SUM AND PRODUCT OF ROOTS OF QUADRATIC EQUATION WORKSHEET

Problems 1-7 : Find the sum and product of roots of the quadratic equation given below.

Problem 1 :

x² - 5x + 6  =  0

Problem 2 :

x² - 6  =  0

Problem 3 :

3x² + x + 1  =  0

Problem 4 :

3x² + 7x  =  2x - 5

Problem 5 :

3x² - 7x + 6  =  6

Problem 6 :

x² + 5x + 1  =  3x² + 6

Problem 7 :

1/(x + 1) + 2/(x - 4)  =  2

Problem 8 :

If the product of roots of the quadratic equation given below is 4, then find the value of m.

2x² + 8x - m³  =  0 

Problem 9 :

If the sum of roots of the quadratic equation given below is 0, then find the value of p. 

x² -(p + 4)x + 5  =  0

Problem 10 :

If the product of roots of the quadratic equation given below is 1, then find the value of m.

x² + (2p - 1)x + p²  =  0

Answers

1. Answer :

Comparing

x² - 5x + 6  =  0

and 

ax² + bx + c  =  0

we get

a  =  1, b  =  -5 and c  =  6

Therefore, 

sum of the roots  =  -b/a  =  -(-5)/1  =  5

product of the roots  =  c/a  =  6/1  =  6

2. Answer :

Comparing

x² - 6  =  0

and 

ax² + bx + c  =  0

we get

a  =  1, b  =  0 and c  =  -6

Therefore, 

sum of the roots  =  -b/a  =  0/1  =  0

product of the roots  =  c/a  =  -6/1  =  -6

3. Answer :

Comparing

3x² + x + 1  =  0

and 

ax² + bx + c  =  0

we get

a  =  3, b  =  1 and c  =  1

Therefore, 

sum of the roots  =  -b/a  =  -1/3

product of the roots  =  c/a  =  1/3

4. Answer :

First write the given quadratic equation in standard form.

3x² +7x  =  2x - 5

3x² + 5x + 5  =  0

Comparing

3x² + 5x + 5  =  0

and 

ax² + bx + c  =  0

we get

a  =  3, b  =  5 and c  =  5

Therefore, 

sum of the roots  =  -b/a  =  -5/3

product of the roots  =  c/a  =  5/3

5. Answer :

First write the given quadratic equation in standard form.

3x² -7x + 6  =  6

3x² - 7x  =  0

Comparing

3x² - 7x  =  0

and 

ax² + bx + c  =  0

we get

a  =  3, b  =  -7 and c  =  0

Therefore, 

sum of the roots  =  -b/a  =  -(-7)/3  =  7/3

product of the roots  =  c/a  =  0/3  =  0

6. Answer :

First write the given quadratic equation in standard form.

x² + 5x + 1  =  3x² + 6

0  =  2x² - 5x + 5

2x² - 5x + 5  =  0

Comparing

2x² - 5x + 5  =  0

and 

ax² + bx + c  =  0

we get

a  =  2, b  =  -5 and c  =  5

Therefore, 

sum of the roots  =  -b/a  =  -(-5)/3  =  5/2

product of the roots  =  c/a  =  5/2

7. Answer :

First write the given quadratic equation in standard form.

1/(x + 1) + 2/(x - 4)  =  2

[1(x - 4) + 2(x + 1)]/[(x + 1)(x - 4)]  =  2

1(x - 4) + 2(x + 1)  =  2(x + 1)(x - 4)

x - 4 + 2x + 2  =  2(x² - 4x + x - 4)

3x - 2  =  2(x² - 3x - 4)

3x - 2  =  2x² - 6x - 8

0  =  2x² - 9x - 6

2x² - 9x - 6  =  0

Comparing

2x² - 9x - 6  =  0

and 

ax² + bx + c  =  0

we get

a  =  2, b  =  -9 and c  =  -6

Therefore, 

sum of the roots  =  -b/a  =  -(-9)/2  =  9/2

product of the roots  =  c/a  =  -6/2  =  -3

8. Answer :

Comparing

2x² + 8x - m³  =  0

and 

ax² + bx + c  =  0

we get

a  =  2, b  =  8 and c  =  -m³

Given : Product of the roots is 4.

Then,

c/a  =  4

-m³/2  =  4

Multiply each side by (-2). 

m³  =  -8

m³  =  (-2)³

m  =  -2

9. Answer :

Comparing

x² -(p + 4)x + 5  =  0

and 

ax² + bx + c  =  0

we get

a  =  1, b  =  -(p + 4) and c  =  5

Given : Sum of the roots is 0.

Then,

-b/a  =  0

-[-(p + 4)]  =  0

(p + 4)  =  0

p + 4  =  0

p  =  -4

10. Answer :

Comparing

x² + (2p - 1)x + p²  =  0

and 

ax² + bx + c  =  0

we get

a  =  1, b  =  (2p - 1) and c  =  p²

Given : Product of the roots is 1.

Then,

c/a  =  1

p²/1  =  1

p²  =  1

Take square root on both sides. 

√p²  =  √1

 p  =  ±1

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.