Questions: Find the real solutions of the following equation. sqrt[15](x^2+2x)=-1

Solution Steps

The given equation is: \[ \sqrt[15]{x^{2}+2 x} = -1 \] The 15th root of a real number is defined for all real numbers, and its output is always a real number. However, the 15th root of a real number is negative only if the radicand (the expression inside the root) is negative.

Since the 15th root of \(x^{2} + 2x\) equals \(-1\), we can write: \[ x^{2} + 2x = (-1)^{15} \] Simplify the right-hand side: \[ x^{2} + 2x = -1 \]

Move all terms to one side to form a quadratic equation: \[ x^{2} + 2x + 1 = 0 \] This simplifies to: \[ (x + 1)^{2} = 0 \]

Take the square root of both sides: \[ x + 1 = 0 \] Solve for \(x\): \[ x = -1 \]

Substitute \(x = -1\) back into the original equation to ensure it satisfies the equation: \[ \sqrt[15]{(-1)^{2} + 2(-1)} = \sqrt[15]{1 - 2} = \sqrt[15]{-1} = -1 \] The solution \(x = -1\) is valid.

The solution set is: \[ \boxed{\{-1\}} \]

Final Answer