Question

(a-b)x^2 + (b-c)x + (c-a)

Answer 1

How to solve your problem

(−)⋅2+(−)+(−)

Simplify

1. Distribute

(−)⋅2+(−)+(−)

2−⋅2+(−)+(−)

2. Distribute

2−⋅2+(−)⋅+(−)

2−⋅2++(−)+(−)

3. Re-order terms so that constants are on the left

2−⋅2++(−)+(−)

2−⋅2+−+(−)

4. Rearrange terms

2−⋅2+−⋅+(−)

2−⋅2+−⋅+(−+)

5 Eliminate redundant parentheses

2−⋅2+−⋅+(−+)

Solution

2−⋅2+−⋅−+

Answer 2

From the given question the correct answer is:

(a-b)x² + (b-c)x + (c-a)=b + c = 2a

Given:

(a-b)x² + (b-c)x + (c-a)

To find:

(a-b)x² + (b-c)x + (c-a)=?

Solution:

Let, (a-b)x² + (b-c)x + (c-a)=0

quadratic equation is (a – b)x2 + (b – c)x + (c – a) = 0

we know that, the root are equal,

we will discriminent of the quadritic equation = 0

Hence (b – c)² = 4(a – b)(c – a)

⇒ b² + c² – 2bc = 4(ac – a² – bc + ab)

⇒ b² + c² – 2bc = 4ac – 4a² – 4bc + 4ab

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

⇒ b²+ c² + 4a² + 2bc – 4ac – 4ab = 0

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

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

⇒ (b + c – 2a)² = 0

⇒ b + c – 2a = 0

∴ b + c = 2a

Hence ,(a-b)x² + (b-c)x + (c-a)=b + c = 2a