Completing the Square help

**Mitch281008** · April 28th 2010, 08:01 PM

Hey guys,

I need help completing this square:

x^2 + 3x + 2

Ive been following tips online, and ive gotten as far as:

x^2 + 3x + 9/4 = 1/4

but i cant get any further because no matter where i go to i dont understand the next step.

can anybody help?

thanks in advance

**Rapha** · April 28th 2010, 09:37 PM

Hi.

Originally Posted by Mitch281008

Hey guys,

I need help completing this square:

x^2 + 3x + 2

Ive been following tips online, and ive gotten as far as:

x^2 + 3x + 9/4 = 1/4

but i cant get any further because no matter where i go to i dont understand the next step.

can anybody help?

thanks in advance

WEll, there only is one step missing.

Do you know that

$(a+b)^2 = a^2+2ab+b^2$ ?

Now consider $x^2 + 3x + 9/4$

What we know is $a^2=x^2, 2ab = 3x$ and $b^2 = 9/4$

We know that $a = x$ and $b^2 = 9/4$ . What is b? b is equal to $\sqrt{ 9/4 } = 3/2$

So we have

$[ (a+b)^2 = ] (x+3/2)^2$

Solving $(x+3/2)^2$ leads to $x^2+2*\frac{3}{2}x + (\frac{3}{2})^2 = x^2+3x+9/4$

Thus our calculation is correct.

**ADARSH** · April 28th 2010, 09:37 PM

Simple steps are

x^2 + 3x + 2

$x^2 + \frac{2*3x}{2} + \frac{9}{4} - \frac{1}{4}$

$x^2 + \frac{2*3x}{2} + (\frac{3}{2})^2 - \frac{1}{4}$

This is of the form

a^2 + 2ab + b^2 -c

it can be written as

(a+b)^2 -c

$(x+\frac{3}{2})^2 - \frac{1}{4}$

Game Over XX

EDIT: Rapha was quicker ..... and correct ofcourse

**Mitch281008** · April 28th 2010, 10:51 PM

ok thanks alot for that. im still not really confident with them, but that does help explain.

cheers

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