(C) Given that $\alpha$ and $\beta$ satisfy the equation $x^2-6x+1=0$. Let $y = \frac{x}{x+1}$. Then $y(x+1) = x$,which implies $yx + y = x$,or $x(1-y

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(C) Given that $\alpha$ and $\beta$ satisfy the equation $x^2-6x+1=0$. Let $y = \frac{x}{x+1}$. Then $y(x+1) = x$,which implies $yx + y = x$,or $x(1-y) = y$,so $x = \frac{y}{1-y}$. Substituting this into the original equation $x^2-6x+1=0$: $(\frac{y}{1-y})^2 - 6(\frac{y}{1-y}) + 1 = 0$. Multiplying by $(1-y)^2$,we get: $y^2 - 6y(1-y) + (1-y)^2 = 0$. $y^2 - 6y + 6y^2 + 1 - 2y + y^2 = 0$. $8y^2 - 8y + 1 = 0$. Replacing $y$ with $x$,the required equation is $8x^2-8x+1=0$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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Let $\alpha \neq \beta$ satisfy $\alpha^2+1=6 \alpha$ and $\beta^2+1=6 \beta$. Then,the quadratic equation whose roots are $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ is

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$8 x^2+8 x+1=0$
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$8 x^2-8 x-1=0$
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$8 x^2-8 x+1=0$
D
$8 x^2+8 x-1=0$

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Let $\alpha \neq \beta$ satisfy $\alpha^2+1=6 \alpha$ and $\beta^2+1=6 \beta$. Then,the quadratic equation whose roots are $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ is

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$\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 3 x^2 - 10 x - 24 = 0$. If $\alpha > \beta > \gamma$ and $\alpha^3 + 3 \beta^2 - 10 \gamma - 24 = 11 k$,then $k = $

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