Question

If $ab+bc+ca=0$, then the roots of the equation $a(b-2c)x^2+b(c-2a)x+c(a-2b)=0$ are

• A. $1$ and $\frac{c}{a}\frac{a+2b}{b+2c}$
• B. $1$ and $\frac{c}{a}\frac{a-2b}{b+2c}$
• C. $1$ and $\frac{c}{a}\frac{a-2b}{b-2c}$
• D. $1$ and $\frac{c}{a}\frac{a+2b}{b-2c}$

Answer 1

The correct option is C $1$ and $\frac{c}{a}\frac{a-2b}{b-2c}$

Given,$a(b-2c)x^2+b(c-2a)x+c(a-2b)=0$
and $ab+bc+ca=0$
If $f\left(x\right)=0$ be the given equation, then
$f\left(1\right)=a(b-2c)+b(c-2a)+c(a-2b)$
$f\left(1\right)=-\sum ab=0(\because \sum ab=0,giveninquestion)$
Hence $1$ is a root of $f\left(x\right)=0$
If the other root be $\alpha$, then
$1\times \alpha$ = Product of roots = $\frac{c(a-2b)}{a(b-2c)}$
Hence $1$ and $\frac{c(a-2b)}{a(b-2c)}$ are the required roots.