If the roots of the equation x^2 + px + q = 0 | vedclass

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(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.Given the roots of $x^2 - xr +

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Both real

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(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.Given the roots of $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,we have $\alpha^4 + \beta^4 = r$ and $\alpha^4\beta^4 = s$.Now,consider the discriminant $D$ of the equation $x^2 - 4qx + 2q^2 - r = 0$:$D = (-4q)^2 - 4(1)(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.Substitute $q = \alpha\beta$ and $r = \alpha^4 + \beta^4$:$D = 8(\alpha\beta)^2 + 4(\alpha^4 + \beta^4) = 4(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 4(\alpha^2 + \beta^2)^2$.Since $(\alpha^2 + \beta^2)^2 \ge 0$ for all real $\alpha, \beta$,it follows that $D \ge 0$.Therefore,the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ are always real.

If the roots of the equation $x^2 + px + q = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,then the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ will be:

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