The roots alpha, beta of the equation x^2-6(k | vedclass

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(D) Given the equation $x^2-6(k-1)x+4(k-2)=0$. Since the roots $\alpha$ and $\beta$ are equal in magnitude but opposite in sign,we have $\alpha + \bet

Vedclass · Accepted Answer

$-18$

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(D) Given the equation $x^2-6(k-1)x+4(k-2)=0$. Since the roots $\alpha$ and $\beta$ are equal in magnitude but opposite in sign,we have $\alpha + \beta = 0$. From the sum of roots formula,$\alpha + \beta = 6(k-1) = 0$,which implies $k = 1$. Substituting $k=1$ into the equation,we get $x^2 - 6(1-1)x + 4(1-2) = 0$,which simplifies to $x^2 - 4 = 0$. Thus,$x^2 = 4$,so $x = \pm 2$. Given $\alpha > \beta$,we have $\alpha = 2$ and $\beta = -2$. Now,consider the second equation $2x^2 - \alpha x + 6\beta(\alpha+1) = 0$. Substituting the values of $\alpha$ and $\beta$: $2x^2 - 2x + 6(-2)(2+1) = 0$. $2x^2 - 2x + 6(-2)(3) = 0 \implies 2x^2 - 2x - 36 = 0$. Dividing by $2$,we get $x^2 - x - 18 = 0$. The product of the roots of a quadratic equation $ax^2+bx+c=0$ is given by $c/a$. Here,the product is $-36/2 = -18$.

The roots $\alpha, \beta$ of the equation $x^2-6(k-1)x+4(k-2)=0$ are equal in magnitude but opposite in sign. If $\alpha > \beta$,then the product of the roots of the equation $2x^2-\alpha x+6\beta(\alpha+1)=0$ is

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