Question

Prove that both the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real but they are equal only when $a=b=c.$

Answer 1

The given equation

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

$\to$ $x^2$ –ax-bx +ab + $x^2$ –bx –cx +bc + $x^2$ -cx –cx –ax +ac = 0

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

Here

a= 3

b= -2 (a+b+c)

c = ( ab+ bc +ac)

We know that D = $b^2$ – 4ac

That is the roots are real only when $b^2$ – 4ac = + value

Hence putting the values we have

$\to$ ${[-2(a+b+c\left)\right]}^2$ $>$ 4 (3) (( ab+ bc +ac)

$\to$ 4 $a^2$ + 4 $b^2$ + 4 $c^2$ $>$ 4ab +4ac +4bc

$\to$ $a^2$ + $b^2$ + $c^2$ $>$ ab + ac +bc

Clearly the roots are real.

For roots to be equal $b^2$= 4 ac. In particular if a=b=c then the roots will be real

So, substituting in $a^2$ + $b^2$ + $c^2$ $>$ ab + ac +bc we have

$a^2$ + $b^2$ + $c^2$ = (for ab$\to$ a.a )+ ( for ac $\to$ c.c) + (for bc $\to$b.b)

Here LHS = RHS