NTSE Stage 2 SAT 2018 Exam Question Paper

Let the common root be t
Then, the equation becomes
$2{\mathit{t}}^2+\mathit{k}\mathit{t}\mathit{–}5=0$......(1)
${\mathit{t}}^2\mathit{–}3\mathit{t}\mathit{–}4=0$.........(2)
Multiply equation (2) by 2 and then subtract from equation (1)
$2{\mathit{t}}^2+\mathit{k}\mathit{t}-5=0$
$2{\mathit{t}}^2-6\mathit{t}-8=0$
$-++$
$\left(\mathit{k}+6\right)\mathit{t}+3=0$
$\mathit{t}=-\frac{3}{\mathit{k}+6}$
Now, put the value of t in equation (1)
$2{\left(\frac{-3}{\mathit{k}+6}\right)}^2+\mathit{k}\left(\frac{-3}{\mathit{k}+6}\right)-5=0$
$\frac{18}{{\left(\mathit{k}+6\right)}^2}+\frac{-3}{\mathit{k}+6}-5=0$
$18-3\mathit{k}\left(\mathit{k}+6\right)-5{\left(\mathit{k}+6\right)}^2=0$
$-8{\mathit{k}}^2-78\mathit{k}-162=0$
$8{\mathit{k}}^2+78\mathit{k}+162=0$
$4{\mathit{k}}^2+39\mathit{x}+81=0$
$4{\mathit{k}}^2+27\mathit{k}+12\mathit{k}+81=0$
$\mathit{k}\left(4\mathit{k}+27\right)+3\left(4\mathit{k}+27\right)=0$
$\left(\mathit{k}+3\right)\left(4\mathit{k}+27\right)=0$
$\mathit{k}=-3,-\frac{27}{4}$