Solve for x.
-3x^2+8=-4
-3x^2 = -12
x^2 = 4
x = ±2
thanks and what about this one ?
Simplify the expression
(3x^-1y^2)^3(5x^2y)0
I'm getting confused because of the negative exponent
anything^0 = 1, so we just have
(3x^-1y^2)^3 = 27 x^-3 y^6 = 27y^6/x^3
negative exponents indicate a swap between numerator and denominator
x^-3 = 1/x^3
1/x^-3 = x^3
recall that x^a/x^b = x^(a-b)
To solve for x in the equation -3x^2 + 8 = -4, we need to isolate the variable x.
Step 1: Move the constant term to the other side of the equation.
Start by subtracting 8 from both sides:
-3x^2 + 8 - 8 = -4 - 8
-3x^2 = -12
Step 2: Divide both sides of the equation by the coefficient of x^2 to make the coefficient equal to 1.
Divide both sides of the equation by -3:
(-3x^2) / -3 = (-12) / -3
x^2 = 4
Step 3: Take the square root of both sides of the equation.
Since x^2 = 4, we can find the square root of both sides:
√(x^2) = ± √4
|x| = ± 2
Step 4: Solve for x.
To solve for x, we consider two cases:
Case 1: x = 2
If we take the positive square root, x = 2.
Case 2: x = -2
If we take the negative square root, x = -2.
Therefore, the solutions for x are x = 2 and x = -2.