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

Find the real solutions of the following equation.
sqrt[15](x^2+2x)=-1
Transcript text: Find the real solutions of the following equation. \[ \sqrt[15]{x^{2}+2 x}=-1 \]
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Solution

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Solution Steps

Step 1: Analyze the equation

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.

Step 2: Set up the equation for the radicand

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

Step 3: Rearrange the equation into standard quadratic form

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

Step 4: Solve the quadratic equation

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

Step 5: Verify the solution

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.

Step 6: Provide the solution set

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

Final Answer

\(\boxed{\{-1\}}\)

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