kjones

06-01-2006, 09:29 PM

Could someone walk me through (baby steps, please) solving this equations:

2x^2 - 20x + 8 = 0

2x^2 - 20x + 8 = 0

View Full Version : Completing the square: 2x^2 - 20x + 8 = 0

kjones

06-01-2006, 09:29 PM

Could someone walk me through (baby steps, please) solving this equations:

2x^2 - 20x + 8 = 0

2x^2 - 20x + 8 = 0

tkhunny

06-02-2006, 07:15 AM

Could someone walk me through (baby steps, please) solving this equations:

2x^2 - 20x + 8 = 0Remove common factors

x^2 - 10x + 4 = 0

Move the constant term

x^2 - 10x = -4

Calculate the new constant term

10/2 = 5

5^2 = 25

Adjust the equation

x^2 - 10x + 25 = -4 + 25

Factor and Simplify

(x-5)^2 = 21

Finish Up

x-5 = sqrt(21) and x-5 = -sqrt(21)

x = 5 + sqrt(21) and x = 5 - sqrt(21)

One for free. You show us the next one.

2x^2 - 20x + 8 = 0Remove common factors

x^2 - 10x + 4 = 0

Move the constant term

x^2 - 10x = -4

Calculate the new constant term

10/2 = 5

5^2 = 25

Adjust the equation

x^2 - 10x + 25 = -4 + 25

Factor and Simplify

(x-5)^2 = 21

Finish Up

x-5 = sqrt(21) and x-5 = -sqrt(21)

x = 5 + sqrt(21) and x = 5 - sqrt(21)

One for free. You show us the next one.

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