A function \( \mathrm{g}(\mathrm{x})=\mathrm{x}+2 \) and \( \mathrm{h}(\mathrm{x})=x^{2} \). Find the value of x for which \( \mathrm{hg}(\mathrm{x})=3 x^{2}+x-1 \) Select one: a. \( \quad x=\frac{-5}{2}, 1 \) b. \( \quad x=\frac{5}{2}, 1 \) c. \( \quad x=\frac{-5}{2},-1 \) d. \( \quad x=\frac{5}{2},-1 \)

To find the value of \( x \) for which \( hg(x) = 3x^2 + x - 1 \), we first need to determine the composition \( hg(x) \). Here, \( g(x) = x + 2 \), so we can substitute this into \( h(x) \): \[ h(g(x)) = h(x + 2) = (x + 2)^2 \] Now, calculating \( (x + 2)^2 \): \[ (x + 2)^2 = x^2 + 4x + 4 \] Next, we set this equal to \( 3x^2 + x - 1 \): \[ x^2 + 4x + 4 = 3x^2 + x - 1 \] Now, rearranging the equation to set it to zero: \[ 0 = 3x^2 + x - 1 - (x^2 + 4x + 4) \] \[ 0 = 3x^2 + x - 1 - x^2 - 4x - 4 \] \[ 0 = 2x^2 - 3x - 5 \] Now, we can use the quadratic formula to solve for \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \] Calculating the discriminant: \[ (-3)^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49 \] Now substituting back into the quadratic formula: \[ x = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4} \] This gives two solutions: \[ x = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1 \] Thus, the values of \( x \) that satisfy the equation are \( x = \frac{5}{2}, -1 \). The correct answer is: d. \( \quad x=\frac{5}{2},-1 \)