# Math Help - Completing the square

1. ## Completing the square

Just have a few questions, I have already attempted them but am having problems on how to square the fractions, therefore my final answers are not correct. Someone please take me through it step by step, especially the squaring fractions..

Co-efficient of x is 1:

1) x^2 - 3x - 5 = 0 ( How to square 3 / 2 )

Co-efficient of x more than 1:

1) 15 - 6 x - 2x^2

2) 4x - x = 8

3) 5x^2 + 8x -2 = 0

2. 1) $\left( \frac{a}{b} \right)^2 = \frac{a^2}{b^2}$

So $\left( \frac{-3}{2} \right) ^2 = \frac{9}{4}$

We want to write this as $(x-\frac{3}{2})^2+C$ where C is a constant that will make this equal to the original expression. We know we're going to add 9/4 when we square that expression, so now we have to subtract the same amount to stay equal. The only other constant left is -5. So $-5-\frac{9}{4}=\frac{-20}{4}-\frac{9}{4}=\frac{-29}{4}$ So the final answer should be $(x-\frac{3}{2})^2-\frac{29}{4}$

3. On the ones where the x^2 coefficient isn't 1, just divide by the coefficient and then you can complete the square like normal. Squaring fractions isn't very bad at all like I showed you. The adding and subtracting them might be worse for you, I'm not sure.

4. I'll help with the last one,

$5x^2 + 8x -2=0$

so divide through my 5 $x^2 + \frac{8}{5}x - \frac{2}{5}= 0$

$x^2 + \frac{8}{5}x = \frac{2}{5}$

now complete the square on the LHS, half the coefficient of x $\frac{8}{5} \div \frac{2}{1} = \frac{4}{5}$

$(x+ \frac{4}{5})^2 - (\frac{4}{5})^2= \frac{2}{5}$

when you square a fraction you just square top and bottom. if thats what your asking?

$(x+ \frac{4}{5})^2 = \frac{2}{5} + \frac{16}{25}$

$(x+\frac{4}{5})^2 = \frac{26}{25}$

$x+\frac{4}{5} = \sqrt\frac{26}{25}$

$x= -\frac{4}{5} \pm \frac{\sqrt26}{5}$

5. Thanks guys, OK so just one more I need someone to go through with me. Please explain all the steps so I don't get 'lost'..

ok,

x^2 + 5x + 2 = 0

Soo, ermm:

(x + 2.5) ^ 2 - (2.5) ^ 2 = 2

Then what ?

6. $x^2 + 5x + 2 = 0$

$x^2 + 5x = -2$

$x^2 + 5x + \left(\frac{5}{2}\right)^2 = -2 + \left(\frac{5}{2}\right)^2$

$\left(x + \frac{5}{2}\right)^2 = -2 + \frac{25}{4}$

$\left(x + \frac{5}{2}\right)^2 = \frac{17}{4}$

$x + \frac{5}{2} = \pm \frac{\sqrt{17}}{2}$

$x = \frac{-5 \pm \sqrt{17}}{2}$