Question

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

• A. $a=2,b=-7$
• B. $a=-\frac{7}{2},b=1$
• C. $a=4,b=-14$
• D. $a=-4,b=1$

Answer 1

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

We have $2x^2-7x+1=0\Rightarrow x=\frac{7\pm \sqrt{7^2-4\times 2}}{2\times 2}=\frac{7\pm \sqrt{41}}{4}=$irrational
And we know that irrational roots of a quadratic occur in pair
Hence both the roots of given quadratics have to be common
$\Rightarrow \frac{2}{a}=\frac{-7}{b}=\frac{1}{2}$
$\therefore a=4,b=-14$