Question

The value of $k$ so that the equations $x^2-x-12=0$ and $k{x}^2+10x+3=0$ may have one root in common, is

• A. $\frac{43}{16}$
• B. $3$
• C. $-3$
• D. $\frac{43}{18}$

Answer 1

Answer: (B), (D)

Let $\alpha$ be the common root
Then, ${\alpha }^2-\alpha -12=0$
and $k{\alpha }^2+10\alpha +3=0$
Solving the two equations, we get
$\frac{{\alpha }^2}{117}=\frac{\alpha }{-12k-3}=\frac{1}{10+k}$
$\Rightarrow (-12k-3{)}^2=117(10+k)$
$\Rightarrow 9(4k+1{)}^2=117(10+k)$
$\Rightarrow (4k+1{)}^2=13(10+k)$
$\Rightarrow 16k^2+8k+1=130+13k$
$\Rightarrow 16k^2-5k-129=0$
$\Rightarrow 16k^2-48k+43k-129=0$
$\therefore k=3$
or $k=\frac{-43}{16}$