Complex Numbers And Quadratic Equations question 171
Question: If $ a<b<c<d $ , then the roots of the equation $ (x-a)(x-c)+2(x-b)(x-d)=0 $ are [IIT 1984]
Options:
A) Real and distinct
B) Real and equal
C) Imaginary
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation can be rewritten as $ 3x^{2}-(a+c+2b+2d)x+(ac+2bd)=0 $ Its discriminant D $ ={{(a+c+2b+2d)}^{2}}-4.3(ac+2bd) $ $ ={{{ (a+2d)+(c+2b) }}^{2}}-12(ac+2bd) $ $ ={{{ (a+2d)-(c+2b) }}^{2}}+4(a+2d)(c+2b)-12(ac+2bd) $ $ ={{{ (a+2d)-(c+2b) }}^{2}}-8ac+8ab+8dc-8bd $ $ ={{{ (a+2d)-(c+2b) }}^{2}}+8(c-b)(d-a) $ which is +ve, since $ a<b<c<d $ . Hence roots are real and distinct.
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