Step-by-step explanation:

We need to prove that 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.

Before doing that, first solve the brackets. So, let's start with brackets!

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

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

We know that-

• Product of (+)(-) is (-)
• Product of (-)(+) is (-)
• Product of (-)(-) is (+)
• Product of (+)(+) is (+)

→ x² - xb - xa + ab + x² -xc - xb + bc + x² - xa - xc + ac = 0

→ 3x² - 2xa - 2xb - 2xc + ab + bc + ac = 0

Take -2x common from (-2xa - 2xb - 2xc)

→ 3x² - 2x(a + b + c) + ab + bc + ac = 0

→ 3x² - 2x(a + b + c) + (ab + bc + ac) = 0

The above equation is in the form ax² + bc + c = 0. Now,

D = b² - 4ac (where value of b is a + b + c and that of c is ab + bc + ac). Substitute the values,

→ D = (2(a + b + c))² - 4(3)(ab + bc + ac) = 0

→ D = 4(a + b + c)² - 12(ab + bc + ac) = 0

→ D = 4(a² + b² + c² + 2ab + 2bc + 2ac) - 12ab - 12bc - 12ac = 0

→ D = 4a² + 4b² + 4c² + 8ab + 8bc + 8ca - 12ab - 12bc - 12ac = 0

→ D = 4(a² + b² + c² + 2ab + 2bc + 2ac - 3ab - 3bc - 3ac) = 0

→ D = 4(a² + b² + c² - ab - bc - ac) = 0

→ D = a² + b² + c² - ab - bc - ac = 0

Multiply by 2 on both sides,

→ D = 2(a² + b² + c² - ab - bc - ac) = 0 × 2

→ D = 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

→ D = (a - b)² + (b - c)² + (c - a)² = 0

Used identity: (a - b)² = a² + b² - 2ab, (b - c)² = b² + c² - 2bc, (c - a)² = a² + c² - 2ac

If D = 0 then,

→ (a - b)² + (b - c)² + (c - a)² = 0

→ (a - b)² = 0

→ a - b = 0

→ a = b

Similarly; a = b = c.

Hence, proved, that if roots of the given equation i.e. (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 are real then, a = b = c.

Verified answer

EXPLANATION.

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

As we know that,

We can write equation as,

⇒ [x² - bx - ax + ab] + [x² - cx - bx + bc] + [² - ax - cx + ac] = 0.

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

⇒ 3x² - 2ax - 2bx - 2cx + ab + bc + ac = 0.

⇒ 3x² - 2(a + b + c)x + (ab + bc + ac) = 0.

Roots are real and equal.

⇒ D = 0  Or  b² - 4ac = 0.

⇒ [(-2)(a + b + c)]² - 4(3)(ab + bc + ac) = 0.

⇒ [4(a + b + c)²] - 12(ab + bc + ac) = 0.

As we know that,

Formula of :

⇒ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Using this formula in the equation, we get.

⇒ [4(a² + b² + c² + 2ab + 2bc + 2ac)] - 12ab - 12bc - 12ac = 0.

⇒ 4a² + 4b² + 4c² + 8ab + 8bc + 8ac - 12ab - 12bc - 12ac = 0.

⇒ 4a² + 4b² + 4c² - 4ab - 4bc - 4ac = 0.

⇒ 4(a² + b² + c² - ab - bc - ac) = 0.

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

⇒ 2[(a - b)² + (b - c)² + (c - a)²] = 0.

⇒ (a - b)² + (b - c)² + (c - a)² = 0.

⇒ a - b = 0.

⇒ a = b.

⇒ b - c = 0.

⇒ b = c.

⇒ c - a = 0.

⇒ c = a.

∴ a = b = c.

Hence Proved.