Question

How do you solve `3x^2-6x-24=0` by completing the square?

Answer 1

Solving a quadratic expression by completing the square means to manipulate the expression in order to write it in the form
`(x+a)^2=b`
So, if `b\ge 0`, you can take the square root at both sides to get
`x+a=\pm\sqrt{b}`
and conclude `x=\pm\sqrt{b}-a`.

First of all, let's divide by `3` both terms to obtain
`x^2-2x-8=0`

Now, we have `(x+a)^2=x^2+2ax+a^2`. Since you equation starts with `x^2-2x`, this means that `2ax=-2x`, and so `a=-1`.
Adding `9` at both sides, we have
`x^2-2x+1=9`
Which is the form we wanted, because now we have
`(x-1)^2=9`
Which leads us to
`x-1=\pm\sqrt{9}= \pm 3` and finally `x=\pm3+1`, which means `x=-3+1=-2` or `x=3+1=4`