SOLUTION: Solve 2x^3 + 3x^2 +hx + k = 0 and find the values of h and k, given that -3 is the first root and the third root is twice the second.

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Question 1183949: Solve 2x^3 + 3x^2 +hx + k = 0 and find the values of h and k, given that -3 is the first root and the third root is twice the second.
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
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Solve 2x^3 + 3x^2 +hx + k = 0 and find the values of h and k, given that -3 is the first root and the third root is twice the second.
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Let x be the second root.  

Then the third root is 2x.


Next, according to Vieta's theorem, the ratio of the coefficient at  x^2  to the leading coefficient,

i.e. the number  ,  is the sum of the three roots of the equation, taken with the opposite sign


     = - .


Simplify and find x


    3 = -2*((-3) + 3x)

    3 = 6 - 6x

    6x = 6 - 3

    6x = 3

    x = 3/6 = 1/2.


Thus, the second root is  1/2;   the third root is 1.


According to Vieta's theorem,  the ratio    is the product of the three roots taken with the opposite sign

     = -  = 3/2;  hence,  k = 3.


According to Vieta's theorem, the ratio    is the sum of in-pairs product of the roots

     =  =  = -4;  hence, h = -8


ANSWER.  k = 3;  h = -8;  the second root is  ;  the third root is 1.

You can check it on your own, that the found polynomial and its roots satisfy all the conditions imposed by the problem.


Solved.



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