Question

If the equation $x^2+ax+bc=0$ and $x^2+bx+ca=0$ have one common root, then their remaining roots are-

• A. $1,b$
• B. $b,a$
• C. $b,c$
• D. $c,a$

Answer 1

The correct option is B $b,a$

Only one root is common: let $\alpha$ be the common root

$\frac{{\alpha }^2}{b_1{c}_2-b_2{c}_1}=\frac{\alpha }{a_2{c}_1-a_1{c}_2}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Above Formula is used to find common root between

$a_1{x}^2+b_{1}x+c_1=0$

$a_2{x}^2+b_{2}x+c_2=0$

Now comparing given equations with above mentioned standard equation, we get

$a_1=1,b_1=a,c_1=bc$ and $a_2=1,b_2=b,c_2=ca$

Now putting above values in the given formula, we get

$\frac{{\alpha }^2}{(a^2-b^2)c}=\frac{\alpha }{(a-b)c}=\frac{1}{b-a}$

Now ${\alpha }^2=-(a+b)$ and $\alpha =c$

Putting $c$ in given equations in the question satisfies both the equations.

Let us assume the roots of first equation given in the question be $\alpha ,x_1$ and of second equation is $\alpha ,x_2$

Now Using $concept$ of $sum$ $and$ $product$ $of$ $roots$

in $First$ equation product of roots $=bc$ and

$\Rightarrow$ $bc=\alpha .x_1$

putting $\alpha =c$

we get

$[x_1=b]$

In $Second$ equation, product of roots $=ac$ and $ac=\alpha .x_2$

Putting $\alpha =c$

$\Rightarrow$ $x_2=a$

Therefore answer is $\left(B\right)$