Question

If the equations $x^2-x-12=0$ and $k{x}^2+10x+3=0$ may have one common root, then $k$ is:

• A. $3$ or $-\frac{43}{16}$
• B. $-3$ or $-\frac{43}{16}$
• C. $3$ or $\frac{13}{16}$
• D. None of these

Answer 1

The correct option is A

$3$ or $-\frac{43}{16}$

Let $\alpha$ be a common roots, then

${\alpha }^2-\alpha -12-0,k{\alpha }^2+10\alpha +3=0$ .

$\frac{{\alpha }^2}{117}=\frac{\alpha }{-12k-3}=\frac{1}{10+k}$

$\Rightarrow$ $(12k+3{)}^2=117(10+k)$

$\Rightarrow 144k^2+9+72k=1170+117k$
$\Rightarrow 144k^2-45k-1161=0$
$\Rightarrow 16k^2-5k-129=0$
$\Rightarrow (k-3)(16k+43)=0$
$\Rightarrow k=3$ or $\frac{-43}{16}$