Question

The values of the parameter $a$ for which the quadratic equations $(1-2a)x^2-6ax-1=0$ and $a{x}^2-x+1=0$ have at least one root in common are

• A. $0,\frac{1}{2}$
• B. $\frac{1}{2},\frac{2}{9}$
• C. $\frac{2}{9}$
• D. $0,\frac{1}{2},\frac{2}{9}$

Answer 1

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

$(1-2a)x^2-6ax-1=0$
$a{x}^2-x+1=0$
$\frac{x^2}{-6a-1}=\frac{-x}{(1-2a)+a}=\frac{1}{-(1-2a)+6a^2}$
$\Rightarrow \frac{x^2}{-(6a+1)}=\frac{-x}{1-a}=\frac{1}{6a^2+2a-1}$
$\frac{x^2}{-(6a+1)}=\frac{-x}{1-a}$
$x=\frac{6a+1}{1-a}........\left(i\right)$
$\frac{-x}{1-a}=\frac{1}{6a^2+2a-1}$
$x=\frac{a-1}{6a^2+2a-1}.......\left(ii\right)$
$from\left(i\right)and\left(ii\right)$
$\frac{6a+1}{1-a}=\frac{a-1}{6a^2+2a-1}$
$\Rightarrow (6a+1)(6a^2+2a-1)=-{(a-1)}^2$
$\Rightarrow 36a^3+12a^2-6a+6a^2+2a-1=-a^2-1+2a$
$\Rightarrow 36a^3+19a^2-6a=0$
$\Rightarrow a(36a^2+19a-6)=0$
$\Rightarrow a(9a-2)(4a+3)=0$
$a=0,\frac{2}{9},\frac{-3}{4}$
$Henceonecommonrootis\frac{2}{9}$