Question

If the equation $\left( {{m}^{2}}+{{n}^{2}} \right){{x}^{2}}-2\left( mp+nq \right)x+{{p}^{2}}+{{q}^{2}}=0$ has equal roots, then
[a] mp = nq
[b] mq = np
[c] mn = pq
[d] $mq=\sqrt{np}$

Answer 1

Hint: Assume that the roots of the equation are $\alpha ,\alpha$. Hence using the sum of roots $=\dfrac{-b}{a}$ and product of roots $=\dfrac{c}{a}$, where a,b and c have their usual meanings form two equations in $\alpha$. One of the equations will be linear in $\alpha$ , and one will be quadratic in $\alpha$. Substitute the value of $\alpha$ from the linear equation in the quadratic equation in $\alpha$ and find the relation between m, n, p and q.
Alternatively, use the fact that when roots are equal, then Discriminant of that quadratic equation is 0.

Complete step-by-step solution -

Let the roots of the quadratic equation be $\alpha .\alpha$
Hence we have
$\begin{aligned} & \alpha +\alpha =\dfrac{2\left( mp+nq \right)}{{{m}^{2}}+{{n}^{2}}} \\\ & \Rightarrow \alpha =\dfrac{mp+nq}{{{m}^{2}}+{{n}^{2}}}\text{ (i)} \\\ \end{aligned}$
Also, we have
$\begin{aligned} & \alpha \times \alpha =\dfrac{{{p}^{2}}+{{q}^{2}}}{{{m}^{2}}+{{n}^{2}}} \\\ & \Rightarrow {{\alpha }^{2}}=\dfrac{{{p}^{2}}+{{q}^{2}}}{{{m}^{2}}+{{n}^{2}}}\text{ (ii)} \\\ \end{aligned}$
Substituting the value of $\alpha$ from equation (ii) in equation (i), we get
${{\left( \dfrac{mp+nq}{{{m}^{2}}+{{n}^{2}}} \right)}^{2}}=\dfrac{{{p}^{2}}+{{q}^{2}}}{{{m}^{2}}+{{n}^{2}}}$
Multiplying both sides by ${{\left( {{m}^{2}}+{{n}^{2}} \right)}^{2}}$, we get
${{\left( mp+nq \right)}^{2}}=\left( {{p}^{2}}+{{q}^{2}} \right)\left( {{m}^{2}}+{{n}^{2}} \right)$
Using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get
${{\left( mp \right)}^{2}}+{{\left( nq \right)}^{2}}+2\left( mp \right)\left( nq \right)={{p}^{2}}{{m}^{2}}+{{p}^{2}}{{n}^{2}}+{{q}^{2}}{{m}^{2}}+{{q}^{2}}{{n}^{2}}$
Using ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$, we get
${{m}^{2}}{{p}^{2}}+{{n}^{2}}{{q}^{2}}+2mnpq={{m}^{2}}{{p}^{2}}+{{n}^{2}}{{q}^{2}}+{{m}^{2}}{{q}^{2}}+{{n}^{2}}{{p}^{2}}$
Subtracting ${{m}^{2}}{{p}^{2}}+{{n}^{2}}{{q}^{2}}$ from both sides, we get
$2mnpq={{m}^{2}}{{q}^{2}}+{{n}^{2}}{{p}^{2}}$
Subtracting 2mnpq from both sides, we get
${{m}^{2}}{{q}^{2}}+{{n}^{2}}{{p}^{2}}-2mnpq=0$
Using ${{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}$, we get
${{\left( mq \right)}^{2}}+{{\left( np \right)}^{2}}-2\left( mq \right)\left( np \right)=0$
Using ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$, we get
${{\left( mq-np \right)}^{2}}=0$
Using zero product property, we get
$mq=np$
Hence option [b] is correct.

Note: Alternative solution:
We know that the roots of the quadratic expression $a{{x}^{2}}+bx+c=0$ are equal when $D={{b}^{2}}-4ac=0$
Here $a={{m}^{2}}+{{n}^{2}},b=-2\left( mp+nq \right)$ and $c={{p}^{2}}+{{q}^{2}}$
Hence we have
$\begin{aligned} & 4{{\left( mp+nq \right)}^{2}}-4\left( {{p}^{2}}+{{q}^{2}} \right)\left( {{m}^{2}}+{{n}^{2}} \right)=0 \\\ & \Rightarrow {{\left( mp+nq \right)}^{2}}-\left( {{p}^{2}}+{{q}^{2}} \right)\left( {{m}^{2}}+{{n}^{2}} \right)=0 \\\ \end{aligned}$
Let ${{z}_{1}}=p-iq$ and ${{z}_{2}}=m+in$
Now we have ${{z}_{1}}{{z}_{2}}=\left( mp+nq \right)+i\left( np-mq \right)$
Taking modulus on both sides, we get
$\begin{aligned} & \left| {{z}_{1}}{{z}_{2}} \right|=\left| \left( mp+nq \right)+i\left( np-mq \right) \right| \\\ & \Rightarrow \sqrt{\left( {{m}^{2}}+{{n}^{2}} \right)\left( {{p}^{2}}+{{q}^{2}} \right)}=\sqrt{{{\left( mp+nq \right)}^{2}}+{{\left( np-mq \right)}^{2}}} \\\ & \Rightarrow \left( {{m}^{2}}+{{n}^{2}} \right)\left( {{p}^{2}}+{{q}^{2}} \right)={{\left( mp+nq \right)}^{2}}+{{\left( np-mq \right)}^{2}} \\\ \end{aligned}$
Hence we have
$\begin{aligned} & {{\left( mp+nq \right)}^{2}}-{{\left( mp+nq \right)}^{2}}-{{\left( np-mq \right)}^{2}}=0 \\\ & \Rightarrow {{\left( np-mq \right)}^{2}}=0 \\\ & \Rightarrow np=mq \\\ \end{aligned}$
Hence option [b] is correct.