a permutation and a combination
  52. Calculate a permutation
  53. Calculate a combination
  54. Calculate basic probabilities
  55. Use the addition rule for probability
  56. Use the complement rule for probability
  57. Calculate a conditional probability
  58. Determine if two events are independent
  59. Use the product rule for probability

SUNY GENERAL EDUCATION LEARNING OUTCOMES:

Students will demonstrate the ability to:

  1. interpret and draw inferences from mathematical models such as formulas, graphs, tables and schematics
  2. represent mathematical information symbolically, visually, numerically and verbally
  3. use arithmetical, algebraic, geometric and statistical methods to solve problems
  4. estimate and check mathematical results for reasonableness and
  5. recognize the limits of mathematical and statistical methods.

GENERAL TOPICS OUTLINE:

  1. Linear Equations and Inequalities- including intercepts, slope, writing equations, graphing, inequalities, linear regression
  2. Functions and Graphs- including identifying functions, evaluating functions, graphing, domains, applications
  3. Quadratic Functions- including graphing, vertex, axis of symmetry, quadratic regression, applications
  4. Mathematics of Finance- including simple interest, compound interest
  5. Systems of Linear Equations and Matrices- including solving 2x2 systems of linear equations algebraically and graphically, solving 3x3 systems, applications
  6. Linear Programming- including graphing systems of linear inequalities, linear programming
  7. Set Theory and combinatorics- including definitions, Venn diagrams, set operations, permutations, combinations, applications
  8. Probability- including addition rule, complement rule, conditional probability, independence, applications

MathHelp.com

However, the only way to know we have the accurate x -intercept, and thus the solution, is to use the algebra, setting the line equation equal to zero, and solving:

If the linear equation were something like y = 47x &ndash 103 , clearly we'll have great difficulty in guessing the solution from the graph. We might guess that the x -intercept is near x = 2 but, while close, this won't be quite right. Algebra would be the only sure solution method.

A quadratic function is messier than a straight line it graphs as a wiggly parabola. If we plot a few non- x -intercept points and then draw a curvy line through them, how do we know if we got the x -intercepts even close to being correct? We don't. The only way we can be sure of our x -intercepts is to set the quadratic equal to zero and solve.

But the whole point of "solving by graphing" is that they don't want us to do the (exact) algebra they want us to guess from the pretty pictures.

So "solving by graphing" tends to be neither "solving" nor "graphing". In a typical exercise, you won't actually graph anything, and you won't actually do any of the solving. Instead, you are told to guess numbers off a printed graph. Or else, if "using technology", you're told to punch some buttons on your graphing calculator and look at the pretty picture and then you're told to punch some other buttons so the software can compute the intercepts.

My guess is that the educators are trying to help you see the connection between x -intercepts of graphs and solutions of equations. But the concept tends to get lost in all the button-pushing.

Okay, enough of my ranting.

To solve by graphing, the book may give us a very neat graph, probably with at least a few points labelled. The book will ask us to state the points on the graph which represent solutions. Otherwise, it will give us a quadratic, and we will be using our graphing calculator to find the answer. Since different calculator models have different key-sequences, I cannot give instruction on how to "use technology" to find the answers you'll need to consult the owner's manual for whatever calculator you're using (or the "Help" file for whatever spreadsheet or other software you're using). I will only give a couple examples of how to solve from a picture that is given to you.

Solve x 2 &ndash 8x + 15 = 0 by using the following graph.

The equation they've given me to solve is:

The picture they've given me shows the graph of the related quadratic function:

The x -intercepts of the graph of the function correspond to where y = 0 . The point here is that I need to look at the picture (hoping that the points really do cross at whole numbers, as it appears), and read the x -intercepts of the graph (and hence the solutions to the equation) from the picture.

The graph appears to cross the x -axis at x = 3 and at x = 5 I have to assume that the graph is accurate, and that what looks like a whole-number value actually is one. So my answer is:

Since they provided the quadratic equation in the above exercise, I can check my solution by using algebra. The given quadratic factors, which gives me:

Now I know that the solutions are whole-number values. The graph can be suggestive of the solutions, but only the algebra is sure and exact.

Solve from the following graph.

