Question

If $a,b,c$ are rational and no two of them are equal, then the quadratic equations $(b-c)x^2+(c-a)x+a-b=0$ and $a(b-c)x^2+b(c-a)x+c(a-b)=0$

• A. have rational roots
• B. will have atleast one rational root
• C. have exactly one root common
• D. have atleast one root common

Answer 1

Answer: (A), (B), (C)

have rational roots
will have atleast one rational root
have exactly one root common

Given,
(i) $(b-c)x^2+(c-a)x+a-b=0$

Sum of all the coefficients $=0$

$\Rightarrow \alpha =1$ is a root.

Product of the roots $=\alpha \beta$.

$=\left(\frac{a-b}{b-c}\right)$

$\beta =\frac{a-b}{b-c}$

(ii) $a(b-c)x^2+b(c-a)x+c(a-b)=0$

Sum of all the co-efficients$=0$.

$\Rightarrow \alpha =1$ is a root.

$\therefore$ 1 is the common root.

Product of the roots$=\alpha \gamma =\frac{c(a-b)}{a(b-c)}$.

$\gamma =\frac{c(a-b)}{a(b-c)}$(rational)

$a\ne c\Rightarrow \gamma \ne \beta$

$\therefore$ They have exactly one root in common and they have rational roots.

Hence, option ${}^{\prime }{A}^{\prime }$, ${}^{\prime }{B}^{\prime }$ and ${}^{\prime }{C}^{\prime }$ are correct.