recall that for any quadrataic in general form
the sum of the roots is -b/a
and the product of the roots is c/a
let the roots be r and 3r
so -b/a = 4r ---- b = -4ar
and c/a = 3r^2 ---- c = 3ar^2
I will assume your b2/ac is supposed to be b^2/(ac)
then b^2/(ac)
= (16 a^2 r^2)/(a(3ar^2))
= 16/3
If one root of ax2+bx+c =2 is 3 times the other root then find b2/ac
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