# Thread: Completing the Square using Fractions

1. ## Completing the Square using Fractions

$2x^2 -7x +3 = 0$

$2x^2 -7x = -3$

$\frac{\2x^2}{2} -\frac{\7}{2}x = -\frac{\3}{2}$

Take one half of - $\frac{\7}{2}x$ and then square that answer which = $\frac{\49}{16}$ and add to both sides

$x^2 -\frac{\7}{2}$ + $\frac{\49}{16}$= - $\frac{\3}{2}$ + $\frac{\49}{16}$

Combine like terms on the right to get

$\frac{\25}{16}$

and then subtract that from both sides to get

$x^2 -\frac{\7}{2} + \frac{\24}{16}=0$

$\frac{\24}{16}$ can be reduced to $\frac{\3}{2}$ so that

$x^2 -\frac{\7}{2} + \frac{\3}{2}=0$

Eliminate the fractions by multiplying it all by 2 and then instead of completing a square I've completed a large useless circle as I arrive back at the original equation.

$2x^2 -7x +3=0$

Now, I also used $\frac{\49}{4}$ as the square of half of $-\frac{\7}{2}$ but eventually it also ends up as
$\frac{\3}{2}$ on the left hand side of the equation.

I did this problem with the quadratic formula in about 45 seconds but I have to learn it this way. If nothing else my LaTex skills have been markedly improved in this post.

Any help is appreciated. Thanks.

2. ## Re: Completing the Square using Fractions

Originally Posted by Ingersoll
$2x^2 -7x +3 = 0$

$2x^2 -7x = -3$

$\frac{\2x^2}{2} -\frac{\7}{2}x = -\frac{\3}{2}$

Take one half of - $\frac{\7}{2}x$ and then square that answer which = $\frac{\49}{16}$ and add to both sides

$x^2 -\frac{7}{2}x +\frac{49}{16}$= $-\frac{3}{2}+\frac{49}{16}$

Combine like terms on the right to get

$\frac{\25}{16}$ ... ok

and then subtract that from both sides to get ... no
$x^2 -\frac{7}{2}x+\frac{49}{16}= \frac{25}{16}$

$\left(x - \frac{7}{4}\right)^2 = \frac{25}{16}$ this is why they call it "completing" the square

"un" square both sides ...

$x - \frac{7}{4} = \pm \frac{5}{4}$

$x = \frac{7 \pm 5}{4}$

btw ... one open tex at the beginning of a line ... one close /tex at the end of a line

3. ## Re: Completing the Square using Fractions

$x^2 -\frac{\7}{2}x +\frac{\49}{16}= -\frac{\3}{2}+ \frac{\49}{16}$

Until this point, you had done everything right. Without doing anything else, the left-hand (only) can be factored into a product of squares. The right is a number, which we can find the square root of. First I'm going to rewrite this after simplifying the right-hand side.

$x^2 -\frac{\7}{2}x$ + $\frac{\49}{16}$= $\frac{\25}{16}$

Now we can factor the left-hand side (AS IT IS RIGHT NOW).

$\left(x-\frac{7}{4}\right)^2=\frac{\25}{16}$

Now that it is in this form, we can take the square root of both sides. We are left with

$x-\frac{7}{4}=\pm \frac{5}{4}$ remember that every number has two square roots...

$x=\frac{7}{4}\pm \frac{5}{4}$

If we subtract, $x=\frac{2}{4}=\frac{1}{2}$. If we add, we have $x=\frac{12}{4}=3$. Thus $x \in \{\frac{1}{2}, 3\}$

I go over the conceptual logic of completing the square in this video.

4. ## Re: Completing the Square using Fractions

Thank you both so much. It's always right in front of my face. Thanks again.