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Q. Roots of the equation $ (x-a)\text{ (}x-b)=ab{{x}^{2}} $ are

Rajasthan PETRajasthan PET 2010

Solution:

Given equation is
$ (x-a)(x-b)=ab{{x}^{2}} $
$ \Rightarrow $ $ {{x}^{2}}-(a+b)x+ab=ab{{x}^{2}} $
$ \Rightarrow $ $ (1-ab){{x}^{2}}-(a+b)x+ab=0 $
Discriminant $ D={{B}^{2}}-4AC $
$ ={{(a+b)}^{2}}-4ab(1-ab) $
$ ={{a}^{2}}+{{b}^{2}}+2ab-4ab+2{{a}^{2}}{{b}^{2}} $
$ ={{(a-b)}^{2}}+{{(2ab)}^{2}}\ge 0 $
Hence, roots are real.