Question

Find the value of $K$, so that the equations $2x^2+Kx-5=0$ and $x^2-3x-4=0$ have one root in common.

Answer 1

Find value of k.

$2x^2+kx-5=0$ & $x^2-3x-4=0$

$\to x^2-3x-4=0$

$x^2-4x+x-4=0$

$x(x-4)+1(x-4)=0$

$(x+1)(x-4)=0$

$x=-1$ and $x=4$

$\to$ one root is common

$\therefore 2(-1{)}^2+k(-1)-5=0$

$\therefore 2-k-5=0$

$\therefore k=-3$

Or

$2(4{)}^2+k(4)-5=0$

$\therefore 32+4k-5=0$

$\therefore k=\frac{-27}{4}$