If the roots of x^3-k x^2+14 x-8=0 are in geo | vedclass

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(B) Given the cubic equation $x^3-k x^2+14 x-8=0$. Let the roots be $\frac{a}{r}, a, ar$ which are in geometric progression. According to the relation

Vedclass · Accepted Answer

$7$

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(B) Given the cubic equation $x^3-k x^2+14 x-8=0$. Let the roots be $\frac{a}{r}, a, ar$ which are in geometric progression. According to the relation between roots and coefficients,the product of the roots is given by $-\frac{d}{a_{coeff}} = -\frac{-8}{1} = 8$. So,$\frac{a}{r} \cdot a \cdot ar = 8$ $\Rightarrow a^3 = 8$ $\Rightarrow a = 2$. Since $a=2$ is a root of the equation,it must satisfy the equation: $(2)^3 - k(2)^2 + 14(2) - 8 = 0$. $8 - 4k + 28 - 8 = 0$. $28 - 4k = 0$. $4k = 28 \Rightarrow k = 7$.

If the roots of $x^3-k x^2+14 x-8=0$ are in geometric progression,then $k$ is equal to

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