Question

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

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

Answer 1

Answer: (C)

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