Question

Solving equation (b-c) x + (c-a) x + (a-b) = 0 following roots are obtained
b-c
(a)
bec
(b) (a-b)(a-c),1 (c)
(d) None
a-b
ob
1
1​

Answer 1

Answer:

x = (a - b)/(b - c) , 1

Note:

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

Solution:

Here,

The given equation is ;

(b - c)x² + (c - a)x + (a - b) = 0

We need to find the roots of the given quadratic equation .

Thus,

Here we go ↓

=> (b - c)x² + (c - a)x + (a - b) = 0

=> (b - c)x² - (a - c)x + (a - b) = 0

=> (b - c)x² - (a - b + b - c)x + (a - b) = 0

=> (b - c)x² - [ (a - b) + (b - c) ]x + (a - b) = 0

=> (b - c)x² - (a - b)x - (b - c)x + (a - b) = 0

=> x[(b - c)x - (a - b)] - [(b - c)x - (a - b)] = 0

=> [ (b - c)x - (a - b) ]•(x - 1) = 0

=> x = (a - b)/(b - c) , 1

Hence,

Required roots are ;

x = (a - b)/(b - c) , 1