Question

If the equations, $2x^2+kx–5=0$ and $x^2–3x–4=0$ have one root in common, then $k$ equals

• A. $3$
• B. $\frac{27}{4}$
• C. $5$
• D. $-3$

Answer 1

The correct option is D $-3$

Since the equation having common root then
$\frac{x^2}{-4k-15}=\frac{x}{-5+8}=\frac{1}{-6-x}\frac{9}{(6+k{)}^2(4k+15)}=\frac{1}{(6+k)}$
$\therefore 4k^2+39k+81=0\Rightarrow k=-3$