To find the zeros of the quadratic function [tex]\( f(x) = 8x^2 - 16x - 15 \)[/tex], we need to solve the equation [tex]\( 8x^2 - 16x - 15 = 0 \)[/tex]. We can use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our quadratic function, [tex]\( a = 8 \)[/tex], [tex]\( b = -16 \)[/tex], and [tex]\( c = -15 \)[/tex].
First, we calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-16)^2 - 4 \cdot 8 \cdot (-15) \][/tex]
[tex]\[ \Delta = 256 + 480 \][/tex]
[tex]\[ \Delta = 736 \][/tex]
Next, we use the quadratic formula with the calculated discriminant:
[tex]\[ x = \frac{-(-16) \pm \sqrt{736}}{2 \cdot 8} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{736}}{16} \][/tex]
To simplify further, notice that:
[tex]\[ \sqrt{736} = \sqrt{16 \cdot 46} = 4\sqrt{46} \][/tex]
So, our equation becomes:
[tex]\[ x = \frac{16 \pm 4\sqrt{46}}{16} \][/tex]
[tex]\[ x = 1 \pm \frac{\sqrt{46}}{4} \][/tex]
Therefore, the zeros of the quadratic function are:
[tex]\[ x = 1 - \frac{\sqrt{46}}{4} \quad \text{and} \quad x = 1 + \frac{\sqrt{46}}{4} \][/tex]
Comparing these solutions to the given options:
[tex]\[ x = 1 - \frac{\sqrt{46}}{4} \quad \text{and} \quad x = 1 + \frac{\sqrt{46}}{4} \][/tex]
These match the correct choice out of the given options. Hence, the correct answer is:
[tex]\[ x = 1 - \sqrt{\frac{23}{8}} \quad \text{and} \quad x = 1 + \sqrt{\frac{23}{8}} \][/tex]
