Given \( h(x)=x^{4}-36 x^{2}+275 \) (a) Evaluate \( h(\sqrt{11}) \) (b) Evaluate \( h(-\sqrt{11}) \) (c) Solve \( h(x)=0 \) Write your answers in exact form.

To solve the function \( h(x) = x^{4} - 36x^{2} + 275 \), let's tackle each part step by step. ### (a) Evaluate \( h(\sqrt{11}) \) Substituting \( x = \sqrt{11} \): \[ h(\sqrt{11}) = (\sqrt{11})^{4} - 36(\sqrt{11})^{2} + 275 \] Calculating \( (\sqrt{11})^{2} = 11 \) and \( (\sqrt{11})^{4} = 121 \): \[ h(\sqrt{11}) = 121 - 36 \cdot 11 + 275 \] \[ = 121 - 396 + 275 \] \[ = 121 + 275 - 396 = 396 - 396 = 0 \] So, \( h(\sqrt{11}) = 0 \). ### (b) Evaluate \( h(-\sqrt{11}) \) Substituting \( x = -\sqrt{11} \): \[ h(-\sqrt{11}) = (-\sqrt{11})^{4} - 36(-\sqrt{11})^{2} + 275 \] Since the fourth power and the square are unaffected by the sign: \[ h(-\sqrt{11}) = 121 - 36 \cdot 11 + 275 \] \[ = 121 - 396 + 275 \] \[ = 121 + 275 - 396 = 396 - 396 = 0 \] So, \( h(-\sqrt{11}) = 0 \). ### (c) Solve \( h(x) = 0 \) We know \( h(x) = x^{4} - 36x^{2} + 275 = 0 \). To solve, let \( y = x^{2} \). Then, we rewrite the equation as: \[ y^{2} - 36y + 275 = 0 \] We will use the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1, b = -36, c = 275 \): First, calculate the discriminant: \[ b^2 - 4ac = (-36)^{2} - 4 \cdot 1 \cdot 275 = 1296 - 1100 = 196 \] Now apply the quadratic formula: \[ y = \frac{36 \pm \sqrt{196}}{2} \] \[ y = \frac{36 \pm 14}{2} \] Calculating the two possible values for \( y \): 1. \( y = \frac{36 + 14}{2} = \frac{50}{2} = 25 \) 2. \( y = \frac{36 - 14}{2} = \frac{22}{2} = 11 \) Now substitute back \( y = x^{2} \): 1. For \( y = 25 \): \[ x^{2} = 25 \Rightarrow x = \pm 5 \] 2. For \( y = 11 \): \[ x^{2} = 11 \Rightarrow x = \pm \sqrt{11} \] So the solutions to \( h(x) = 0 \) are: \[ x = 5, -5, \sqrt{11}, -\sqrt{11} \]