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 $\alpha$. If $p$ and $q$ are rational, then uncommon root of the equation $x^2-px+2q=0$ and $x^2-qx+2p=0$ is equal to

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

Answer 1

The correct option is A $-q$

$x^2-px+2q=0$
$x^2-qx+2p=0$
Subtracting the above equations, we get
$x(q-p)+2(q-p)=0$
$(q-p)[x+2]=0$ ...(i)
Hence,
$x=-2$ is the common root.
Substituting, we get
$4+2p+2q=0$
$2(p+q)=-4$
$p+q=-2$
Hence the equation, becomes
$x^2+(q+2)x+2q=0$
$(x+q)(x+2)=0$
Hence, uncommon root is $-q$.