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