# Completing the Square

• Apr 21st 2013, 10:32 AM
gramgilb
Completing the Square
Hey i am attempting to CTS 4x2 +24x-18=0

i divide by four to make the a coefficient zero but then get lost at the 18/4 and dont know what do with that.

Any help is appreciated
• Apr 21st 2013, 10:36 AM
dokrbb
Re: Completing the Square
you are trying to what... CTS?
• Apr 21st 2013, 10:44 AM
gramgilb
Re: Completing the Square
Ya i know the answer from the answer sheet but everytime i try i get the wrong answer.
The answer is supposed to be x=1/2(-6 plus/minus 3 √6) bu this way seems like it was divided by two and the a coefficient was 2...
• Apr 21st 2013, 10:52 AM
Plato
Re: Completing the Square
Quote:

Originally Posted by gramgilb
Ya i know the answer from the answer sheet but everytime i try i get the wrong answer.
The answer is supposed to be x=1/2(-6 plus/minus 3 √6) bu this way seems like it was divided by two and the a coefficient was 2...

Exapand $\displaystyle (2x+6)^2-54$ and see what you get.
• Apr 21st 2013, 12:52 PM
gramgilb
Re: Completing the Square
ya i got it ty
• Apr 21st 2013, 08:08 PM
ibdutt
Re: Completing the Square
Foe such questions adopt the folloing algorithm.
Step - 1: Make coefficient = 1. in this case : divide by 4: x^2 + 6x - 9/2 = 0
step - 2: transfer the constant term to the RHS: we will get x^2 + 6x = 9/2
Step 3: Add the square of half the coefficient of x on both the sides of the equation. in this case the coefficient of x = 6; Half the coefficient of x = 3 and square of this is 9. Thus we have: x^2 + 6x + 9= 9/2 + 9
That gives ( x + 3 )^2 = 27/2
• Apr 21st 2013, 10:32 PM
agentmulder
Re: Completing the Square
Quote:

Originally Posted by gramgilb
Hey i am attempting to CTS 4x2 +24x-18=0

i divide by four to make the a coefficient zero but then get lost at the 18/4 and dont know what do with that.

Any help is appreciated

For this particular problem here is a possible approach.

$\displaystyle 4x^2 + 24x = 18$

fnd $\displaystyle \sqrt{4} = 2$ , now divide 24 by $\displaystyle (2 \sqrt{4}) = 4$ , $\displaystyle \frac{24}{4} = 6$ , then square 6, 6^2 = 36

now look where sqrt(4) = 2 , 6 and 36 go...

$\displaystyle (2x + 6)^2 - 36 = 18$

$\displaystyle (2x + 6)^2 = 54$

The square has been completed... extract the roots...

$\displaystyle 2x + 6 = \pm \sqrt{54}$

$\displaystyle 2x = -6 \pm 3 \sqrt{6}$

$\displaystyle x = \frac{-6 \pm 3 \sqrt{6}}{2}$

:)
• Apr 22nd 2013, 12:40 AM
Prove It
Re: Completing the Square
Quote:

Originally Posted by gramgilb
Hey i am attempting to CTS 4x2 +24x-18=0

i divide by four to make the a coefficient zero but then get lost at the 18/4 and dont know what do with that.

Any help is appreciated

Using the standard method...

\displaystyle \displaystyle \begin{align*} 4x^2 + 24x - 18 &= 0 \\ 4 \left( x^2 + 6x - \frac{9}{2} \right) &= 0 \\ 4 \left( x^2 + 6x + 3^2 - 3^2 - \frac{9}{2} \right) &= 0 \\ 4 \left[ \left( x + 3 \right) ^2 - 9 - \frac{9}{2} \right] &= 0 \\ 4 \left[ \left( x + 3 \right) ^2 - \frac{27}{2} \right] &= 0 \\ 4 \left( x + 3 \right) ^2 - 54 &= 0 \end{align*}

Go from here...