The two solutions of the equation x^2 + bx + 18 = 0 are in the ratio of 2 to 1 for some values of b. What is the largest possible value of b?
What I did (assuming b is an integer):
By Vieta's formulas
\(-b=x+y\)
\(18=xy\)
and since they are in a ratio of 2 to 1.
Let us pretend Y is the larger solution.
\(2x = y\)
So we substitute.
\(-b=3x\)
\(18=2x^2\)
Solve \(18=2x^2\)
\(9=x^2\)
\(x=3, -3\).
We have \(-b=3x\)
So the largest value should be:
\(-b=3(-3)\)
\(b=9\)