Question

If the equation $2x^2+kr-5=0$ and $x^2-3x-4=0$ have a common root, then the value of $k$ is

• A. $-2$
• B. $-3$
• C. $\frac{-27}{4}$
• D. $-\frac{1}{4}$

Answer 1

Answer: (B), (C)

The correct options are
B $-3$
C $\frac{-27}{4}$

$2x^2+kx-5=0$
$let\alpha ,\beta betherootsthenc$
$\alpha +\beta =\frac{-k}{2}$
$\alpha \beta =\frac{-5}{2}$
$x^2-3x-4=0$
$\Rightarrow (x-4)(x+1)=0$
$x=4,-1$
$\therefore \alpha =-1,\beta =4$
$ifcommonrootis-1$
$(-1)\beta =\frac{-5}{2}$
$\beta =\frac{5}{2}$
$-1+\frac{5}{2}=\frac{-k}{2}$
$\Rightarrow \frac{-2+5}{2}=\frac{-k}{2}$
$k=-3$
$ifcommonrootis4$
$\left(4\right)\beta =\frac{-5}{2}$
$\beta =\frac{-5}{8}$
$4-\frac{5}{8}=\frac{-k}{2}$
$\Rightarrow \frac{32-5}{8}=\frac{-k}{2}$
$k=-\frac{27}{4}$