What are the zeros of the quadratic function [tex][tex]$f(x)=8x^2-16x-15$[/tex][/tex]?

A. [tex][tex]$x = -1 - \sqrt{2}$[/tex][/tex] and [tex][tex]$x = -1 + \sqrt{2}$[/tex][/tex]
B. [tex][tex]$x = -1 - \sqrt{\frac{15}{8}}$[/tex][/tex] and [tex][tex]$x = -1 + \sqrt{\frac{15}{8}}$[/tex][/tex]
C. [tex][tex]$x = 1 - \sqrt{\frac{23}{8}}$[/tex][/tex] and [tex][tex]$x = 1 + \sqrt{\frac{23}{8}}$[/tex][/tex]
D. [tex][tex]$x = 1 - \sqrt{7}$[/tex][/tex] and [tex][tex]$x = 1 + \sqrt{7}$[/tex][/tex]



Answer :

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]