Question

if root of equation (b-c)x^2+(c-a)x+(a-b)=0.then show that a,b,c are in AP​

Answer 1

Step-by-step explanation:

hope it will help u...............

Answer 2

$\large\bf\underline{Given:-}$

• Roots of equation are equal

$\implies$which means D(Discriminant) = 0

$\large\bf\underline {To \: verify:-}$

a,b and c are in AP.

$\huge\bf\underline{Solution:-}$

Given equation is,

(b-c)x² + (c-a)x + (a-b)= 0

D = b² - 4ac

D = (c-a)² - 4 (b-c)(a-b) =0

c² + a² - 2ac - 4(ab - b² - 4ac + bc) = 0

$\implies$ c² + a² - 2ac - 4ab +4b² + 4ac - 4bc = 0

$\implies$ a² + 4b² + c² - 4ab - 4bc - 2ac = 0

$\implies$ ( a - 2b + c )² = 0

$\implies$ a - 2b + c = 0

$\implies$ a + c = 2b

This can also be written as :

$\implies$ ( b - a ) = ( c - b )

So, this is the required condition for a,b and c.

Therefore, a,b and c are in A.P

( hence proved )