3x^2 + 8 = -4
no solution. I suspect a typo. Assuming an 8x, we have
(x+2)(3x+2) = 0
there isn't any 8x it just say solve for x
well, shoot. Then just subtract 8 from both sides to get
3x^2 = -12
x^2 = -4
there is no real solution, since x^2 is never negative for real x.
If you allow complex solutions, then x = 2i or -2i. I believe this is from Algebra I, and should not be posing a difficulty for someone in Pre-Cal...
To find the value of x in the equation 3x^2 + 8 = -4, we need to solve for x. Here's how you can do it step by step:
Step 1: Subtract 8 from both sides of the equation:
3x^2 + 8 - 8 = -4 - 8
3x^2 = -12
Step 2: Divide both sides of the equation by 3 to isolate x^2:
3x^2 / 3 = -12 / 3
x^2 = -4
Step 3: Take the square root of both sides of the equation to solve for x:
√(x^2) = √(-4)
x = ±√(-4)
At this point, we encounter an issue since the square root of a negative number (√(-4)) is not a real number. Therefore, the equation has no real solutions.