Question

If both the roots of the quadratic equation $x^2-(2n+18)x-n-11=0,n\in Z$ are rational, then the value(s) of $n$ is/are

• A. $-8$
• B. $-10$
• C. $-11$
• D. $-12$

Answer 1

Answer: (A), (C)

$-11$

$x^2-(2n+18)x-n-11=0$
$D=(2n+18{)}^2+4(n+11)$
As the roots are rational, $D$ should be perfect square,
$(2n+18{)}^2+4(n+11)=m^2\Rightarrow (2n+18{)}^2+2\cdot (2n+18)\cdot 1+1^2+7=m^2\Rightarrow \left(\right(2n+18)+1{)}^2+7=m^2$
This is possible iff
$(2n+18+1{)}^2=9$ so that, we get $\text{L.H.S.}=9+7=16$ which is a perfect square.
$\Rightarrow 2n+19=\pm 3\Rightarrow 2n=-19\pm 3\therefore n=-8,-11$