How do you solve 3x^2-6x-24=03x2−6x−24=0 by completing the square?

1 Answer
KillerBunny
Apr 3, 2015

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(x+a)2=b
So, if b\ge 0b≥0, you can take the square root at both sides to get
x+a=\pm\sqrt{b}x+a=±√b
and conclude x=\pm\sqrt{b}-ax=±√b−a.

First of all, let's divide by 33 both terms to obtain
x^2-2x-8=0x2−2x−8=0

Now, we have (x+a)^2=x^2+2ax+a^2(x+a)2=x2+2ax+a2. Since you equation starts with x^2-2xx2−2x, this means that 2ax=-2x2ax=−2x, and so a=-1a=−1.
Adding 99 at both sides, we have
x^2-2x+1=9x2−2x+1=9
Which is the form we wanted, because now we have
(x-1)^2=9(x−1)2=9
Which leads us to
x-1=\pm\sqrt{9}= \pm 3x−1=±√9=±3 and finally x=\pm3+1x=±3+1, which means x=-3+1=-2x=−3+1=−2 or x=3+1=4x=3+1=4

Related questions
See all questions in Completing the Square
Impact of this question
3391 views around the world
You can reuse this answer
Creative Commons License