Question

solve for z.
$32z^{2}-18z + 1 = 0$
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.
z =

Explanation:

Step1: Identify quadratic coefficients

For $32z^2 - 18z + 1 = 0$, we have $a=32$, $b=-18$, $c=1$.

Step2: Apply quadratic formula

The quadratic formula is $z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute values:

$$ z=\frac{-(-18)\pm\sqrt{(-18)^2-4(32)(1)}}{2(32)} $$

Step3: Calculate discriminant

Compute $b^2-4ac$:

$$ (-18)^2-4(32)(1)=324-128=196 $$

Step4: Simplify square root

$\sqrt{196}=14$

Step5: Compute two solutions

First solution (plus sign):

$$ z=\frac{18+14}{64}=\frac{32}{64}=\frac{1}{2} $$

Second solution (minus sign):

$$ z=\frac{18-14}{64}=\frac{4}{64}=\frac{1}{16} $$

Answer:

$\frac{1}{2}, \frac{1}{16}$