Hi all, I would appreciate some help on how to solve the following problem. If x^2+kx+54=0 has one root that is twice the other root, what are the values of k? Thanks a lot!
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Originally Posted by asamimasa Hi all, I would appreciate some help on how to solve the following problem. If x^2+kx+54=0 has one root that is twice the other root, what are the values of k? Thanks a lot! The equation has the solutions: $\displaystyle x = t~\vee~ x = 2t$ Then $\displaystyle (x-t)(x-2t)=0~\implies~x^2-3t+2t^2 = 0$ Thus: $\displaystyle 2t^2 = 54 ~\implies~t = 3\sqrt{3}~\vee~t = -3\sqrt{3}$ And therefore $\displaystyle k = -9\sqrt{3}~\vee~k=9\sqrt{3}$
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