what are the zeros of the quadratic function $f(x)=8x^{2}-16x - 15$?\n$x=-1-sqrt{2}$ and $x=-1+sqrt{2}$\n$x=…

what are the zeros of the quadratic function $f(x)=8x^{2}-16x - 15$?\n$x=-1-sqrt{2}$ and $x=-1+sqrt{2}$\n$x=-1-sqrt{\frac{15}{8}}$ and $x=-1+sqrt{\frac{15}{8}}$\n$x=1-sqrt{\frac{23}{8}}$ and $x=1+sqrt{\frac{23}{8}}$\n$x=1-sqrt{7}$ and $x=1+sqrt{7}$

what are the zeros of the quadratic function $f(x)=8x^{2}-16x - 15$?\n$x=-1-sqrt{2}$ and $x=-1+sqrt{2}$\n$x=-1-sqrt{\frac{15}{8}}$ and $x=-1+sqrt{\frac{15}{8}}$\n$x=1-sqrt{\frac{23}{8}}$ and $x=1+sqrt{\frac{23}{8}}$\n$x=1-sqrt{7}$ and $x=1+sqrt{7}$

Answer

Explanation:

Step1: Recall quadratic - formula

For a quadratic function (ax^{2}+bx + c = 0), the solutions are given by (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). For (f(x)=8x^{2}-16x - 15), we have (a = 8), (b=-16), and (c=-15).

Step2: Calculate the discriminant (\Delta=b^{2}-4ac)

Substitute (a = 8), (b=-16), (c=-15) into the discriminant formula. (\Delta=(-16)^{2}-4\times8\times(-15)=256 + 480=736).

Step3: Find the roots using the quadratic - formula

(x=\frac{-(-16)\pm\sqrt{736}}{2\times8}=\frac{16\pm\sqrt{16\times46}}{16}=\frac{16\pm4\sqrt{46}}{16}=\frac{4\pm\sqrt{46}}{4}). Another way: [ \begin{align*} x&=\frac{16\pm\sqrt{736}}{16}\ &=\frac{16\pm4\sqrt{46}}{16}\ & = 1\pm\frac{\sqrt{46}}{4}=1\pm\sqrt{\frac{46}{16}}=1\pm\sqrt{\frac{23}{8}} \end{align*} ]

Answer:

C. (x = 1-\sqrt{\frac{23}{8}}) and (x = 1+\sqrt{\frac{23}{8}})