Question

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

1. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
2. $x = \frac{-6 \pm \sqrt{(6)^2 - 4(9)(1)}}{2(9)}$
3. $x = \frac{-6 \pm \sqrt{36 - 36}}{18}$
4. $x = \square$

Explanation:

Step1: Simplify the square root

We have \( \sqrt{36 - 36}=\sqrt{0} = 0 \). So the equation becomes \( x=\frac{-6\pm0}{18} \).

Step2: Calculate the value of x

Since \( \pm0 \) doesn't change the numerator, we have \( x=\frac{-6}{18} \). Simplifying the fraction \( \frac{-6}{18} \) by dividing both the numerator and the denominator by their greatest common divisor, which is 6, we get \( x =-\frac{1}{3} \). (Note: Since the square root term is 0, both the 'plus' and 'minus' cases will give the same result, so there's a repeated root here.)

Answer:

\( -\frac{1}{3} \) (or \( \frac{-1}{3} \), and since it's a repeated root, we can also note that the zero of the quadratic function is \( x =-\frac{1}{3} \) with multiplicity 2, but as per the problem's request to write the answer as a fraction, \( -\frac{1}{3} \) is the solution for \( x \) in this step.)