Question 119488
Solve for x:
{{{5x^2-30x+40 = 0}}} First, you can factor a 5 from this quadratic equation.
{{{5(x^2-6x+8) = 0}}} Now apply the "zero product principle": If{{{a*b = 0}}}then either{{{a = 0}}}or{{{b = 0}}} or both.
So...
If {{{5(x^2-6x+8) = 0}}}then either {{{5 = 0}}} (obviously not true!) or {{{x^2-6x+8 = 0}}}. Ok, so...
{{{x^2-6x+8 = 0}}} Solve this by factoring:
{{{(x-2)(x-4) = 0}}} ...again, apply the "zero product principle"...
If {{{x-2 = 0}}} then {{{x = 2}}} or
If {{{x-4 = 0}}} then {{{x = 4}}}
The solutions are, as you have stated, x = 2 or x = 4.