what are the zeros of the quadratic function f(x) = 8x² - 16x - 15?\no x=-1 - √2 and x=-1 + √2\no x=-1…

Explanation:

Step1: Recall quadratic - formula

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

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

Substitute the values of $a$, $b$, and $c$ into the discriminant formula: [ \begin{align*} \Delta&=(-16)^{2}-4\times8\times(-15)\ &=256 + 480\ &=736 \end{align*} ]

Step3: Find the zeros using the quadratic - formula

[ \begin{align*} 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}\ & = 1\pm\frac{\sqrt{23}}{2} \end{align*} ] [ \begin{align*} x&=\frac{16\pm\sqrt{736}}{16}=\frac{16\pm4\sqrt{46}}{16}=1\pm\frac{\sqrt{23}}{2}=1\pm\sqrt{\frac{23}{4}} \end{align*} ] [ \begin{align*} x&=\frac{-(-16)\pm\sqrt{(-16)^{2}-4\times8\times(-15)}}{2\times8}\ &=\frac{16\pm\sqrt{256 + 480}}{16}\ &=\frac{16\pm\sqrt{736}}{16}\ &=\frac{16\pm4\sqrt{46}}{16}\ &=1\pm\frac{\sqrt{23}}{2}=1\pm\sqrt{\frac{23}{8}\times\frac{4}{4}}=1\pm\sqrt{\frac{23}{8}} \end{align*} ]

Answer:

$x = 1-\sqrt{\frac{23}{8}}$ and $x = 1+\sqrt{\frac{23}{8}}$