Question

which is a zero of the quadratic function $f(x)=9x^2 - 54x - 19$? $x = 6\frac{1}{3}$ $x = 3\frac{1}{3}$ $x = \frac{1}{3}$ $x = 9\frac{1}{3}$

Explanation:

Step1: Set function to 0

$9x^2 - 54x - 19 = 0$

Step2: Use quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=9$, $b=-54$, $c=-19$.

$$ x=\frac{54\pm\sqrt{(-54)^2-4(9)(-19)}}{2(9)} $$

Step3: Calculate discriminant

$\sqrt{2916 + 684}=\sqrt{3600}=60$

Step4: Compute two roots

$x=\frac{54+60}{18}=\frac{114}{18}=6\frac{1}{3}$, $x=\frac{54-60}{18}=\frac{-6}{18}=-\frac{1}{3}$

Answer:

$x=6\frac{1}{3}$