## Precalculus (6th Edition)

Francisco was wrong because the quadratic formula works even when $c=0$.
RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-8, c=0$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-8) \pm \sqrt{(-8)^2-4(1)(0)}}{2(1)} \\x=\dfrac{8\pm\sqrt{64-0}}{2} \\x=\dfrac{8\pm\sqrt{64}}{2} \\x=\dfrac{8\pm8}{2} \\x_1=\dfrac{8-8}{2}=0 \\x_2=\dfrac{8+8}{2}=8$ Thus, the solution set is $\color{blue}{\left\{0, 8\right\}}$. Francisco was wrong because the quadratic formula actually works for regardless of whether $c=0$ or not.