For this picture, they labelled a bunch of points. Partly, this was to be helpful, because the x -intercepts are messy, so I could not have guessed their values without the labels. But mostly this was in hopes of confusing me, in case I had forgotten that only the x -intercepts, not the vertices or y -intercepts, correspond to "solutions".

Point B is the y -intercept (because x = 0 for this point), so I can ignore this point. Point C appears to be the vertex, so I can ignore this point, also. Points A and D are on the x -axis (because y = 0 for these points). So I can assume that the x -values of these graphed points give me the solution values for the related quadratic equation.

Because they provided the equation in addition to the graph of the related function, it is possible to check the answer by using algebra. But the intended point here was to confirm that the student knows which points are the x -intercepts, and knows that these intercepts on the graph are the solutions to the related equation.

Find the solutions to the quadratic equation from the following graph:

A = (&ndash2.1429, 0) , B = (2.8, 0) , C = (0.3286, &ndash3.0540) , D = (0, &ndash3)

They haven't given me a quadratic equation to solve, so I can't check my work algebraically. They have only given me the picture of a parabola created by the related quadratic function, from which I am supposed to approximate the x -intercepts, which really is a different question. But I know what they mean.

I can ignore the point which is probably the vertex (Point C). I can ignore the point which is the y -intercept (Point D). So I'll pay attention only to the x -intercepts, being those points where y is equal to zero. So my answer is:

Solving quadratics by graphing is silly in terms of "real life", and requires that the solutions be the simple factoring-type solutions such as " x = 3 ", rather than something like " x = &ndash4 + sqrt(7) ". In other words, they either have to "give" you the answers (b labelling the graph), or they have to ask you for solutions that you could have found easily by factoring.

About the only thing you can gain from this topic is reinforcing your understanding of the connection between solutions of equations and x -intercepts of graphs of functions that is, the fact that the solutions to "(some polynomial) equals (zero)" correspond to the x -intercepts of the graph of " y equals (that same polynomial)". If you come away with an understanding of that concept, then you will know when best to use your graphing calculator or other graphing software to help you solve general polynomials namely, when they aren't factorable. And you'll understand how to make initial guesses and approximations to solutions by looking at the graph, knowledge which can be very helpful in later classes, when you may be working with software to find approximate "numerical" solutions.

But in practice, given a quadratic equation to solve in your algebra class, you should not start by drawing a graph. Which raises the question: For any given quadratic, which method should one use to solve it?


Solving quadratic equations COLLECTION (Bundle)

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A collection of EIGHT FULL LESSONS, which could definitely be extended to at least 10-11 lessons for the right classes, on solving quadratic equations by factorising, the quadratic formula or completing the square.

Contents of download:

  • Normal PowerPoint lessons with which you can use a clicker / mouse / keyboard to continue animations and show fully animated and worked solutions.
  • At least one printable worksheets for students with examples for each lesson. These are designed to speed up the lesson (no copying down questions etc). The worksheet could also be used independent of the PowerPoint lesson!

We are learning about:

  • Lesson 4.1.1h - Factorising quadratic equations (a = 1)
  • Lesson 4.1.2h - Factorising harder quadratic equations (a ≠ 1)
  • Lesson 4.2.1h - The quadratic formula - part 1
  • Lesson 4.2.2h - The quadratic formula - part 2
  • Lesson 4.3.1h - Completing the square - part 1 (a = 1)
  • Lesson 4.3.2h - Completing the square - part 2 (a ≠ 1)
  • Lesson 4.4.1h - Forming and solving quadratic equations (geometric problems)
  • Lesson 4.4.2h - Forming and solving quadratic equations (worded problems)

Main: Lessons consist of examples with notes and instructions, following on to increasingly difficult exercises with problem solving tasks. Lessons can start at any section of the PPT examples judged against the ability of the students in your class. All solutions given on PPT.


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Class: 11th
Subject: Mathematics
Chapter : Ch-10 Quadratic Equations of Section -A
Board ISC
Writer OP Malhotra
Publications S.Chand Publications 2020-21

Quadratic Equations OP Malhotra S.Chand ISC Class-11 Maths Solutions

Argand Plane
Any complex number z = x + iy can be represented geometrically by a point (x, y) in a plane, called argand plane or gaussian plane. A purely number x, i.e. (x + 0i) is represented by the point (x, 0) on X-axis. Therefore, X-axis is called real axis. A purely imaginary number iy i.e. (0 + iy) is represented by the point (0, y) on the y-axis. Therefore, the y-axis is called the imaginary axis.

