Question

Let the quadratic equation be $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$, where $a,b,c\in R$ and $a\ne c$. If $a+b+c=0$, then the roots are

• A. real, equal in magnitude but opposite in sign
• B. real and equal
• C. not real
• D. real and distinct

Answer 1

The correct option is D real and distinct

Given equation is
$(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$
We know that
$a+b+c=0$
So,
$(-2a)x^2+(-2b)x+(-2c)=0$
$\Rightarrow a{x}^2+bx+c=0$

$D=b^2-4ac=(-c-a{)}^2-4ac=(c-a{)}^2>0(\because a\ne c)\therefore D>0$
$\Rightarrow D>0$ as $a\ne c$

Hence, the roots are real and distinct.


Alternate solution:
$(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$
By observation, one root is
$x=1$
$b+c-a+c+a-b+a+b-c=a+b+c=0$
Let the other root be $\alpha$,
Now, the product of roots
$\alpha \times 1=\frac{a+b-c}{b+c-a}\Rightarrow \alpha =\frac{-2c}{-2a}=\frac{c}{a}$

As $c\ne a$, so the equation has real and distinct roots.