Question

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

Answer 1

General form is Ax² + Bx + C= 0
if A + B + C = 0 
then zeroes of polynomial are 1 and C/A
here A + B + C = b -c +c - a + a - b = 0
therefore zeroes are 1 and (a - b) / (b - c)
its given that they are equal
∴ (a - b) / (b - c) = 1
⇒a - b = b - c
2b = a + c
hence proved
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Answer 2

SOLUTION :  

Given :  (b - c) x² + (c - a)x + (a - b) = 0 .

On comparing the given equation with ax² + bx + c = 0

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

Discriminant(D) = b² -2ac  

D = (c - a)² - 4(b - c)(a - b)  

D = c² + a² - 2ac – 4(ab - b² - ac + cb)  

[(a - b)² = a² + b² - 2ab]

D = c² + a² - 2ac – 4ab + 4b² + 4ac - 4cb  

D = c² + a² - 2ac + 4ac – 4ab + 4b² - 4cb  

D = c² + a² + 2ac  – 4ab + 4b² - 4cb  

D = a² + 4b² + c² + 2ac  - 4ab - 4cb  

As We know that , (a + c - 2b)² = a² + 4b² + c² + 2ac  - 4ab - 4cb  

D = (a + c - 2b)²  

Real and equal root are Given  

Discriminant(D) = 0

0 = (a + c - 2b)²

Put Square root on both sides  

0 = √(a + c - 2b)²

0 = (a + c - 2b)

a + c =  2b

Hence , 2b = a + C  

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