Need to know the answer to this equation.
(x+4)^2=-13
Need to know the answer to this equation.
(x+4)^2=-13
(x + 4)^{2} = -13
Take the square root of both sides of the equation:
√(x + 4)^{2} = √(-13)
x + 4 = ± √(-13)
The square root of a negative number is not a real number. Recall that the imaginary number, i, is defined as the square root of -1. That is, i = √(-1).
So, since the square root of -13 is equal to the product of the square root of -1 and the square root of 13, we can generate a solution to x that is in the form of a complex number:
x + 4 = ± √(-13)
x + 4 = ± √(-1)·√(13)
x + 4 = ± i√(13)
Solve for x by subtracting 4 from both sides of the equation:
x = -4 ± i√(13)
Thus, there are two solutions for x: x = -4 + i√(13) and x = -4 - i√(13)
There's no real number x that satisfies that equation. The square of a real number is always a non-negative number, so (x+4)^{2} must be positive or zero, it cannot be a negative number like -13.
Hope this helps!