Question

The quadratic equation
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ has equal roots, if

• A. $a\ne b,b=c$
• B. $a=b,b\ne c$
• C. $a\ne b,b\ne c$
• D. $a=b=c$
• E. $a+b+c=0$

Answer 1

Answer: (D)

$a=b=c$

Given, quadratic equation is,

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$

$\Rightarrow x^2-(a+b)x+ab+x^2-(b+c)x+bc+x^2-(c+a)x+ac=0$

$\Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$

Since, quadratic equation has equal roots.

Therefore, $D=B^2-4AC=0$

$\Rightarrow 4(a+b+c{)}^2-4\cdot 3\cdot (ab+bc+ca)=0$

$\Rightarrow (a+b+c{)}^2-3(ab+bc+ca)=0$

$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$

$\Rightarrow \frac{1}{2}\{2a^2+2b^2+2c^2-2ab-2bc-2ca\}=0$

$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$
Only possible when $a=b=c$.