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(D) Let the roots of the equation $x^3-13x^2+15x+189=0$ be $\alpha, \alpha+2, \beta$. From the relation between roots and coefficients: $1) \alpha + (

Vedclass · Accepted Answer

$-3, -7, 9$

Vedclass · Answer

(D) Let the roots of the equation $x^3-13x^2+15x+189=0$ be $\alpha, \alpha+2, \beta$. From the relation between roots and coefficients: $1) \alpha + (\alpha+2) + \beta = 13 \implies 2\alpha + \beta = 11 \implies \beta = 11 - 2\alpha$. $2) \alpha(\alpha+2) + (\alpha+2)\beta + \alpha\beta = 15$. $3) \alpha(\alpha+2)\beta = -189$. Substituting $\beta = 11 - 2\alpha$ into the product equation: $\alpha(\alpha+2)(11-2\alpha) = -189$. Testing the options,if the roots are $-3, 7, 9$: Sum: $-3 + 7 + 9 = 13$ (Matches coefficient of $x^2$). Product: $(-3) 	imes 7 	imes 9 = -189$ (Matches constant term). Difference between $7$ and $9$ is $2$. Thus,the roots are $-3, 7, 9$.

The difference between two roots of the equation $x^3-13x^2+15x+189=0$ is $2$. Then the roots of the equation are:

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