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: x2+2x15=−1 \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 x2+2xx^{2} + 2x equals −1-1, we can write: x2+2x=(−1)15 x^{2} + 2x = (-1)^{15} Simplify the right-hand side: x2+2x=−1 x^{2} + 2x = -1

Step 3: Rearrange the equation into standard quadratic form

Move all terms to one side to form a quadratic equation: x2+2x+1=0 x^{2} + 2x + 1 = 0 This simplifies to: (x+1)2=0 (x + 1)^{2} = 0

Step 4: Solve the quadratic equation

Take the square root of both sides: x+1=0 x + 1 = 0 Solve for xx: x=−1 x = -1

Step 5: Verify the solution

Substitute x=−1x = -1 back into the original equation to ensure it satisfies the equation: (−1)2+2(−1)15=1−215=−115=−1 \sqrt[15]{(-1)^{2} + 2(-1)} = \sqrt[15]{1 - 2} = \sqrt[15]{-1} = -1 The solution x=−1x = -1 is valid.

Step 6: Provide the solution set

The solution set is: {−1} \boxed{\{-1\}}

Final Answer

{−1}\boxed{\{-1\}}

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