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# toioo

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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 ?

ant101  Dec 31, 2017
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### 2+0 Answers

#1
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For this problem, we make use of the correspondence between sums/products of roots and coefficients of a polynomial.

Denote the two roots of the equation $$\alpha$$  and $$\beta$$ . We know that $$\alpha\beta = 48$$ , and $$\alpha/\beta = 3 \implies \alpha = 3\beta$$ .

So $$b = -\alpha - \beta = -4\beta$$ . To maximize $$b$$, we want to make $$\beta$$ negative and as large as possible. Given the relationship that $$\alpha = 3\beta$$ and that $$\alpha*\beta = 48$$, we see that $$\beta=4$$ or $$-4$$ . Clearly $$-4$$  maximizes $$b$$ , and $$b = \boxed{16}$$ .

azsun  Dec 31, 2017
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x^2  + bx  + 48  =   0

So........let  R  be  one root.......and  the other root  =  3R

And the product of the roots  =   48

So....this implies that

R * 3R  =  48

3R^2  =  48        divide both sides by 3

R^2  = 16            take both roots

R  = ±4

And... the sum of the roots  = - b

So

R +  3R  = - b

So....either ...

4 + 12  = - b

b  = -16

Or

- 4  +  -12  =  -b

16  =  b

So....the roots  -4 and -12    make   b  a  maximum    =  16

CPhill  Dec 31, 2017

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