Completing the Square using Fractions

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

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

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

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

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

Combine like terms on the right to get

$\displaystyle \frac{\25}{16}$

and then subtract that from both sides to get

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

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

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

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.

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

Now, I also used $\displaystyle \frac{\49}{4}$ as the square of half of $\displaystyle -\frac{\7}{2}$ but eventually it also ends up as

$\displaystyle \frac{\3}{2}$ 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.

Re: Completing the Square using Fractions

Quote:

Originally Posted by

**Ingersoll** $\displaystyle 2x^2 -7x +3 = 0$

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

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

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

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

Combine like terms on the right to get

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

and then subtract that from both sides to get ... no

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

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

"un" square both sides ...

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

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

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

Re: Completing the Square using Fractions

Your error occurs right about here:

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

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.

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

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

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

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

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

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

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

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

Re: Completing the Square using Fractions

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