Question

The value of ${}^{\prime }{k}^{\prime }$ so that the equations $2x^2+kx-5=0$ & $x^2-3x-4=0$ may have one root common

• A. $3,-\frac{27}{4}$
• B. $-3,-\frac{27}{4}$
• C. $\frac{27}{4},3$
• D. $\frac{3}{7},\frac{8}{7}$

Answer 1

The correct option is B $-3,-\frac{27}{4}$

$x^2-3x-4=0$
$(x-4)(x+1)=0$
$x=4$ and $x=-1$
Hence, the common roots is either $-1$, or $4$
Now,
Substituting, $x=-1$ in the first equation we get,
$2-k-5=0$
$k=-3$
Substituting, $x=4$ in the first equation, we get,
$2\left(16\right)+4k-5=0$
$32-5=-4k$
$k=\frac{-27}{4}$.