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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?

 Nov 13, 2019
 #1
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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:
\(-b=3(-3)\)

\(b=9\)

.
 Nov 13, 2019

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