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
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.
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
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
Certainly! It appears that you made a minor mistake in simplifying the equation. Let's go through the steps again together:
Starting from 4x^2 + 16x = 16, we want to rearrange the equation to solve for x.
Step 1: Subtract 16 from both sides of the equation:
4x^2 + 16x - 16 = 0
Step 2: Divide the entire equation by 4 to simplify:
x^2 + 4x - 4 = 0
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0. To find the solution for x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Comparing with our equation, we have a = 1, b = 4, and c = -4.
Substituting these values into the quadratic formula, we can calculate the solutions for x:
x = (-4 ± √(4^2 - 4*(1)*(-4))) / (2*1)
x = (-4 ± √(16 + 16)) / 2
x = (-4 ± √32) / 2
x = (-4 ± √(16 * 2)) / 2
x = (-4 ± 4√2) / 2
Now, let's simplify the solutions:
x₁ = (-4 + 4√2) / 2
x₁ = -2 + 2√2
x₂ = (-4 - 4√2) / 2
x₂ = -2 - 2√2
Therefore, the correct solution to the equation 2(x+2)^2 - 4 = 28 is:
3) -2 and 6