Question

Find the value of a for which quadratic equations (1 - 2a) $x^2$ - 6ax - 1 = 0 and $a{x}^2$ - x + 1 = 0 has at least one root in common.

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

Answer 1

The correct option is D $a=\frac{2}{9}$

Do not solve to find the common root, put the options in the given equations to check which satisfy.
For a = 0, and a = $\frac{1}{2}$, one quadratic equation becomes linear.
So, a can't be equal to 0 or $\frac{1}{2}$.
Hence, $a=\frac{2}{9}$ is the only option available.