Question

If the roots of the quadratic equation (a-b)x^2+(b-c)x+(c-a)=0, a unequal to b are equal, then prove that b+c=2a

Answer 1

◘ Given ◘

The roots of the quadratic equation (a - b)x² + (b - c)x + (c - a) = 0, are equal.

◘ To Prove ◘

That b + c = 2a.

◘ Proof ◘

As the zeros are equal, so D = 0.

→ D = b² - 4ac

In the given equation,

• a = (a - b)
• b = (b - c)
• c = (c - a)

Now,

Substituting the values :-

(b - c)² - 4(a - b)(c - a) = 0  

⇒ b² + c² - 2bc - 4ac + 4a² + 4bc - 4ab = 0

⇒ b² + c² + 4a² + 4bc - 4ac - 4ab = 0

⇒ b² + c² + (-2a)² + 2bc + 2c(-2a) + 2(-2a)b = 0

⇒ (b + c - 2a)² = 0

⇒ b + c - 2a = 0

⇒ b + c = 2a

Hence, proved.

Answer 2

Given

If the roots of the quadratic equation (a-b)x^2+(b-c)x+(c-a)=0.

To prove

b + c = 2a

Proof

According to the quadratic formula

→ D = b² - 4ac

• a = (a - b)
• b = (b - c)
• c = (c - a)

→ (b - c)² - 4[(a - b)(c - a)] = 0

Apply identity : (a - b)² = a² + b² - 2ab

→ (b)² + (c)² - 2 × b × c - 4[a(c - a)-b(c - a)] = 0

→ b² + c² - 2bc - 4[ac - a² -bc + ba] = 0

→ b² + c² - 2bc - 4ac + 4a² + 4bc - 4ba = 0

→ 4a² + b² + c² - 4ac - 2bc + 4bc - 4ba = 0

→ 4a² + b² + c² - 4ab + 2bc - 4ac = 0

Apply identity : (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

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

→ (- 2a + b + c)² = 0

→ - 2a + b + c = 0

→ b + c = 2a

Hence,

• b + c = 2a -------proved