solve for the zeros of the quadratic function $f(x)=9x^{2}+6x + 1$. write the answer as a fraction.\n1…

solve for the zeros of the quadratic function $f(x)=9x^{2}+6x + 1$. write the answer as a fraction.\n1. $x=\frac{-bpmsqrt{b^{2}-4ac}}{2a}$\n2. $x=\frac{-6pmsqrt{(6)^{2}-4(9)(1)}}{2(9)}$\n3. $x=\frac{-6pmsqrt{36 - 36}}{18}$\n4. $x=square$

solve for the zeros of the quadratic function $f(x)=9x^{2}+6x + 1$. write the answer as a fraction.\n1. $x=\frac{-bpmsqrt{b^{2}-4ac}}{2a}$\n2. $x=\frac{-6pmsqrt{(6)^{2}-4(9)(1)}}{2(9)}$\n3. $x=\frac{-6pmsqrt{36 - 36}}{18}$\n4. $x=square$

Answer

Explanation:

Step1: Simplify the numerator

Since $\sqrt{36 - 36}=\sqrt{0}=0$, the numerator of the fraction $-6\pm\sqrt{36 - 36}$ becomes $- 6\pm0$.

Step2: Calculate the value of $x$

$x=\frac{-6\pm0}{18}$. Whether we take the plus - or minus sign, the result is $x =-\frac{6}{18}=-\frac{1}{3}$.

Answer:

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