Question

The value(s) of $k$ for which the quadratic equations $(1-2k)x^2-6kx-1=0$ and $k{x}^2-x+1=0$ have at least one root in common, is (are)

• A. $\left\{\frac{1}{2}\right\}$
• B. $\{\frac{1}{3},\frac{2}{9}\}$
• C. $\left\{\frac{2}{9}\right\}$
• D. $\{\frac{1}{2},\frac{2}{9}\}$

Answer 1

The correct option is C $\left\{\frac{2}{9}\right\}$

Let common root be $\alpha .$
$(1-2k){\alpha }^2-6k\alpha -1=0$
$k{\alpha }^2-\alpha +1=0$
Using condition of one root common, we get
$\frac{{\alpha }^2}{-6k-1}=\frac{\alpha }{-k-(1-2k)}=\frac{1}{-(1-2k)+6k^2}$
$\Rightarrow \frac{{\alpha }^2}{-(6k+1)}=\frac{\alpha }{k-1}=\frac{1}{6k^2+2k-1}$
$\Rightarrow {\alpha }^2=\frac{-(6k+1)}{6k^2+2k-1},\alpha =\frac{k-1}{6k^2+2k-1}$

Hence, $(k-1{)}^2=-(6k+1\left)\right(6k^2+2k-1)$
$\Rightarrow -k^2+2k-1=36k^3+12k^2-6k+6k^2+2k-1$
$\Rightarrow 36k^3+19k^2-6k=0$
$\Rightarrow k(36k^2+19k-6)=0$
As $k\ne 0,$ then
$36k^2+19k-6=0$
$\Rightarrow 36k^2+27k-8k-6=0$
$\Rightarrow 9k(4k+3)-2(4k+3)=0$
$\Rightarrow (4k+3)(9k-2)=0$
$\therefore k=\frac{-3}{4},k=\frac{2}{9}$