Question

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

Answer 1

the answer is given in photo

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  

HOPE THIS ANSWER WILL HELP YOU….