Question

If the roots of the quadratic equation (-a)(x-b) + (x-b)(x-c)+(x-c)(x-a) = 0 are equal. Then,
show a=b=c​

Answer 1

Answer:

Root of quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = 0 are equal

Means D = b² - 4ac = 0 for this equation,

first we should rearrange the equation ,

(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

⇒3x² - 2(a + b + c)x + (ab + bc + ca)

D = {2(a + b + c)}² - 4(ab + bc + ca).3 = 0

⇒4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0

⇒ a² + b² + c² - ab - bc - ca = 0

⇒2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

⇒(a - b)² + (b - c)² + (c - a)² = 0

This is possible only when , a = b = c

Hence, proved , if roots of given equation are equal then, a = b = c

Answer 2

Appropriate 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.

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

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