Question

If the equations $2x^2-7x+1=0$ and $a{x}^2+bx+2=0$ ($a,b$ are rational) have a common root, then

• A. $a=2,b=-7$
• B. $a=-\frac{7}{2},b=1$
• C. $a=4,b=-14$
• D. none of these

Answer 1

The correct option is C $a=4,b=-14$

Given,
$2x^2-7x+1=0$ and $a{x}^2+bx+2=0$
Let $\alpha$ be the common root.
$\Rightarrow 2{\alpha }^2-7\alpha +1=0$
$a{\alpha }^2+b\alpha +2=0$
Solving using Cramer's rule,
$\frac{{\alpha }^2}{-14-b}=\frac{\alpha }{a-4}=\frac{1}{2b+7a}$
$\alpha =\frac{-b-14}{a-4}=\frac{a-4}{2b+7a}$
From the given options only 'C' satisfies these conditions.