Need to know the answer to this equation.

(x+4)^2=-13

Need to know the answer to this equation.

(x+4)^2=-13

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(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!