If alpha, beta are the roots of the equation | vedclass

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(C) Since $\alpha$ and $\beta$ are the roots of the equation $x^2+5x+2=0$,we have $\alpha^2+5\alpha+2=0$ and $\beta^2+5\beta+2=0$. From these,we get $

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$\frac{21}{4}$

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(C) Since $\alpha$ and $\beta$ are the roots of the equation $x^2+5x+2=0$,we have $\alpha^2+5\alpha+2=0$ and $\beta^2+5\beta+2=0$. From these,we get $5\alpha+2 = -\alpha^2$ and $5\beta+2 = -\beta^2$. Also,from the properties of roots,$\alpha+\beta = -5$ and $\alpha\beta = 2$. Now,substituting these into the expression: $\left(\frac{\alpha}{2+5\alpha}\right)^2+\left(\frac{\beta}{2+5\beta}\right)^2 = \left(\frac{\alpha}{-\alpha^2}\right)^2+\left(\frac{\beta}{-\beta^2}\right)^2$ $= \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2+\beta^2}{(\alpha\beta)^2}$ $= \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$ $= \frac{(-5)^2-2(2)}{(2)^2} = \frac{25-4}{4} = \frac{21}{4}$.

If $\alpha, \beta$ are the roots of the equation $x^2+5x+2=0$,then $\left(\frac{\alpha}{2+5\alpha}\right)^2+\left(\frac{\beta}{2+5\beta}\right)^2=$

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