Question

The solution set of $pq{x^2} - {\left( {p + q} \right)^2}x - {\left( {p + q} \right)^2} = 0$ 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. \[\left\{ {\dfrac{{\left[ {p + q} \right]}}{p},\dfrac{{\left[ {p + q} \right]}}{q}} \right\}\]
E. \[\left\{ {\dfrac{{\left[ {p - q} \right]}}{p},\dfrac{{\left[ {p - q} \right]}}{q}} \right\}\]

Answer 1

Hint: We solve this equation by comparing the equation with the standard form of quadratic equation given by \[a{x^2} + bx + c = 0\] and then applying quadratic formula to find the values of x that is \[x = \dfrac{{\left( { - b \pm \sqrt D } \right)}}{{2a}}\] where \[D = {b^2} - 4ac\]. $D$ denotes the discriminant of the quadratic formula and it describes the nature of the roots of the equation.

Complete step by step answer:
First we compare the given equation $pq{x^2} - {\left( {p + q} \right)^2}x - {\left( {p + q} \right)^2} = 0$ and \[a{x^2} + bx + c = 0\].
So, we get, \[a = pq,b = - {\left( {p + q} \right)^2},c = \;{\left( {p + q} \right)^2}\].
Now, we find out the value of discriminant D by using the formula \[D = {b^2} - 4ac\]. So, we get,
\[D = {\left[ { - {{\left( {p + q} \right)}^2}} \right]^2} - 4pq{\left( {p + q} \right)^2}\]
Then, we know the quadratic formula to solve the quadratic equation given to us. So, we get,
\[ \Rightarrow x = \dfrac{{\left( { - b \pm \sqrt D } \right)}}{{2a}}\]
Substituting the values of a, b and Discriminant in the formula, we get,
\[ \Rightarrow x = \dfrac{{\left( { - \left\{ { - {{\left( {p + q} \right)}^2}} \right\} \pm \sqrt {{{\left[ { - {{\left( {p + q} \right)}^2}} \right]}^2} - 4pq{{\left( {p + q} \right)}^2}} } \right)}}{{2pq}}\]

Now, we take \[{\left( {p + q} \right)^2}\] common from within the square root. So, we get,
\[ \Rightarrow x = \dfrac{{\left( {{{\left( {p + q} \right)}^2} \pm \left( {p + q} \right)\sqrt {{{\left( {p + q} \right)}^2} - 4pq} } \right)}}{{2pq}}\]
Now, expanding the square of binomial, we get,
\[ \Rightarrow x = \dfrac{{\left( {{{\left( {p + q} \right)}^2} \pm \left( {p + q} \right)\sqrt {{p^2} + {q^2} + 2pq - 4pq} } \right)}}{{2pq}}\]
\[ \Rightarrow x = \dfrac{{\left( {{{\left( {p + q} \right)}^2} \pm \left( {p + q} \right)\sqrt {{p^2} + {q^2} - 2pq} } \right)}}{{2pq}}\]
Condensing the square of the binomial using the whole square identity ${\left( {a - b} \right)^2} = {a^2} - 2ab - {b^2}$. So, we get,
\[ \Rightarrow x = \dfrac{{\left( {{{\left( {p + q} \right)}^2} \pm \left( {p + q} \right)\sqrt {{{\left( {p - q} \right)}^2}} } \right)}}{{2pq}}\]
Simplifying the expression, we get,
\[ \Rightarrow x = \dfrac{{\left( {{{\left( {p + q} \right)}^2} \pm \left( {p + q} \right)\left( {p - q} \right)} \right)}}{{2pq}}\]
\[ \Rightarrow x = \dfrac{{\left( {{{\left( {p + q} \right)}^2} \pm \left( {{p^2} - {q^2}} \right)} \right)}}{{2pq}}\]

Now, we can find both the roots by considering the positive and negative signs one by one.
So, considering positive sign, we get,
\[ \Rightarrow x = \dfrac{{\left( {\left( {{p^2} + 2pq + {q^2}} \right) + \left( {{p^2} - {q^2}} \right)} \right)}}{{2pq}}\]
\[ \Rightarrow x = \dfrac{{{p^2} + 2pq + {q^2} + {p^2} - {q^2}}}{{2pq}}\]
Cancelling like terms with opposite signs,
\[ \Rightarrow x = \dfrac{{{p^2} + 2pq + {p^2}}}{{2pq}}\]
Factoring out the common factors,
\[ \Rightarrow x = \dfrac{{2p\left( {p + q} \right)}}{{2pq}}\]
\[ \Rightarrow x = \dfrac{{\left( {p + q} \right)}}{q}\]
So, \[x = \dfrac{{\left( {p + q} \right)}}{q}\]
Similarly, for negative sign, we get,
\[ \Rightarrow x = \dfrac{{\left( {\left( {{p^2} + 2pq + {q^2}} \right) - \left( {{p^2} - {q^2}} \right)} \right)}}{{2pq}}\]
\[ \Rightarrow x = \dfrac{{\left( {{p^2} + 2pq + {q^2} - {p^2} + {q^2}} \right)}}{{2pq}}\]
Simplifying the expression, we get,
\[ \Rightarrow x = \dfrac{{\left( {2pq + 2{q^2}} \right)}}{{2pq}}\]
Factoring out common terms, we get,
\[ \Rightarrow x = \dfrac{{2q\left( {p + q} \right)}}{{2pq}}\]
\[ \therefore x = \dfrac{{\left( {p + q} \right)}}{p}\]
So, we get, solution set of the equation as \[\left\{ {\dfrac{{\left[ {p + q} \right]}}{p},\dfrac{{\left[ {p + q} \right]}}{q}} \right\}\].

Hence, option D is the correct answer.

Note: Given problem deals with solving the quadratic equation to find the value of x by using a discriminant formula. The main thing to do with such questions is to remember the discriminant formula and be careful while opening the brackets. Take care while doing the calculations. Algebraic identities must be remembered in order to solve the problem.