Question

The number of real solutions of the equation $x^2 - 10 |x| - 56 = 0$ is:

• A. \( 3 \)
• B. \( 4 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

When solving absolute value equations, split the equation into separate cases based on the definition of absolute value. Then solve each case individually.

Answer 1

The Correct Option is C

Given: $x^2 - 10 |x| - 56 = 0$ Goal: Find the number of real solutions. Steps: \begin{enumerate} \item Consider two cases: \begin{itemize} \item Case 1: $x \ge 0$ The equation becomes $x^2 - 10x - 56 = 0$. We can use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-56)}}{2 \cdot 1} = \frac{10 \pm \sqrt{324}}{2} = \frac{10 \pm 18}{2},\]so $x = 14$ or $x = -4$. Since $x \ge 0$, $x = 14$. \item Case 2: $x<0$ The equation becomes $x^2 + 10x - 56 = 0$. We can use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-56)}}{2 \cdot 1} = \frac{-10 \pm \sqrt{324}}{2} = \frac{-10 \pm 18}{2},\]so $x = 4$ or $x = -14$. Since $x<0$, $x = -14$. \end{itemize} \item Count the solutions: We have two solutions: $x = 14$ and $x = -14$. \end{enumerate} Answer: (c) 2