Question

The solution set of the equation $pq{x^2} - {(p + q)^2}x + {(p + q)^2} = 0$ is
a.$\left\{ {\dfrac{p}{q},\dfrac{q}{p}} \right\}$
b.$\left\{ {pq,\dfrac{p}{q}} \right\}$
c.$\left\{ {\dfrac{q}{p},pq} \right\}$
d.$\left\{ {\dfrac{{p + q}}{p},\dfrac{{p + q}}{q}} \right\}$

Answer 1

Hint: We are given a quadratic equation and we need to solve it to find the solution so we will solve it using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step-by-step answer:
We are given a quadratic equation $pq{x^2} - {(p + q)^2}x + {(p + q)^2} = 0$and we can solve it using the formula method
The formula to solve a quadratic equation $a{x^2} - bx + c = 0$is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now in our given equation ,
$ \Rightarrow a = pq,b = - {(p + q)^2},c = {(p + q)^2}$
Substituting in the formula , we get
\[
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \sqrt {{{\left( {{{(p + q)}^2}} \right)}^2} - 4pq{{(p + q)}^2}} }}{{2pq}} \\
    \\
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \sqrt {{{(p + q)}^2}\left( {{{(p + q)}^2} - 4pq} \right)} }}{{2pq}} \\
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \sqrt {{{(p + q)}^2}\left( {{p^2} + {q^2} + 2pq - 4pq} \right)} }}{{2pq}} \\
    \\
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \sqrt {{{(p + q)}^2}\left( {{p^2} + {q^2} - 2pq} \right)} }}{{2pq}} \\
    \\
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \sqrt {{{(p + q)}^2}{{\left( {p - q} \right)}^2}} }}{{2pq}} \\
    \\
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \sqrt {{{\left( {{p^2} - {q^2}} \right)}^2}} }}{{2pq}} \\
    \\
   \Rightarrow x = \dfrac{{{{(p + q)}^2} \pm \left( {{p^2} - {q^2}} \right)}}{{2pq}} \\
\]
Now taking the positive sign we get
\[
   \Rightarrow x = \dfrac{{{{(p + q)}^2} + \left( {{p^2} - {q^2}} \right)}}{{2pq}} \\
   \Rightarrow x = \dfrac{{{p^2} + {q^2} + 2pq + {p^2} - {q^2}}}{{2pq}} = \dfrac{{2{p^2} + 2pq}}{{2pq}} \\
   \Rightarrow x = \dfrac{{2p(p + q)}}{{2pq}} = \dfrac{{p + q}}{q} \\
\]
Now taking the negative sign
\[
   \Rightarrow x = \dfrac{{{{(p + q)}^2} - \left( {{p^2} - {q^2}} \right)}}{{2pq}} \\
   \Rightarrow x = \dfrac{{{p^2} + {q^2} + 2pq - {p^2} + {q^2}}}{{2pq}} = \dfrac{{2{q^2} + 2pq}}{{2pq}} \\
   \Rightarrow x = \dfrac{{2q(p + q)}}{{2pq}} = \dfrac{{p + q}}{p} \\
\]
Hence the solution set is $\left\{ {\dfrac{{p + q}}{p},\dfrac{{p + q}}{q}} \right\}$
The correct option is d.

Note: Alternative method
The given equation is $pq{x^2} - {(p + q)^2}x + {(p + q)^2} = 0$
We know that the
Sum of the roots =$\dfrac{{ - b}}{a}$ =$\dfrac{{{{\left( {p + q} \right)}^2}}}{{pq}}$
Product of the roots ,$ = \dfrac{c}{a} = \dfrac{{{{\left( {p + q} \right)}^2}}}{{pq}}$
From the product we can write as
 $ \Rightarrow \dfrac{{{{\left( {p + q} \right)}^2}}}{{pq}} = \dfrac{{p + q}}{p}*\dfrac{{p + q}}{q}$
Hence from this we get that $\left\{ {\dfrac{{p + q}}{p},\dfrac{{p + q}}{q}} \right\}$are the roots