Question

If the roots of the equation (b - c)x2 + (c -a)x + (a - b)= 0 are equal. Then prove that 2 b = a + c.

PLZ HELP!!!​

Answer 1

Step-by-step explanation:

Given Equation is (b - c)x² + (c - a)x + (a - b) = 0

Here a = (b - c), b = (c - a), c = (a - b).

Now,

Given that the roots are equal.

∴ b² - 4ac = 0

⇒ (c - a)² - 4(b - c)(a - b) = 0

⇒ c² + a² - 2ac - 4[ab - b² - ac + cb] = 0

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

⇒ c² + a² + 2ac - 4ab - 4cb + 4b² = 0

⇒ (a + c - 2b)² = 0

⇒ a + c = 2b

Hope it helps!

Answer 2

Solution:

We have been given that a quadratic Equation (b - c)x² + (c -a)x + (a - b)= 0 and also roots of the equation is Equal i.e Discriminant (D) = 0

Define Discriminant of the Equation:

For Equal roots , We have:

• Discriminant = 0

⇒ b² - 4ac = 0

⇒ ( c - a)² - 4(b - c)(a - b) = 0

⇒ [c² + a² - 2ac - 4{b(a - b) - c(a - b)}] = 0

⇒ [ c² + a² - 2ac- 4{ab - b² - ca + bc ] = 0

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

⇒ ( a + c - 2b)² = 0

⇒ a + c - 2b = 0

⇒ a + c = 2b

Hence, It's Proved.