Question

if the roots of the quadratic equation:
(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a)=0
are equal, then show that a=b=c​

Answer 1

Answer:

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

(x-a)(x-b)=x²-(a+b)x+ab. (Because (x-a)(x-b)=x²-(a+b)x+ab where x=x, a=a & b=b).

(x-b)(x-c)=x²-(b+c)x+bc. (Because (x-a)(x-b)=x²-(a+b)x+ab where x=x, a=b & b=c).

(x-c)(x-a)=x²-(c+a)x+ca. (Because (x-a)(x-b)=x²-(a+b)x+ab where x=x, a=c & b=a).

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=(x²-(a+b)x+ab)+(x²-(b+c)x+bc)+(x²-(c+a)x+ca). (Because (x-a)(x-b)=x²-(a+b)x+ab, (x-b)(x-c)=x²-(b+c)x+bc & (x-c)(x-a)=x²-(c+a)x+ca).

3x²-2x(a+b+c)+(ab+bc+ca).

Let D be discriminant.

D=4(a+b+c)²−12(ab+bc+ca).

=4[(a+b+c)²−3(ab+bc+ca)].

=2[2a²+2b²+2c²−2ab−2bc−2ca].

=2[(a−b)²+(b−c)²+(c−a)²].

If D=0, then.

(a−b)²+(b−c)²+(c−a)²=0.

a−b=0;b=c;c=a.

a=b.

(or).

a=b=c.

I think this is your answer.

Answer 2

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\sf \: (x - a)(x - b) + (x - b)(x - c) + (x - a)(x - c) = 0$

$\sf \:  {x}^{2} - (a + b)x + ab +  {x}^{2} - (b + c)x + bc +  {x}^{2} - (c + a)x + ac = 0$

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

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

 

So, on comparing with Ax² + Bx + C = 0, we get

$\:\boxed{\begin{aligned}& \qquad \:\sf \:A=3 \qquad \: \\ \\& \qquad \:\sf \: B= - 2(a + b + c)\qquad\\ \\& \qquad \:\sf \: C=ab + bc + ca\qquad\end{aligned}} \qquad \:$

As it is given that, equation has real and equal roots.

$\sf \: Discriminant, D = 0$

$\sf \:  {B}^{2} - 4AC = 0$

$\sf \:  {[  - 2(a + b + c)]}^{2} - 4(3)(ab + bc + ca) = 0$

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

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

$\sf \: {a}^{2} +  {b}^{2} +  {c}^{2} + 2ab + 2bc + 2ca   - 3ab - 3bc - 3ca = 0$

$\sf \: {a}^{2} +  {b}^{2} +  {c}^{2}    - ab - bc - ca = 0$

can be rewritten as

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

$\sf \:2{a}^{2} +  2{b}^{2} +  2{c}^{2}    - 2ab - 2bc - 2ca= 0$

$\sf \: {a}^{2}  + {a}^{2} +   {b}^{2}  + {b}^{2} +   {c}^{2}  + {c}^{2}    - 2ab - 2bc - 2ca= 0$

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

$\sf \:  {(a - b)}^{2} +  {(b - c)}^{2} +  {(c - a)}^{2} = 0$

$\implies\sf \: a - b = 0 \:  \: and \:  \: b - c = 0 \:  \: and \:  \: c - a = 0$

$\implies\sf \: a = b \:  \: and \:  \: b = c \:  \: and \:  \: c  = a$

$\implies\sf \: a = b  = c\:$

$\rule{190pt}{2pt}$

Concept Used :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac