Question

Let $m, n$ be positive integers and the quadratic equation $4 x^2+m x+n=0$ has two distinct real roots $p$ and $q ( p < q )$. Also the quadratic equations $x ^2- px +2 q =0$ and $x ^2- qx +2 p =0$ have a common root say $\alpha$.
If $p$ and $q$ are rational, then uncommon root of the equation $x^2-p x+2 q=0$ is equal to

• A. 1
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

Answer 1

Answer: (C)

One root of $x^2-p x+2 q=0$ is $x=-2$ and product of roots $=2 q$
$\Rightarrow $ uncommon root $=- q = x$
Given $p$ and $q$ are roots of equation $4 x ^2+8 x + n =0$, where $n =1,2$ or 3
If $n=1$, then $4 x^2+8 x+1=0 \Rightarrow$ Roots are not rational as $D$ is not perfect square.
If $n=2$, then $4 x^2+8 x+2=0 \Rightarrow 2 x^2+4 x+1=0 \Rightarrow$ Roots are not rational as $D$ is not perfect square.
If $n=3$, then $4 x^2+8 x+3=0 \Rightarrow 4 x^2+6 x+2 x+3=0 \Rightarrow(2 x+3)(2 x+1)=0$
$\therefore p =\frac{-3}{2}$ and $q =\frac{-1}{2}$
Hence roots are rational.
$\therefore $ Uncommon root of the equation $x ^2- px +2 q =0$ is $x =- q \Rightarrow x =\frac{1}{2}$