The equation 5x^2 + px + q = 0, where p and q are constants, has roots t and t+4. Show that p^2 = 20q + 400.

We know that if we have a polinomial of the form ax^2 + bx + c = 0, then:sum of the roots = -b/a and product of the roots = c/a Therefore: t + (t + 4) = -p/5 and t(t + 4) = q/5 Therefore: 2t + 4 = -p/5 so t + 2 = -p/10 and t^2 + 4t = q/5 so (t + 2)^2 = q/5 +4 Sub. the first relation into the second one: p^2/100 = q/5 + 4p^2 = 100(q/5 + 4) = 20q + 400 Therefore: p^2 = 20q + 400