Question 223003
The key to this and similar problems is to understand that if some number, let's call it "r", is a root of a polynomial, let's call it P(x), then (x-r) is a factor of P(x) (and vice versa).<br>
So a polynomial with roots of 2/3 and 4 would be:
{{{P(x) = (x - 2/3)(x-4)}}}
Multiplying this out we get:
{{{P(x) = x^2 -4x - (2/3)x + 8/3}}}
Adding like terms we get:
{{{P(x) = x^2 - (14/3)x + 8/3}}}
So our quadratic equation is:
{{{x^2 - (14/3)x + 8/3 = 0}}}
If we don't want fractions, multiply both sides by 3:
{{{3x^2 - 14x + 8 = 0 }}}