Solve: 48+2x^2=16
I think I saw this same question yesterday.
If as written:
48 + 2x^2 = 16
2x^2 = -32
x^2 = -16
x = ± 4i , where i = √-1
If you meant:
48x + 2x^2 = 16
x^2 + 24x - 8 = 0
x + (-24 ± √608)/2
If you meant:
.....
Hello. It is the first one.
I get up to x^2= -16, but how does x = ± 4i work? I assume it's opposite of ^ in this case to remove ^2. What is ± in this case?
To solve the given equation 48 + 2x^2 = 16, we need to isolate the variable x and solve for its value.
Step 1: Subtract 48 from both sides of the equation to remove it from the left side:
48 + 2x^2 - 48 = 16 - 48
2x^2 = -32
Step 2: Divide both sides of the equation by 2 to isolate x^2:
(2x^2)/2 = -32/2
x^2 = -16
Step 3: Take the square root of both sides of the equation to solve for x:
√(x^2) = √(-16)
x = ±√(-16)
At this point, we encounter a problem because we are taking the square root of a negative number. The square root of a negative number is not a real number, so there are no real solutions to this equation. If you're working in the complex number system, you can simplify √(-16) as ±4i, where i is the imaginary unit.