Asked by sara
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
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
Answers
Answered by
Nonetheless
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.
(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.
Answered by
sara
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
Answered by
Nonetheless
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
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