## Factoring Worksheet1 Solution7

In this page factoring worksheet1 solution7 we are going to see solution of some practice questions from factoring worksheet1.

Question 3:

Solve by completing the square method 2x² + 5 x - 3 = 0

Solution:

2x² + 5 x - 3 = 0

Now we are going to divide the whole equation by 2

x² + (5/2) x – (3/2) = 0

x² + 2 (x) (5/4) + (5/4)² - (5/4)² – (3/2) = 0

[x + (5/4)]² - (5/4)² – (3/2) = 0

[x + (5/4)]² - (25/16) – (3/2) = 0

[x + (5/4)]²  = (25/16) +(3/2)

[x + (5/4)]² = (25/16) +(3/2) x (8/8)

[x + (5/4)]² = (25/16) + (24/16)

[x + (5/4)]² = (25 + 24)/16

[x + (5/4)]² = 49/16

[x + (5/4)] = (49/16)

x + (5/4) = ± 7/4

x + (5/4) = 7/4                                    x + (5/4) = - 7/4

x = (7/4) – (5/4)                                   x = (-7/4) – (5/4)

x = (7-5)/4                                              x = (-7 – 5)/4

x = 2/4                                                     x = -12/4

x = 1/2                                                                    x = -3

Verification:

2x² + 5 x - 3 = 0

if x = 1/2

2(1/2)² + 5 (1/2) – 3 = 0

(1/2) + (5/2) - 3 = 0

(1 + 5 - 6)/2 = 0

(6 - 6)/2 = 0

0/2 = 0

0 = 0

if x = -3

2x² + 5 x - 3 = 0

2 (-3)² + 5 (-3) - 3 = 0

2 (9) - 15 - 3 = 0

18 - 15 - 3 = 0

18 - 18 = 0

0 = 0

Question 4:

Solve by completing the square method 4x² + 4 b x – (a² - b²) = 0

Solution:

4x² + 4 b x – (a² - b²) = 0

Now we are going to divide the whole equation by 4

x² +  b x – [(a² - b²)/4] = 0

x² +  2 (x) (b/2) + (b/2)² – (b/2)² – [(a² - b²)/4] = 0

[x + (b/2)]² – (b²/4) – [(a² - b²)/4] = 0

[x + (b/2)]² = (b²/4) + [(a² - b²)/4]

[x + (b/2)]² = [(b² + a² - b²)/4]

[x + (b/2)]² = [a²/4]

[x + (b/2)] = √(a²/4)

[x + (b/2)] = ±(a/2)

[x + (b/2)] = (a/2)                            [x + (b/2)] = -(a/2)

x = (a/2) – (b/2)                           x = (-a/2) – (b/2)

x = (a - b)/2                                  x = -(a +b)/2

factoring worksheet1 solution7 factoring worksheet1 solution7