Question

If the equations $a{x}^2+bx+c=0$ and $c{x}^2+bx+a=0,a\ne c$ have a negative common root, then the value of $a-b+c$ is

• A. $0$
• B. $2$
• C. $1$
• D. none of these

Answer 1

The correct option is A $0$

Given,
$a{x}^2+bx+c=0$ and $c{x}^2+bx+a=0$ have a negative common root.
Let $\alpha$ be the common root.
$\Rightarrow a{\alpha }^2+b\alpha +c=0\to \left(1\right)$
$c{\alpha }^2+b\alpha +a=0\to \left(2\right)$
Solving using Cramer's rule,
$\frac{{\alpha }^2}{ab-bc}=\frac{\alpha }{c^2-a^2}=\frac{1}{ab-bc}$
$\Rightarrow \alpha =\frac{b(a-c)}{c^2-a^2},\alpha =\frac{c^2-a^2}{b(a-c)}$
$\alpha =\frac{-b}{a+c},\alpha =\frac{-(a+c)}{b}$
and ${\alpha }^2=\frac{ab-bc}{ab-bc}=1$
$\Rightarrow \alpha =\pm 1$
But it is given that the common root is negative.
$\Rightarrow \alpha =-1=\frac{-b}{a+c}$
$\Rightarrow -b=-a-c$
$a-b+c=0$
Hence, option 'A' is correct.