Hello there, I have a question on how to complete the x on a quadratic equation which has the coefficient of a. It cannot be solved using factorization. The equation is:
$\displaystyle 8x^2 + 4x - 1 = 0$
Thank You
I assume you mean to complete the square...
$\displaystyle 8x^2 + 4x - 1 = 0$
$\displaystyle 8\left(x^2 + \frac{1}{2}x - \frac{1}{8}\right) = 0$
$\displaystyle 8\left[x^2 + \frac{1}{2}x + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{8}\right] = 0$
$\displaystyle 8\left[\left(x + \frac{1}{4}\right)^2 - \frac{3}{16}\right] = 0$
$\displaystyle 8\left(x + \frac{1}{4}\right)^2 - \frac{3}{2} = 0$.
Go from here.
Thank you but I'm little bit confused on how do you exactly have
$\displaystyle - \frac{3}{2}$
From
$\displaystyle \frac{3}{16}$
I've also noticed that the second from the last line did not show up, so I edit your formatting a little bit. I'm not sure if what I'm doing is right. Can you verify?
Is it possible to do this:
$\displaystyle 8(x+1/4)^2 = 3/2 $
$\displaystyle 8(x+1/4) = root(3/2) $
$\displaystyle 8x + 2 = root(3/2)$
$\displaystyle 8x = root(3/2) - 2 $
$\displaystyle x = (root(3/2) - 2) / 8$
But it gives up quite weird answer...
I'm sorry also because of bad formattig I do not know how to operate Latex properly.
Very close. There are two square roots to every nonnegative number, one positive and one negative.
So that means it should read...
$\displaystyle \left(x + \frac{1}{4}\right)^2 = \frac{3}{16}$
$\displaystyle x + \frac{1}{4} = \pm \sqrt{\frac{3}{16}}$
$\displaystyle x + \frac{1}{4} = \pm \frac{\sqrt{3}}{4}$
$\displaystyle x = -\frac{1}{4} \pm \frac{\sqrt{3}}{4}$.
So your two solutions are $\displaystyle x = \frac{-1 - \sqrt{3}}{4}$ and $\displaystyle x = \frac{-1 + \sqrt{3}}{4}$.