Question

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 roots are equal,prove that a=b=c​

Answer 1

Given :-

• The roots of the equation (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 are equal

To prove :-

• a = b = c

Concept Used :-

Let us consider a quadratic equation ax² + bx + c = 0, then equation has real and equal roots,

• If Discriminant, D = 0, i.e. b² - 4ac = 0.

Solution :-

The given equation is

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

can be rewritten as

x² - ax - bx + ab + x² - bx - cx + bc + x² - ax - cx + ac = 0

⇛ 3x² - 2(a + b + c)x + (ab + bc + ca) = 0 --------(1)

∴ On comparing with Ax² + Bx + C = 0, we get

• A = 3

• B = - 2(a + b + c)

• C = ab + bc + ca

Now,

• It is given that

Equation (1) has equal roots.

So,

• Discriminant, D = 0

• B² - 4AC = 0

$\rm :\implies\: \bigg( - 2(a + b + c) \bigg)^{2} - 4 \times 3 \times (ba + bc + ca) = 0$

$\rm :\implies\:4 {(a + b + c)}^{2} - 12(ba + bc + ca) = 0$

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

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

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

☆ Multiply by 2, on both sides, we get

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

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

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

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

As, we know sum of squares can be 0 only, if

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

$\rm :\implies\:a = b = c$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$