How to Solve Quadratic Equations by Factorization

To solve quadratics by factoring, we use something called “the Zero-Product Property”. This property says something that seems fairly obvious, but only after it’s been pointed out to us namely:

Zero-Product Property :

If we multiply two (or more) things together and the result is equal to zero, then we know that at least one of those things that we multiplied must also have been equal to zero. Put another way, the only way for us to get zero when we multiply two (or more) factors together is for one of the factors to have been zero.

So, if we multiply two (or more) factors and get a zero result, then we know that at least one of the factors was itself equal to zero. In particular, we can set each of the factors equal to zero, and solve the resulting equation for one solution of the original equation.

We can only draw the helpful conclusion about the factors (namely, that one of those factors must have been equal to zero, so we can set the factors equal to zero) if the product itself equals zero. If the product of factors is equal to anything non-zero, then we can not make any claim about the values of the factors.


10.3: Re-examining the Quadratic Formula - Mathematics

In the past several lessons, we have dealt with the solution of various types of linear equations. We started out by deriving various formulae to solve equations that fall into specific types using the Paravartya Yojayet sutra. We then proceeded to applications of the powerful Sunyam Samyasamuccaye sutra. We then covered various kinds of mergers, and finally we covered applications of the Anurupye Sunyam Anyat sutra to the solution of simultaneous linear equations.

You can find all my previous posts about Vedic Mathematics below:

In this lesson, we will deal with quadratic equations, which are polynomial equations of the second degree. What this means is that when the equation is expanded out with no fractional terms, there exists at least one term in which the unknown quantity is raised to the second power.

The general form of a quadratic equation is ax^2 + bx + c. The quadratic formula can be used to solve equations in the standard form. The quadratic formula is usually written as below:


In this lesson, however, we will not deal with quadratic equations that are written in the standard form. It may take quite a bit of effort in cross-multiplication, collection of like terms, etc., to get these equations into the standard form. The solution of these equations using the quadratic formula, therefore, is quite cumbersome, not to mention, error-prone.

Rather than go through all that labor, we will identify these special types of equations and apply certain simple procedures to solve them quickly and easily.

The first special type of quadratic equations we will consider are like the one below:

In the traditional method, we would go through the process of getting the equation into the standard form using the steps below:

x + 1/x = 10/3 becomes
(x^2 + 1)/x = 10/3 becomes
3x^2 + 3 = 10x becomes
3x^2 - 10x + 3 = 0

We would then solve it using the quadratic formula by setting a = 3, b = -10 and c = 3. This would lead to the solutions x = 3 and x = 1/3.

However, we don't have to go through all that trouble to solve this kind of equation. All we have to do is observe that 10/3 = 3 + 1/3. Thus, we would immediately have figured out that we can rewrite the given equation as below:

The symmetry of the equation above immediately reveals the answers, x = 3 and x = 1/3.

The same method can be used to solve many different problems such as the examples below:

x + 1/x = 26/5 => x + 1/x = 5 + 1/5 => x = 5, 1/5
x + 1/x = 50/7 => x + 1/x = 7 + 1/7 => x = 7, 1/7
x + 1/x = -17/4 => x + 1/x = -4 - 1/4 => x = -4, -1/4

We are not restricted to x + 1/x on the left-hand side either. Consider the equation:

Since the right-hand side of the equation can be expanded to 7 + 1/7, by the symmetry of the equation, we can equate 2x + 3 to either 7 or 1/7 (or equivalently, by equating (2x + 3) and 1/(2x + 3) to 7). We then get the solutions x = 2 and x = -10/7 to the given equation.

Similarly, consider the equation:

The right-hand side of the equation can be expanded to 5 + 1/5. Thus, we can use the symmetry of the resulting equation to derive the following linear equations:

5x/(2x + 3) = 5
5x/(2x + 3) = 1/5 or alternatively, (2x + 3)/5x = 5

These two equations can then be solved to give us x = -3 or x = 23/3.

