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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 ?

 Dec 31, 2017
 #1
avatar+183 
+5

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 \(b = \boxed{16}\) .

 Dec 31, 2017
 #2
avatar+100571 
+4

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

 

 

 

cool cool cool

 Dec 31, 2017

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