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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 \(b\)?

 Mar 20, 2020
 #1
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It factors to (X+4)(X+12) = 0  so the largest value is –4

 Mar 20, 2020
 #2
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As guest above showed...this can factor into   (x+4)(x+12)     or    (x-4)(x-12)     to keep the root ratio 3:1

   this would make b =  16       or    -16

      the largest would be    b=  16

 Mar 20, 2020

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