Question

If $a, b, p, q$ are nonzero real numbers, then how many common roots would two equation $2{a^2}{x^2} - 2abx + {b^2} = 0$ and ${p^2}{x^2} + 2pqx + {q^2} = 0$ have?

Answer 1

Hint:
As we know that any quadratic equation has $2$roots. So there are three possibilities of common roots:
(1) The equations have two common roots,
(2) The equations have one common root and
(3)The equations have no common root.
We have to check all the possibilities one-by-one.

Complete step by step solution:
 The given equations are
$2{a^2}{x^2} - 2abx + {b^2} = 0$ ….. (1)
${p^2}{x^2} + 2pqx + {q^2} = 0$ ….. (2)
On comparing these equations with standard form of quadratic equations, i.e., ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$; we get-
${a_1} = 2{a^2},{b_1} = - 2ab,{c_1} = {b^2}$
${a_2} = {p^2},{b_2} = 2pq,{c_2} = {q^2}$
We know that any quadratic equation has $2$roots. So there are three possibilities of common roots:
(1) The equations have two common roots,
(2) The equations have one common root and
(3)The equations have no common root.
Step 1: Let the equations have two common roots .
Therefore, by the condition of two common roots, we have:
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$
$ \Rightarrow $$\dfrac{{2{a^2}}}{{{p^2}}} = \dfrac{{ - 2ab}}{{2pq}} = \dfrac{{{b^2}}}{{{q^2}}}$
Taking, $\dfrac{{2{a^2}}}{{{p^2}}} = \dfrac{{ - 2ab}}{{2pq}}$
$ \Rightarrow \dfrac{{2a}}{p} = \dfrac{{ - b}}{q}$
$ \Rightarrow - 2aq = bp$ …. (1)
Similarly, $\dfrac{{ - 2ab}}{{2pq}} = \dfrac{{{b^2}}}{{{q^2}}}$
$ \Rightarrow \dfrac{{ - a}}{p} = \dfrac{b}{q}$
$ \Rightarrow - aq = bp$ …. (2)
From (1) and (2), we get-
$ - 2aq = - aq$
$ \Rightarrow - 2aq + aq = 0$
$ \Rightarrow - aq = 0$
$ \Rightarrow aq = 0$
$ \Rightarrow a = 0$ or $q = 0$
Which is not possible; since $a,b,p,q$ are non-zero real numbers.
Thus our assumption s wrong.
Therefore, the equations not have two common roots.
Step 2: Let the equations have one common root, i.e., $\alpha $.
Therefore, by the condition of one common root, we have:
$\dfrac{{{\alpha ^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{\alpha }{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
$ \Rightarrow $$\dfrac{{{\alpha ^2}}}{{ - 2ab{q^2} - 2pq{b^2}}} = \dfrac{\alpha }{{2{p^2}{b^2} - 2{a^2}{q^2}}} = \dfrac{1}{{4{a^2}pq + 4ab{p^2}}}$
Taking, $\dfrac{{{\alpha ^2}}}{{ - 2ab{q^2} - 2pq{b^2}}} = \dfrac{1}{{4{a^2}pq + 4ab{p^2}}}$
$ \Rightarrow {\alpha ^2} = \dfrac{{ - 2ab{q^2} - 2pq{b^2}}}{{4{a^2}pq + 4ab{p^2}}}$
$ \Rightarrow {\alpha ^2} = \dfrac{{ - 2bq\left( {aq + pb} \right)}}{{4ap\left( {aq + pb} \right)}}$
$ \Rightarrow {\alpha ^2} = \dfrac{{ - bq}}{{2ap}}$ …. (3)
Similarly, $\dfrac{\alpha }{{2{p^2}{b^2} - 2{a^2}{q^2}}} = \dfrac{1}{{4{a^2}pq + 4ab{p^2}}}$
$ \Rightarrow \alpha = \dfrac{{2{p^2}{b^2} - 2{a^2}{q^2}}}{{4{a^2}pq + 4ab{p^2}}}$
$ \Rightarrow \alpha = \dfrac{{2\left( {{p^2}{b^2} - {a^2}{q^2}} \right)}}{{4ap\left( {aq + pb} \right)}}$
$ \Rightarrow \alpha = \dfrac{{\left( {pb - aq} \right)\left( {pb + aq} \right)}}{{2ap\left( {aq + pb} \right)}}$
$ \Rightarrow \alpha = \dfrac{{\left( {pb - aq} \right)}}{{2ap}}$ …. (4)
From equation (3) and (4);
$\dfrac{{ - bq}}{{2ap}} = {\left( {\dfrac{{pb - aq}}{{2ap}}} \right)^2}$
$ \Rightarrow $$ - bq = \dfrac{{{p^2}{b^2} + {a^2}{q^2} - 2abpq}}{{2ap}}$
$ \Rightarrow $$ - 2abpq = {p^2}{b^2} + {a^2}{q^2} - 2abpq$
$ \Rightarrow $${p^2}{b^2} + {a^2}{q^2} = 0$
Which is not possible; since $a,b,p,q$ are non-zero real numbers.
Thus our assumption is wrong.
Therefore, the equations do not have one common root.
Step3: Since equations have neither two nor one common root. So, the only possibility remaining here is that the given equations have no common root.

Hence, the given equations have no common root.

Note:
An another method to solve this question is described below:
We have also found the no. of common roots by finding the nature of roots $\left( {D = {b^2} - 4ac} \right)$ of the given quadratic equations.
For equation $2{a^2}{x^2} - 2abx + {b^2} = 0$
${D_1} = {\left( { - 2ab} \right)^2} - 4\left( {2{a^2}} \right)\left( {{b^2}} \right) = 4{a^2}{b^2} - 8{a^2}{b^2} = - 4{a^2}{b^2}$$ < 0$
Therefore, roots are imaginary.
For equation ${p^2}{x^2} + 2pqx + {q^2} = 0$
${D_2} = {\left( {2pq} \right)^2} - 4\left( {{p^2}} \right)\left( {{q^2}} \right) = 4{p^2}{q^2} - 4{p^2}{q^2} = 0$
Therefore, roots are real and equal.
Hence, the given equations have no common root.