If the roots of the equations px^2 + 2qx + r | vedclass

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(B) For the equation $px^2 + 2qx + r = 0$ to have real roots,the discriminant $D_1 \ge 0$.$D_1 = (2q)^2 - 4(p)(r) = 4q^2 - 4pr \ge 0 \implies q^2 \ge

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$q^2 = pr$

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(B) For the equation $px^2 + 2qx + r = 0$ to have real roots,the discriminant $D_1 \ge 0$.$D_1 = (2q)^2 - 4(p)(r) = 4q^2 - 4pr \ge 0 \implies q^2 \ge pr$ ... $(i)$For the equation $qx^2 - 2\sqrt{pr}x + q = 0$ to have real roots,the discriminant $D_2 \ge 0$.$D_2 = (-2\sqrt{pr})^2 - 4(q)(q) = 4pr - 4q^2 \ge 0 \implies pr \ge q^2$ ... $(ii)$From $(i)$ and $(ii)$,we have $q^2 \ge pr$ and $pr \ge q^2$,which implies $q^2 = pr$.

If the roots of the equations $px^2 + 2qx + r = 0$ and $qx^2 - 2\sqrt{pr}x + q = 0$ are real,then

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