Question

Consider two quadratic equations, $p{x}^2-2qx+p=0...\left(i\right)$and $q{x}^2-2px+q=0...\left(ii\right)\left(\mathrm{both}p\mathrm{and}q\mathrm{are}\mathrm{real}\right).$If the roots of the equation $\left(i\right)$ are real and unequal, then the roots of the equation $\left(ii\right)$ are

• A. Rational and equal
• B. Real and unequal
• C. Real and equal
• D. imaginary

Answer 1

The correct option is D imaginary

Given : $p{x}^2-2qx+p=\mathrm{0...}\left(i\right);q{x}^2-2px+q=\mathrm{0...}\left(ii\right);$
Roots of $\left(i\right)$ are real and equal.
On comparing with standard form of quadratic expression $\text{y}=a{x}^2+bx+c$
we get, $a_1=p,b_1=-2q,c_1=p$ and $a_2=q,b_2=-2p,c_2=q$

Let $D_1,D_2$ be the respective discrimanant values.
$\text{Where}{D}_1=(-2q{)}^2-4.p.p=4(q^2-p^2)$
As the roots of $\left(i\right)$ are real and unequal therefore the discriminant of $\left(i\right)$ must be greater than zero.
$\therefore D_1=4(q^2-p^2)>0$
$\Rightarrow p^2-q^2<0...\left(iii\right)$
And
$\Rightarrow D_2=(-2p{)}^2-4.q.q=4(p^2-q^2)$
From $\left(iii\right)$ we come to know that $D_2$ is always less than zero, therefore the roots of the equation $\left(ii\right)$ are imaginary.