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 \(b?\)
+14995 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.
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\(x=-\frac{b}{2}\pm \sqrt{(\frac{b}{2})^2-18}\\ \ \)
b = - 9 : x1 = 6 x2 = 3 x2 : x1 = 1 : 2
b= 9 : x1 = - 6 x2 = - 3 x2 : x1 = 1 : 2
The largest possible value of b is 9 (for x1: x2 = 1: 2).
\(b\in \mathbb R\ |b\le-6\sqrt{2}\ and\ 6\sqrt{2}\le b\)
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