Question

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

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For \(f(x)=8x^2-16x - 15\), we have \(a = 8\), \(b=-16\), \(c=-15\).

Step2: Calculate Discriminant (\(D\))

Discriminant \(D=b^2-4ac\). Substitute \(a = 8\), \(b=-16\), \(c=-15\):
\(D=(-16)^2-4\times8\times(-15)=256 + 480 = 736\)

Step3: Simplify Discriminant

Simplify \(\sqrt{D}=\sqrt{736}=\sqrt{16\times46}=\sqrt{16\times2\times23}=4\sqrt{46}\)? Wait, no, wait: Wait, \(736 = 16\times46\)? Wait, \(16\times46 = 736\), but also \(736=8\times92=8\times4\times23 = 32\times23\)? Wait, no, let's recalculate \(b^2-4ac\) again. \(b=-16\), so \(b^2 = 256\). \(4ac=4\times8\times(-15)=-480\), so \(-4ac = 480\). So \(D=256 + 480=736\). Wait, but \(736=16\times46\)? No, \(16\times46 = 736\), but \(46 = 2\times23\), so \(\sqrt{736}=\sqrt{16\times46}=4\sqrt{46}\)? Wait, no, maybe I made a mistake. Wait, the quadratic is \(8x^2-16x - 15\). Let's complete the square instead.

Step4: Complete the Square

\(f(x)=8x^2-16x - 15=8(x^2 - 2x)-15\). Complete the square inside the parentheses: \(x^2 - 2x=(x - 1)^2-1\). So \(f(x)=8[(x - 1)^2-1]-15=8(x - 1)^2-8 - 15=8(x - 1)^2-23\). Set \(f(x)=0\): \(8(x - 1)^2-23 = 0\). Then \(8(x - 1)^2=23\), \((x - 1)^2=\frac{23}{8}\), \(x - 1=\pm\sqrt{\frac{23}{8}}\), so \(x=1\pm\sqrt{\frac{23}{8}}\).

Answer:

\(x = 1-\sqrt{\frac{23}{8}}\) and \(x = 1+\sqrt{\frac{23}{8}}\) (the third option: \(x=1-\sqrt{\frac{23}{8}}\) and \(x=1+\sqrt{\frac{23}{8}}\))