Question

Consider the following quadratic equation. Find the real zeros, if any, of this function. Reduce all fractions to lowest term.

Answer 1

The given function is

`y=x^2-8x+12\text{ }`

To find the zeros, we use the quadratic formula.

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where a = 1, b = -8, and c = 12.

`\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(12)}_{}}{2(1)}=\frac{8\pm\sqrt[]{64-48}}{2} \\ x=\frac{8\pm\sqrt[]{16}}{2}=(8\pm4)/(2)=4\pm2 \\ x_1=4+2=6 \\ x_2=4-2=2 \end{gathered}`

Therefore, the zeros x = 6 and x = 2.