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?

Guest Nov 13, 2019

#1**+2 **

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:

\(-b=3(-3)\)

\(b=9\)

.CalculatorUser Nov 13, 2019