Hi, need help with completing the square

**Harvey** · October 22nd 2012, 03:45 AM

Hi I'm Harvey in London and I'm back at college after almost two decades of no academic studies and have my 1st assignment due in a week. I've managed to get all the questions answered but have been struggling with a particular on completing the square to solver a quadratic and have two issues with it:

Firstly I keep getting surd form answers for my x values, is this ok if I am to sketch the curve for it???

Secondly, that's if it's not ok, then can someone please tell me where I'm going wrong.

I have attached the question and my method of solving in a word doc to this message,

Please please can help me out here; I don't know how many hours I spent on this one question already

P.S. as this is an assignment question, obviously, I'm not asking for the answers just help with methodology

**Soroban** · October 22nd 2012, 09:52 AM

Hello, Harvey!

Solve by completing the square: . $y \;=\;1 + 3x - 72x^2$

We have: . $y \;=\;\text{-}72x^2 + 3x + 1$

Factor: n . $y \;=\; \text{-}72\left(x^2 - \tfrac{1}{24}x - \tfrac{1}{72}\right)$

Then: . . . $y\;=\; \text{-}72\left(x^2 - \tfrac{1}{24}x {\color{red}\:+\: \tfrac{1}{2304}} - \tfrac{1}{72} {\color{red}\:-\: \tfrac{1}{2304}}\right)$

n . . . . . . $y \;=\;\text{-}72\left[\left(x-\tfrac{1}{48}\right)^2 - \tfrac{33}{2304}\right]$

n . . . . . . $y \;=\;\text{-}72\left(x-\tfrac{1}{48}\right)^2 + \tfrac{33}{32}$

Got it?

**MarkFL** · October 22nd 2012, 11:42 AM

I opened the document, and it appears you want to solve by completing the square:

$-2x^2+3x+1=0$

There are two methods I know of to do this. The first is the method I was taught in school:

$-2\left(x^2-\frac{3}{2}x \right)=-1$

$x^2-\frac{3}{2}x=\frac{1}{2}$

$x^2-\frac{3}{2}x+\left(\frac{1}{2}\cdot\frac{3}{2} \right)^2=\frac{1}{2}+\left(\frac{1}{2}\cdot\frac{ 3}{2} \right)^2$

$\left(x-\frac{3}{4} \right)^2=\frac{17}{16}$

$x-\frac{3}{4}=\pm\frac{\sqrt{17}}{4}$

$x=\frac{3\pm\sqrt{17}}{4}$

The second method I picked up on another forum:

$-2x^2+3x+1=0$

$2x^2-3x=1$

Multiply through by $4(2)=8$

$16x^2-24x=8$

Add $(-3)^2=9$ to both sides:

$16x^2-24x+9=17$

$(4x-3)^2=17$

$4x-3=\pm\sqrt{17}$

$x=\frac{3\pm\sqrt{17}}{4}$

**Harvey** · October 22nd 2012, 03:30 PM

Thanks ever so much guys, you're both great people!! It turns out that my problem with the surds wasn't a problem at all and that in order to treat them as intercepts of the parabola on x-axis I just need to convert to decimal forms, duh. Thanks a lot for caring enough to post a reply and it's brilliant to have another way of completing the square, I shall show that to my maths teacher.

Keep calculating

Cheers

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