# Solving Quadratic Equations by Completing the Square

• Jul 15th 2010, 06:24 PM
Solving Quadratic Equations by Completing the Square
Hello!

I have been doing quadratic equations and I am a little uncertain about the following question from my textbook. Thank you in advance to anyone who may be able to help me!

Solve 2x^2 + 4x - 6 = 0 by first finding a common factor and using the 'completing the square' method.
• Jul 15th 2010, 06:29 PM
skeeter
Quote:

Hello!

I have been doing quadratic equations and I am a little uncertain about the following question from my textbook. Thank you in advance to anyone who may be able to help me!

Solve 2x^2 + 4x - 6 = 0 by first finding a common factor and using the 'completing the square' method.

divide every term by 2 ...

$\displaystyle x^2 + 2x - 3 = 0$

$\displaystyle x^2 + 2x = 3$

$\displaystyle x^2 + 2x + 1 = 3 + 1$

$\displaystyle (x+1)^2 = 4$

$\displaystyle x+1 = \pm 2$

$\displaystyle x = -1 \pm 2$

$\displaystyle x = -3$

$\displaystyle x = 1$
• Jul 15th 2010, 06:31 PM
eumyang
It's easier to complete the square if the $\displaystyle x^2$ coefficient is 1, so first factor out the 2:
\displaystyle \begin{aligned} 2x^2 + 4x - 6 &= 0 \\ 2(x^2 + 2x - 3) &= 0 \\ x^2 + 2x - 3 &= 0 \end{aligned}
What I did at the last step was divide both sides by 2. Now move the constant term to the other side and complete the square:
\displaystyle \begin{aligned} x^2 + 2x &= 3 \\ x^2 + 2x + 1 &= 3 + 1 \\ (x + 1)^2 &= 4 \\ x + 1 &= \pm 2 \\ x + 1 &= 2 \rightarrow x = 1 \\ x + 1 &= -2 \rightarrow x = -3 \end{aligned}

EDIT: Ack, too slow! ^o^
• Jul 15th 2010, 06:34 PM
tonio
Quote:

$\displaystyle 0=2x^2 + 4x - 6=2(x^2+2x-3)=2(x-1)(x+3)...$ , or also
$\displaystyle 0=2x^2 + 4x - 6=2(x^2+2x-3)=2\left[(x+1)^2-4\right]=2(x+1-2)(x+1+2) ...$