Question

If $x^2+ax+bc=0$ and $x^2+bx+ca=0$ $(c\ne 0)$ have a common root. Then other root satisfies the equation

• A. $x^2+cx+ab=0$
• B. $x^2+ax+bc=0$
• C. $x^2+bc+ac=0$
• D. $x^2+abx+c=0$

Answer 1

The correct option is A $x^2+cx+ab=0$

$x^2+ax+bc=0$ ... (i)
$x^2+bx+ac=0$ ... (ii)
Subtracting (ii) from (i), we get
$x(a-b)+c(b-a)=0$
$(x-c)(a-b)=0$
Hence, the common root is $c$.
Substituting in (i) we get
$c^2+ac+bc=0$
$c=0$ or $a+b+c=0$
Now considering $a+b+c=0$
In the first equation if one of the roots is $c$, then another roots has to be ${}^{\prime }{b}^{\prime }$ since the product of roots is $bc$.
Similarly, other root of equation ii is ${}^{\prime }{a}^{\prime }$.
Hence, the non-common roots must satisfy
$(x-a)(x-b)=0$
$x^2-(a+b)x+ab=0$
But we know that $a+b+c=0$
$-(a+b)=c$
Hence, simplifying, we get
$x^2+cx+ab=0$.