## Factoring Worksheet1 Solution6

In this page factoring worksheet1 solution6 we are going to see solution of some practice questions from factoring worksheet1.

Question 1:

Solve by completing the square method x² + 6 x – 7 = 0

Solution:

x² + 6 x – 7 = 0

we have to make the second term as the multiple of 2

x² + 2 (x) (3) + 3² - 3² – 7 = 0

(x + 3)² – 9 – 7 = 0

(x + 3)² – 16 = 0

(x + 3)² = 16

(x + 3) = 16

x + 3 = ±4

x + 3 = 4                      x + 3 = -4

x = 4 – 3                        x = -4 – 3

x = 1                                x = -7

Verification:

x² + 6 x – 7 = 0

if x = 1

(1)² + 6(1) – 7 = 0

1 + 6 - 7 = 0

7 - 7 = 0

0 = 0

if x = -7

x² + 6 x – 7 = 0

(-7)² + 6 (-7)– 7 = 0

49 - 42 - 7 = 0

49 - 49 = 0

0 = 0

Question 2:

Solve by completing the square method x² + 3 x + 1 = 0

Solution:

x² + 3 x + 1 = 0

we have to make the second term as the multiple of 2.Since 3 cannot be written  as the multiple of 2 we have to multiply and divide 3 by 2.

x² + 2 (x) (3/2) + (3/2)² - (3/2)² + 1 = 0

[x + (3/2)]² – (9/4) + 1 = 0

[x + (3/2)]² = (9/4) – 1

[x + (3/2)]² = (9 – 4)/4

[x + (3/2)]² = 5/4

[x + (3/2)]  = (5/4)

[x + (3/2)]  = ± √5/2

x + (3/2) = √5/2                                          x + (3/2) = -√5/2

x = (√5/2) – (3/2)                                              x = -(√5/2) –(3/2)

x = (√5 -3)/2                                                      x = (-√5 -3)/2 factoring worksheet1 solution6 factoring worksheet1 solution6 