Now, consider the equation:

(x + 3)/(3x + 5) + (3x + 5)/(x + 3) = 17/4

The right-hand side of the equation can be expanded to 4 + 1/4. This then lets us solve the equation by deriving the linear equations below:

(x + 3)/(3x + 5) = 4
(x + 3)/(3x + 5) = 1/4 or alternatively, (3x + 5)/(x + 3) = 4

These equations can then be solved to give us x = -7 and x = -17/11.

Now, consider an equation of the type below:

We can rewrite the equation as below:

We may be tempted to conclude from the symmetry of the equation on both sides of the equal-to sign that x = 2 or x = 1/2. That would be wrong. In equations such as the above where the terms are connected by "-" signs instead of "+" signs, the solutions are x = 2 and x = -1/2. Only with x = -1/2 is it possible to get -1/x = 2, and therefore x - 1/x = 2 - 1/2. This is important to remember.

We will illustrate this with a few examples as below:

x - 1/x = 8/3 becomes
x - 1/x = 3 - 1/3, which then leads to x = 3 and x = -1/3 as the solutions.

x - 1/x = 63/8 becomes
x - 1/x = 8 - 1/8, which then leads to x = 8 and x = -1/8 as the solutions.

x - 1/x = -24/5 becomes
x - 1/x = -5 + 1/5 which then leads to x = -5 and x = 1/5 as the solutions.

The technique is equally applicable to cases where the left-hand side consists of other terms than x and 1/x. The following examples illustrate a few examples of these cases:

(3x + 2) - 1/(3x + 2) = 63/8 => (3x + 2) - 1/(3x + 2) = 8 - 1/8 =>
3x + 2 = 8, 3x + 2 = -1/8 => x = 2, x = -17/24
2x/(5x + 1) - (5x + 1)/2x = -15/4 => 2x/(5x + 1) - (5x + 1)/2x = 1/4 - 4 =>
2x/(5x + 1) = 1/4, 2x/(5x + 1) = -4 => x = 1/3, -2/11
(4x + 3)/(3x + 4) - (3x + 4)/(4x + 3) = 24/5 => (4x + 3)/(3x + 4) - (3x + 4)/(4x + 3) = 5 - 1/5 =>
(4x + 3)/(3x + 4) = 5, (4x + 3)/(3x + 4) = -1/5 => x = -17/11, -19/23

Sometimes, the equation may have undergone some transformations that hide its true nature. Consider the equation below:

The equation can actually be simplified as below:

x + (x + 1)/x = 7/2 becomes
x + x/x + 1/x = 7/2 becomes
x + 1/x = 7/2 - 1 becomes
x + 1/x = 5/2 becomes
x + 1/x = 2 + 1/2 which directly leads to the solutions x = 2 and x = 1/2.

Many of these transformations are difficult to peal back on sight to reveal the true nature of the equation. But it may be worthwhile to examine the equation and try a few transformations to see if any simplifications are possible before giving up and solving it using the traditional method.

Sometimes, the right-hand side is not as easily decomposed to a pair of reciprocals as in the above examples. Consider the equation below:

The right-hand side may look as if it can not be converted into a pair of reciprocals since at first glance, it breaks apart as 2 and 1/12. But a closer look will reveal that 25/12 = 3/4 + 4/3. Thus, we can actually rewrite the equation above as:

This then leads immediately to the solutions x = 3/4 and x = 4/3 by the symmetry of the equation on both sides of the equal-to sign.

Let us tackle a few more equations like the above:

x + 1/x = 13/6 => x + 1/x = 2/3 + 3/2 => x = 2/3, 3/2
x + 1/x = 29/10 => x + 1/x = 2/5 + 5/2 => x = 2/5, 5/2

Similarly, we can solve the equations below also by recognizing the right-hand side to be a difference of reciprocals:

x - 1/x = 11/30 => x - 1/x = 6/5 - 5/6 => x = 6/5, -5/6
x - 1/x = -40/21 => x - 1/x = 3/7 - 7/3 => x = 3/7, -7/3

This same technique can be extended to equations where the left-hand side is a sum or difference of any two reciprocal quantities, not just x and 1/x. I will leave those extensions to the reader in the interest of keeping this lesson from growing any longer than it already has become!

Assuming that the left-hand side is a pair of reciprocals, connected by either "+" or "-", how do we verify whether the right-hand side can be expressed as a pair of reciprocals with the same sign between them? We will explore this question in greater detail in the next lesson.

