Questions: Solve the following quadratic-like equation. [ (y^2-5)^2+4(y^2-5)-32=0 ]

Solution Steps

Let \( u = x^2 - 5 \). This substitution simplifies the given equation.

Substitute \( u \) into the equation: \[ u^2 + 4u - 32 = 0 \]

Solve the quadratic equation \( u^2 + 4u - 32 = 0 \) using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ u = \frac{-4 \pm \sqrt{16 + 128}}{2} \] \[ u = \frac{-4 \pm \sqrt{144}}{2} \] \[ u = \frac{-4 \pm 12}{2} \] \[ u = 4 \quad \text{or} \quad u = -8 \]

Substitute back \( u = x^2 - 5 \):

1. For \( u = 4 \): \[ x^2 - 5 = 4 \] \[ x^2 = 9 \] \[ x = \pm 3 \]

2. For \( u = -8 \): \[ x^2 - 5 = -8 \] \[ x^2 = -3 \] \[ x = \pm \sqrt{-3} \] \[ x = \pm i\sqrt{3} \]

Final Answer