complete the square

**whiteowl** · June 8th 2008, 02:42 PM

I know these are simple but these 2 are eating me up. I am having a real problem with complete the square when it envolves fractions for some reason. These 2 problems must be solved completing the square and please show me all the steps so that I can see where I have been messing up. Thanx
X^2+7x+12=0
2X^2+X-10=0

**galactus** · June 8th 2008, 03:17 PM

Use the general formula and you won't have trouble.

$a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}$

Enter in your values and you have it built. Try the first one: a=1, b=7, c=12

**masters** · June 8th 2008, 03:31 PM

Originally Posted by whiteowl

I know these are simple but these 2 are eating me up. I am having a real problem with complete the square when it envolves fractions for some reason. These 2 problems must be solved completing the square and please show me all the steps so that I can see where I have been messing up. Thanx
X^2+7x+12=0
2X^2+X-10=0

1. $x^2+7x+12=0$

Step1: Move +12 to right side of the equation.
$x^2+7x+ \;\;=-12+ \;\;$

Step 2: Take half the coefficient of x, square it, and add it to both sides.
$x^2+7x+\frac{49}{4}=-12+\frac{49}{4}$

Step 3: Left side is now a perfect square trinomial.
$(x+\frac{7}{2})^2=\frac{-48}{4}+\frac{49}{4}$

Step 4: Combine right side
$(x+\frac{7}{2})^2=\frac{1}{4}$

Step 5: Take square root of both sides.
$x+\frac{7}{2}=\pm \frac{1}{2}$

Step 6: Finish up.
$x+\frac{7}{2}=\frac{1}{2} \; \; or \; \; x+\frac{7}{2}=\frac{-1}{2}$

Step 7: Finally
$x=-3 \; \; or \; \; x=-4$

**masters** · June 8th 2008, 03:48 PM

Originally Posted by whiteowl

I know these are simple but these 2 are eating me up. I am having a real problem with complete the square when it envolves fractions for some reason. These 2 problems must be solved completing the square and please show me all the steps so that I can see where I have been messing up. Thanx
X^2+7x+12=0
2X^2+X-10=0

Take a look at Glactus' method. You might like it better.

2. $2x^2+x-10=0$

Before you begin with the steps I showed you, you must make the coefficient of $x^2$ to be 1. Divide everything by 2.

$x^2+\frac{1}{2}x-5=0$

Step 1: Move -5 to right side of the equation.
$x^2+\frac{1}{2}+ \; \; =5+$

Step 2: See if you can finish the other steps.

**whiteowl** · June 8th 2008, 07:10 PM

Thanx for the help. When you see it worked out it seems so simple. Thanx again

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