Question

Solve the following quadratic equation by completing the square.
3x^2 -36x +96 =0

Answer 1

completing the square
you have something in this form
\[ax^2+bx+c = 0\]
subtract c from both sides
\[ax^2+bx = -c\]
add (\[(b/2)^2\] to both sides
so you have
\[ax^2+bx+(b/2)^2=-c\]
then you factor the left side of the eqn and solve from there

Answer 2

so for an example we take this
3x^2 -36x +96 =0

we want in form ax^2+bx+c so we divide 3x^2-36x+96 by 3
so we have \[x^2-12x+32=0\] (we just divided the whole eqn by 3)
bring c to the other side
\[x^2-12x=-32\]
now b = -12
\[(b/2)^2=(-12/2)^2\]
=36
so add 36 to both sides
\[x^2-12x+36=-32+36\]
\[x^2-12x+36=4\]
factor left side of eqn into
\[(x-6)^2=4\]
take positive/negative square root of 4
\[x-6 =2 \]
and \[x-6 =-2 \]
x = 8 and x = 4 plug into the original function to see if it works

Answer 3

TY :)

Answer 4

np