If the roots of the equation x^3-7x^2+14x-8=0 | vedclass

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(D) Given the equation $x^3-7x^2+14x-8=0$. Let the roots be $\frac{a}{r}, a, ar$. From the product of the roots: $\frac{a}{r} \cdot a \cdot ar = 8$ $\

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$3$

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(D) Given the equation $x^3-7x^2+14x-8=0$. Let the roots be $\frac{a}{r}, a, ar$. From the product of the roots: $\frac{a}{r} \cdot a \cdot ar = 8$ $\Rightarrow a^3 = 8$ $\Rightarrow a = 2$. From the sum of the roots: $\frac{a}{r} + a + ar = 7$. Substituting $a=2$: $\frac{2}{r} + 2 + 2r = 7$ $\Rightarrow \frac{2}{r} + 2r = 5$ $\Rightarrow 2 + 2r^2 = 5r$ $\Rightarrow 2r^2 - 5r + 2 = 0$. Solving the quadratic equation: $2r^2 - 4r - r + 2 = 0$ $\Rightarrow 2r(r-2) - 1(r-2) = 0$ $\Rightarrow (2r-1)(r-2) = 0$. So,$r = 2$ or $r = \frac{1}{2}$. If $r=2$,the roots are $\frac{2}{2}, 2, 2(2)$,which are $1, 2, 4$. If $r=\frac{1}{2}$,the roots are $\frac{2}{1/2}, 2, 2(1/2)$,which are $4, 2, 1$. In both cases,the roots are $1, 2, 4$. The largest root is $4$ and the smallest root is $1$. The difference is $4 - 1 = 3$.

If the roots of the equation $x^3-7x^2+14x-8=0$ are in geometric progression,then the difference between the largest and the smallest roots is

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