If the ratio of two roots of the equation x^3 | vedclass

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Question

(A) Let the roots be $3\alpha, 2\alpha, \beta$. (Since the ratio of two roots is $3 : 2$.)From the sum of roots: $3\alpha + 2\alpha + \beta = -(-9)/1

Vedclass · Accepted Answer

$6, 4, -1$

Vedclass · Answer

(A) Let the roots be $3\alpha, 2\alpha, \beta$. (Since the ratio of two roots is $3 : 2$.)From the sum of roots: $3\alpha + 2\alpha + \beta = -(-9)/1 = 9$$\Rightarrow 5\alpha + \beta = 9 \dots (i)$From the sum of product of roots taken two at a time: $(3\alpha)(2\alpha) + (2\alpha)(\beta) + (3\alpha)(\beta) = 14$$\Rightarrow 6\alpha^2 + 5\alpha\beta = 14 \dots (ii)$From the product of roots: $(3\alpha)(2\alpha)(\beta) = -24$$\Rightarrow 6\alpha^2\beta = -24$$\Rightarrow \alpha^2\beta = -4 \dots (iii)$From $(i)$,$\beta = 9 - 5\alpha$. Substituting this into $(ii)$:$6\alpha^2 + 5\alpha(9 - 5\alpha) = 14$$6\alpha^2 + 45\alpha - 25\alpha^2 = 14$$-19\alpha^2 + 45\alpha - 14 = 0$$19\alpha^2 - 45\alpha + 14 = 0$$(19\alpha - 7)(\alpha - 2) = 0$So,$\alpha = 2$ or $\alpha = 7/19$.If $\alpha = 2$,then $\beta = 9 - 5(2) = -1$.Check with $(iii)$: $(2)^2(-1) = 4(-1) = -4$. This satisfies the equation.Thus,the roots are $3(2), 2(2), -1$,which are $6, 4, -1$.

If the ratio of two roots of the equation $x^3 - 9x^2 + 14x + 24 = 0$ is $3 : 2$,find the roots of the equation.

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