Question

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

• A. $\left\{\frac{p}{q},\frac{q}{p}\right\}$
• B. $\left\{pq,\frac{p}{q}\right\}$
• C. $\left\{\frac{q}{p},pq\right\}$
• D. $\left\{\frac{p+q}{p},\frac{p+q}{q}\right\}$
• E. $\left\{\frac{p-q}{p},\frac{p-q}{q}\right\}$

Answer 1

The correct option is D

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

Explanation for the correct option:

Find the solution set of the given equation.

In the question, an equation $pq{x}^2-{(p+q)}^{2}x+{(p+q)}^2=0$ is given.

We know that the solution of the equation $a{x}^2+bx+c=0$ can be given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

So,

$$\begin{gathered} x=\frac{{(p+q)}^2\pm \sqrt{{\left[-{(p+q)}^2\right]}^2-4pq{(p+q)}^2}}{2pq} \\ \Rightarrow x=\frac{{(p+q)}^2\pm \sqrt{{(p+q)}^4-4pq{(p+q)}^2}}{2pq} \\ \Rightarrow x=\frac{{(p+q)}^2\pm \sqrt{{(p+q)}^2\left({(p+q)}^2-4pq\right)}}{2pq} \\ \Rightarrow x=\frac{{(p+q)}^2\pm (p+q)\sqrt{{(p-q)}^2}}{2pq} \\ \Rightarrow x=\frac{{(p+q)}^2\pm (p+q)(p-q)}{2pq} \\ \Rightarrow x=\frac{p^2+q^2+2pq\pm (p^2-q^2)}{2pq} \\ \Rightarrow x=\left\{\frac{p^2+q^2+2pq-(p^2-q^2)}{2pq},\frac{p^2+q^2+2pq+(p^2-q^2)}{2pq}\right\} \\ \Rightarrow x=\left\{\frac{q^2+2pq+q^2}{2pq},\frac{p^2+2pq+p^2}{2pq}\right\} \\ \Rightarrow x=\left\{\frac{2q^2+2pq}{2pq},\frac{2p^2+2pq}{2pq}\right\} \\ \Rightarrow x=\left\{\frac{q+p}{p},\frac{p+q}{q}\right\} \end{gathered}$${Since, ${\left(a+b\right)}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ab}\mathbf{=}{\left(a-b\right)}^{\mathbf{2}}$ and $\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}$}

Therefore, the solution set of the given equation is $\left\{\frac{p+q}{p},\frac{p+q}{q}\right\}$.

Hence, option $D$ is the correct answer.