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.
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Answer 1

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

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