Question

If one root of the quadratic equation $2x^2+kx-6=0$ is $2$, find the value of k. Also, find the other root.

• A. $k=-1$ and the other root is $=\frac{3}{2}$
  
• B. $k=1$ and the other root is $=\frac{-3}{2}$
• C. $k=1$ and the other root is $=\frac{3}{2}$
• D. $k=-1$ and the other root is $=\frac{-3}{2}$

Answer 1

The correct option is D $k=-1$ and the other root is $=\frac{-3}{2}$

$2x^2+kx-6=0$

$\alpha +2=\frac{-k}{2}$

$2\alpha =\frac{-6}{2}$

$\alpha =\frac{-6}{4}=-\frac{3}{2}$

$\therefore 2-\frac{6}{4}=-\frac{k}{2}$

$\therefore \frac{4}{2}=-\frac{k}{2}$

$\therefore k=-1$

$2x^2+kx-6=0$

$2x^2-x-6=0$

$2x^2-4x+3x-6=0$

$2x(x-2)+3(x-2)$

$(2x+3)(x-2)$

$x=-\frac{3}{2}\text{or}2$