Solving Quadratic Equations by Completing the Square

**JadeKiara** · July 15th 2010, 06:24 PM

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.

**skeeter** · July 15th 2010, 06:29 PM

Originally Posted by JadeKiara

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 ...

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

$x^2 + 2x = 3$

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

$(x+1)^2 = 4$

$x+1 = \pm 2$

$x = -1 \pm 2$

$x = -3$

$x = 1$

**eumyang** · July 15th 2010, 06:31 PM

It's easier to complete the square if the $x^2$ coefficient is 1, so first factor out the 2:
$\begin{aligned} 2x^2 + 4x - 6 &= 0 \\ 2(x^2 + 2x - 3) &= 0 \\ x^2 + 2x - 3 &= 0 \end{aligned}$
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:
$\begin{aligned} x^2 + 2x &= 3 \\ x^2 + 2x + 1 &= 3 + 1 \\ (x + 1)^2 &= 4 \\ x + 1 &= \pm 2 \\ x + 1 &= 2 \rightarrow x = 1 \\ x + 1 &= -2 \rightarrow x = -3 \end{aligned}$

EDIT: Ack, too slow! ^o^

**tonio** · July 15th 2010, 06:34 PM

Originally Posted by JadeKiara

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.

$0=2x^2 + 4x - 6=2(x^2+2x-3)=2(x-1)(x+3)...$ , or also

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

The first method is the direct one, the trinomial decomposition. The second one uses the completion of the square method.
Tonio

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