The two solutions of the equation \(x^2+bx+48=0\) are in the ratio of 3 to 1 for some values of \(b\). What is the largest possible value of \(b\)?
It factors to (X+4)(X+12) = 0 so the largest value is –4.
As guest above showed...this can factor into (x+4)(x+12) or (x-4)(x-12) to keep the root ratio 3:1
this would make b = 16 or -16
the largest would be b= 16