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(C) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 5 = 0$.Since $\alpha$ and $\beta$ satisfy the equation,we have:$\alpha^2

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$x^2 - 4x + 4 = 0$

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(C) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 5 = 0$.Since $\alpha$ and $\beta$ satisfy the equation,we have:$\alpha^2 - 3\alpha + 5 = 0 \implies \alpha^2 - 3\alpha = -5$$\beta^2 - 3\beta + 5 = 0 \implies \beta^2 - 3\beta = -5$Now,substitute these values into the required roots:Root $1 = \alpha^2 - 3\alpha + 7 = -5 + 7 = 2$Root $2 = \beta^2 - 3\beta + 7 = -5 + 7 = 2$The required quadratic equation with roots $2$ and $2$ is given by:$x^2 - (\sum \ of \ roots)x + (product \ of \ roots) = 0$$x^2 - (2 + 2)x + (2 	imes 2) = 0$$x^2 - 4x + 4 = 0$

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 3x + 5 = 0$,then find the quadratic equation whose roots are $(\alpha^2 - 3\alpha + 7)$ and $(\beta^2 - 3\beta + 7)$.

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