CBSE BOARD X, asked by entrepreneur4, 1 year ago

If the roots of the equation

(b-c)x² (c-a)x (a-b)=0 are equal then prove that 2b=ac


Bipashanayak: Actually q. Is little wrong

Answers

Answered by Bipashanayak
2
Hi,

We know that,

If the quadratic equation ax²+bx+c=0

whose roots are equal then it's

deteminant is equal to zero.

(a-b)x²+(b-c)x+(c-a)=0

Deteminant =0

(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

Therefore,

b+c=2a

Hence proved.

I hope this helps you.

:)
Answered by mathsdude85
1

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….

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