The two roots of the equation x^3 - 9x^2 + 14 | vedclass

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Question

(A) Let the roots of the cubic equation be $3\alpha, 2\alpha, \beta$.From the properties of roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$:$1$.

Vedclass · Accepted Answer

$6, 4, -1$

Vedclass · Answer

(A) Let the roots of the cubic equation be $3\alpha, 2\alpha, \beta$.From the properties of roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$:$1$. Sum of roots: $3\alpha + 2\alpha + \beta = -(-9)/1 = 9 \implies 5\alpha + \beta = 9 \implies \beta = 9 - 5\alpha$ $(i)$.$2$. Product of roots taken two at a time: $(3\alpha)(2\alpha) + (2\alpha)(\beta) + (3\alpha)(\beta) = 14 \implies 6\alpha^2 + 5\alpha\beta = 14$ $(ii)$.$3$. Product of roots: $(3\alpha)(2\alpha)(\beta) = -24/1 = -24 \implies 6\alpha^2\beta = -24 \implies \alpha^2\beta = -4$ $(iii)$.Substitute $\beta = 9 - 5\alpha$ 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 \implies 19\alpha^2 - 45\alpha + 14 = 0$.Solving the quadratic equation: $(19\alpha - 7)(\alpha - 2) = 0$.So,$\alpha = 2$ or $\alpha = 7/19$.If $\alpha = 2$,then $\beta = 9 - 5(2) = -1$. Check with $(iii)$: $(2)^2(-1) = -4$,which is correct.The roots are $3(2), 2(2), -1$,which are $6, 4, -1$.

The two roots of the equation $x^3 - 9x^2 + 14x + 24 = 0$ are in the ratio $3 : 2$. The roots are:

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