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solve this equation (x+4)^2=-13

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2 Answers

     (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!