Question

If the quadratic equations $3x^2+ax+1=0$ and $2x^2+bx+1=0$ have a common root, where $a,b\in R$, then the value of $5ab-2a^2-3b^2$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

If $\alpha$ is common root, then
$\frac{{\alpha }^2}{b_1{c}_2-b_2{c}_1}=\frac{\alpha }{c_1{a}_2-a_1{c}_2}=\frac{-1}{a_1{b}_2-a_2{b}_1}$

for the assumed equations say $a_1{x}^2+b_{1}x+c_1=0$ and $a_2{x}^2+b_{2}x+c_2$
Therefore for the given question:

$\frac{{\alpha }^2}{a-b}=\frac{\alpha }{2-3}=\frac{-1}{3b-2a}$
$\Rightarrow {\alpha }^2=1;\alpha =1.$
$a=-2;b=-1.$
Value of $5ab-2a^2-3b^2$ is
$5(-2)(-1)-2(-2{)}^2-3(-1{)}^2$
$10-8-3=-1$.