Question

Two quadratic equation $x^2-bx+6=0$ and $x^2-6x+c=0$ have a common root. If the remaining roots of the first and second equations are positive integers and are in the ratio $3:4$ respectively, then the common root is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B), (D)

The correct options are
B $2$
D $4$

${\alpha }^2-b\alpha +6=0$
${\alpha }^2-6\alpha +c=0$
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$(6-b)\alpha +6-c=0$
$\alpha =\frac{c-6}{6-b}$
$\alpha .3a=6\Rightarrow a=\frac{2}{\alpha }$
$\alpha +4a=6$
$\alpha +\frac{4\times 2}{\alpha }=6$
${\alpha }^2+8-6\alpha =0$
${\alpha }^2-6\alpha +8=0$
$(\alpha -4)(\alpha -2)=0$
$\alpha =4,2$