Solve by graphing: \( \frac{-4}{2 x+5}=x^{2}-1 \) \( \left\{-\frac{8}{5}, \frac{6}{5}, \frac{29}{10}\right\} \) \( \left\{\frac{19}{10}\right\} \) \( \varnothing \) \( \left\{-\frac{29}{10},-\frac{6}{5}, \frac{8}{5}\right\} \)

To solve the equation by graphing: \[ \frac{-4}{2x + 5} = x^2 - 1 \] **Step 1: Rearrange the Equation** Bring all terms to one side to set the equation to zero: \[ x^2 - 1 + \frac{4}{2x + 5} = 0 \] **Step 2: Multiply Both Sides by \(2x + 5\) (Assuming \(2x + 5 \neq 0\))** \[ -4 = (x^2 - 1)(2x + 5) \] Expanding the right side: \[ -4 = 2x^3 + 5x^2 - 2x - 5 \] Bring all terms to one side: \[ 2x^3 + 5x^2 - 2x - 1 = 0 \] **Step 3: Analyze the Cubic Equation** To find the real roots of \(2x^3 + 5x^2 - 2x - 1 = 0\): - **Possible Rational Roots**: \(\pm1, \pm\frac{1}{2}\) Testing \(x = -\frac{1}{2}\): \[ 2\left(-\frac{1}{2}\right)^3 + 5\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) - 1 = 0 \] So, \(x = -\frac{1}{2}\) is a root. However, substitute \(x = -\frac{1}{2}\) back into the original equation to verify: \[ \frac{-4}{2\left(-\frac{1}{2}\right) + 5} = \left(-\frac{1}{2}\right)^2 - 1 \] \[ \frac{-4}{-1 + 5} = \frac{1}{4} - 1 \] \[ -1 \neq -\frac{3}{4} \] This shows that \(x = -\frac{1}{2}\) is an **extraneous solution**. **Step 4: Conclusion** Since the only potential real root does not satisfy the original equation, there are **no valid solutions**. **Final Answer:** \(\varnothing\)