Question

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Answer 1

Answer:

a = b = c

Step-by-step explanation:

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

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

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

⇒ 3x² - 2ax - 2bx - 2cx + (ab + bc + ca) = 0

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

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

On comparing with ax² + bx + c = 0, we get

a = 3, b = -2(a + b + c), c = (ab + bc + ca)

Given that the equation has real roots.

∴ D = 0

⇒ b² - 4ac = 0

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

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

⇒ 4[(a² + b² + c² + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca)] = 0

⇒ 4[(a² + b² + c² - ab - bc - ca)] = 0

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

⇒ 2[(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²)] = 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, b - c = 0, c - a = 0

⇒ a = b, b = c, c = a

⇒ a = b = c

Hope it helps!

Answer 2

Expanding and rearranging, the equation is:

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

iii) So its discriminant is: D = b² - 4ac

= 4(a + b + c)² - 12(ab + bc + ca)

= 4{(a + b + c)² - 3(ab + bc + ca)}

= 4(a² + b² + c² - ab - bc - ca) = 2(2a² + 2b² + 2c² - 2ab - 2bc - 2ca)

= 2{(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²)}

= 2{(a - b)² + (b - c)² + (c - a)²}

iv) For equal roots condition, D = 0

==> (a - b)² + (b - c)² + (c - a)² = 0

For all a, b & c real, (a - b)² + (b - c)² + (c - a)² = 0, if and only if every term is zero.

==> a - b = 0 = b - c = c - a

==> a = b = c