Not quite. You cannot factor x^2-14=0 and get (x-7)(x+2)=0 nor can you factor x^2-6=0 and get (x-2)(x+3). Try to expand them and you'll see they are not the same.
(x-7)(x+2) = x^2-5x-14
(x-2)(x+3) = x^2+x-6
Your work was find until you tried to factor them. There is a much simpler(?) solution. Try to do it and I'll be happy to check your answer if needed.
|x^2-10|=4
x^2-10=4 or x^2-10=-4
X^2-14=0 x^2-6=0
(x-7)(x+2)=0 (x-2)(x+3)=0
x=7 x=-2 x=2 x=-3
solution set {-3,-2,2,7} (?????)
3 answers
Is this right?
x^2-14=0 x^2-6=0
x^2=14 x^2=6
sqrtx^2=sqrt14 sqrtx^2=sqrt6
x=+-sqrt14 x=+-sqrt6
solution set is:
{+-sqrt6, +-sqrt14}
x^2-14=0 x^2-6=0
x^2=14 x^2=6
sqrtx^2=sqrt14 sqrtx^2=sqrt6
x=+-sqrt14 x=+-sqrt6
solution set is:
{+-sqrt6, +-sqrt14}
Yes.