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The real roots of 2x^3 - 7x^2 + kx - 2 = 0 are in geometric progression.  Find these three roots.

Dec 18, 2019

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The real roots of $$2x^3 - 7x^2 + kx - 2 = 0$$ are in geometric progression.
Find these three roots.

$$\begin{array}{|rcll|} \hline 2(x-a_1)(x-a_1r)(x-a_1r^2) &=& 2x^3 - 7x^2 + kx - 2 = 0 \\ \ldots \\ 2\left(x^3-a_1(r^2+r+1)x^2+a_1^2(r^3+r^2+r)x-a_1^3r^3 \right) &=& 2x^3 - 7x^2 + kx - 2 = 0 \\ 2\left(x^3-a_1(r^2+r+1)x^2+a_1^2r(r^2+r+1)x-a_1^3r^3 \right) &=& 2x^3 - 7x^2 + kx - 2 = 0 \\ 2x^3\underbrace{-2a_1(r^2+r+1)}_{=-7}x^2+\underbrace{2a_1^2r(r^2+r+1)}_{=k}x\underbrace{-2a_1^3r^3}_{=-2} &=& 2x^3 - 7x^2 + kx - 2 = 0 \\ \text{compare} \\ -2a_1^3r^3=-2 \\ a_1^3r^3= 1 \\ a_1 r = 1 \\ \mathbf{a_1=\dfrac{1}{r}} \\\\ -2a_1(r^2+r+1)=-7 \\ 2a_1(r^2+r+1)=7 \\ 2\dfrac{1}{r}(r^2+r+1)=7 \\ \mathbf{r^2+r+1=\dfrac{7r}{2}} \\\\ k = 2a_1^2r(r^2+r+1) \\ k = 2\dfrac{1}{r^2}r(r^2+r+1) \\ k = 2\dfrac{1}{r}(r^2+r+1) \\ k = 2\dfrac{1}{r}*\dfrac{7r}{2} \\ \mathbf{k = 7} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{2x^3 - 7x^2 + 7x - 2} &=& \mathbf{0} \\ x_1 &=& \dfrac{1}{2} \\ x_2 &=& 1 \\ x_3 &=& 2 \\ \text{geometric progression:}~ \dfrac{1}{2},\ \dfrac{1}{2}*2^1,\ \dfrac{1}{2}*2^2 \\ \hline \end{array}$$

Dec 18, 2019