Question

Show that the roots of equation (x-a)(x-b) = abx^2; a,b belong R are always real . When are the equal​

Answer 1

Given :   (x-a)(x-b) = abx^2  a,b belong R  

To Find : When roots are real

Solution:

(x-a)(x-b) = abx²

=> x²  -ax  - bx  + ab = abx²

=> x²(ab - 1)  + (a + b)x  - ab = 0

roots are real  

(a + b)² - 4(ab - 1)(-ab) ≥ 0

=> a² + b² + 2ab   + 4a²b²  - 4ab   ≥ 0

=> (a - b)² + 4a²b²  ≥ 0

=>  (a - b)² + (2ab)²  ≥ 0

This is true for all values of a & b

Hence Roots are always real

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