The solution set of the equation pqx^2 - (p + | vedclass

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(D) Given the quadratic equation $pqx^2 - (p + q)^2x + (p + q)^2 = 0$.Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = pq$,$b =

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$\left\{ \frac{p + q}{p}, \frac{p + q}{q} \right\}$

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(D) Given the quadratic equation $pqx^2 - (p + q)^2x + (p + q)^2 = 0$.Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = pq$,$b = -(p + q)^2$,and $c = (p + q)^2$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:Discriminant $D = b^2 - 4ac = (-(p + q)^2)^2 - 4(pq)(p + q)^2 = (p + q)^4 - 4pq(p + q)^2$.$D = (p + q)^2 [(p + q)^2 - 4pq] = (p + q)^2 [p^2 + 2pq + q^2 - 4pq] = (p + q)^2 (p - q)^2$.So,$\sqrt{D} = (p + q)(p - q) = p^2 - q^2$.$x = \frac{(p + q)^2 \pm (p^2 - q^2)}{2pq}$.Case $1$: $x = \frac{p^2 + 2pq + q^2 + p^2 - q^2}{2pq} = \frac{2p^2 + 2pq}{2pq} = \frac{2p(p + q)}{2pq} = \frac{p + q}{q}$.Case $2$: $x = \frac{p^2 + 2pq + q^2 - p^2 + q^2}{2pq} = \frac{2pq + 2q^2}{2pq} = \frac{2q(p + q)}{2pq} = \frac{p + q}{p}$.Thus,the solution set is $\left\{ \frac{p + q}{p}, \frac{p + q}{q} \right\}$.

The solution set of the equation $pqx^2 - (p + q)^2x + (p + q)^2 = 0$ is

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