Question

Let $m,n$ be positive integers and the quadratic equation $4x^2+mx+n=0$ has two distinct real roots $p$ and $q$ $(p\le q)$. Also, the quadratic equations $x^2-px+2q=0$ and $x^2-qx+2p=0$ have a common root say $\alpha$.

Number of possible ordered pairs $(m,n)$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$4x^2+m+n=0$ has distinct real roots $p,q$.

$\Rightarrow p+q=\frac{-m}{4}$,$pq=\frac{n}{4}$

Discriminant>0

$\Rightarrow m^2-16n>0\to \left(1\right)$

$x^2-px+2q=0$,$x^2-qx+2p=0$ has a common root.

$\Rightarrow {\alpha }^2-p\alpha +2q=0\to \left(2\right)$

${\alpha }^2-q\alpha +2p=0\to \left(3\right)$

$\left(2\right)-\left(3\right)$

$\Rightarrow \alpha (q-p)+2(q-p)=0$

$\alpha =-2(\because p\pm q)$

On subtituting the value of $\alpha =-2$ in $\left(2\right)$

$\Rightarrow 4+2p+2q=0$

$\Rightarrow p+q=-2$

$\frac{-m}{4}=-2$

$m=8$

From $\left(1\right)$,$m^2-16n>0$

$\Rightarrow n<\frac{64}{16}$

$\Rightarrow n<4$

As $n\in Z^{+}$, $n=1,2,3$ are possible values so,no of ordered pairs $(m,n)=3$.

Hence, option C is correct.