In the meantime, I hope you have found this lesson useful and interesting. I also hope you will apply the techniques explained in this lesson on real problems so that you become familiar not only with the technique itself (which is actually quite trivial), but also with the fractions that result either from the addition of numbers with their reciprocals, or the differences between numbers and their reciprocals. That will enable one to apply this technique where appropriate, on sight and mentally, to solve the types of quadratic equations we have dealt with in this lesson. Good luck, and happy computing!


Lesson 20

Decide whether each number is rational or irrational.

Problem 2

Here are the solutions to some quadratic equations. Select all solutions that are rational.

Problem 3

Solve each equation. Then, determine if the solutions are rational or irrational.

Problem 4

Here is a graph of the equation (y=81(x-3)^2-4) .

Based on the graph, what are the solutions to the equation (81(x-3)^2=4) ?

Expand Image

Problem 5

Match each equation to an equivalent equation with a perfect square on one side.

Problem 6

To derive the quadratic formula, we can multiply (ax^2+bx+c=0) by an expression so that the coefficient of (x^2) a perfect square and the coefficient of (x) an even number.

  1. Which expression, (a) , (2a) , or (4a) , would you multiply (ax^2+bx+c=0) by to get started deriving the quadratic formula?
  2. What does the equation (ax^2+bx+c=0) look like when you multiply both sides by your answer?

Problem 7

Here is a graph the represents (y=x^2) .

On the same coordinate plane, sketch and label the graph that represents each equation:

Expand Image

Description: <p>A curve in an x y plane, origin O. Horizontal axis, scale negative 8 to 8, by 2’s. Vertical axis, scale negative 12 to 12, by 2’s. A curve, labeled y equals x squared, passes through the points negative 2 comma 4, 0 comma 0, and 2 comma 4.</p>

Problem 8

Which quadratic expression is in vertex form?

Problem 9

Function (f) is defined by the expression (frac<5>) .

The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics.

This book includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.


The Quadratic Formula Explained

Quadratic equations are an important stepping stone in algebra, geometry, and calculus. Quadratic formulas are simply polynomial equations of the second degree: ax^2+bx+c=0. This represents a graph of a concave shape which graphs formulas representing exponential growth and decay, maximizing or minimizing at a point.

While much of the theory behind a quadratic equation is solely for those reaching into the upper levels of calculus, it is still important to know how to solve one as early as the freshman year of high school. Below, I'll describe each component of the quadratic formula and how to apply it.

If we toy with the quadratic formula, we start to see the problem in solving it. We end up with x on each side of the equation, or we end up trying to take the root of a complex problem. We could delve into how to reach the final "x b" Variable

First, identify "b." This variable is the second coefficient in the quadratic formula. In our case, it's 8. The first thing we'll plug in will be this. This leaves us with ( -8 +/- Sq (8^2 + 4ac ). 8^2 is 64, so we can further simplify ( -8 +/- Sq (64+4ac) ).

Step Two: The "c" Variable

Next, we will need to plug in the "c" coefficient from our original formula. This value is 12, since c represents the coefficient without an x next to it. So, our formula is now ( -8 +/- Sq (64+(4)a(12) ). We further simplify to get ( -8 +/- Sq (64+48a) ).

Step Three: The "a" Variable

Now, let's plug in our last value, "a."

In words, negative eight plus or minus the square root of two hundred and fifty six.

This is the top of our equation.

This part is simple. Plug in "a," in our case 4, and evaluate:

Now, put the whole equation together:

We can simplify the square root of 256 to be 16.

x, therefore equals either

( -8 + 16 ) / 8, or 1
or
(-8 - 16 ) /8 or -3

That's it! Because the physical nature of a quadratic equation is a concave, it will have two different x values, meaning it will cross the x axis twice.

If you struggle with math in a school or university setting, consider contacting a home tutor. A tutor can help you understand the more difficult concepts in any subject, and will provide a measurable benefit in grade point average.

Cias Hart is a veteran and resident of Arizona. If you're in the greater metro area and are or have a student who is struggling with math concepts, contact a Scottsdale home tutor. A math tutor can help you or your student bring their GPA up in the subject and prepare them for more advanced studies later.