On the first line, the 2 should get distributed only across the first set of brackets:
(2x+4)(x+2)-4=28
Work it down again similar to what you did, then get all your terms to one side of the equation. Solve by factoring (if possible) or by using the quadratic formula.
This is a multiple choice question.
What is the solution of the equation 2(x+2)^2-4=28
1)2 only
2)6 only
3)-2 and 6
4)2 and -6
I did (2x+4)(2x+4)-4=28
4x^2+8x+8x+16-4=28
4x^2+16x+12=28
4x^2+16x=16
This is where I get stuck, can you help? thank you
3 answers
ok, corrected is
2x^2+4x+4x+8-4=28
2x^2+8x+4=28
2x^2 +8x-24=0
none of them still work
2x^2+4x+4x+8-4=28
2x^2+8x+4=28
2x^2 +8x-24=0
none of them still work
That's correct.
Then,
2x^2+8x-24=0
2(x^2+4x-12)=0
2(x+6)(x-2)=0
x = -6, x = 2
Then,
2x^2+8x-24=0
2(x^2+4x-12)=0
2(x+6)(x-2)=0
x = -6, x = 2