Question

If the equations $x^2+cx+ab=0$ and $x^2+bx+ca=0$ have a common root, then

• A. $a+b+c=0$
• B. $a+b-c=0$
• C. the other roots are given by equation $x^2+ax+bc=0$
• D. the other roots are given by equation $x^2-ax+bc=0$

Answer 1

Answer: (A), (C)

The correct options are
A $a+b+c=0$
C the other roots are given by equation $x^2+ax+bc=0$

Let the equations $x^2+cx+ab=0$ ...............(1)
$x^2+bx+ca=0$ ..............(2)
have the common roots $\alpha$ then
${\alpha }^2+c\alpha +ab=0{\alpha }^2+b\alpha +ca=0$
solving by cross multiplication law
$\frac{{\alpha }^2}{c^{2}a-{ab}^2}=\frac{\alpha }{ab-ca}=\frac{1}{b-c}$
from first two relations.
$\alpha =\frac{c^{2}a-{ab}^2}{ab-ca}=\frac{a(c+b)(c-b)}{a(b-c)}$.........(3)
from last two realtions
$\frac{\alpha }{ab-ca}=\frac{1}{b-c}$ ......(4)
$\therefore \alpha =a$
from $\left(3\right)$ and $\left(4\right),$ we get
$a+b+c=0$
Let the other root of the equation (1) and (2) be $\beta$ and $\gamma$ respectively,
then $\alpha \beta =ab$ and $\alpha \gamma =ca$ But $\alpha =a,\therefore \beta =b$ and$\gamma =c$
Now the equation whose roots are $\beta$ and $\gamma$ is
$x^2-(\beta +\gamma )x+\beta \gamma =0\Rightarrow x^2-(b+c)x+bc=0\Rightarrow x^2-(-a)x+bc=0(\therefore a+b+c)\Rightarrow x^2+ax+bc=0$