Complex Numbers And Quadratic Equations question 812
Question: If $ a,b,c $ are real and $ x^{3}-3b^{2}x+2c^{3} $ is divisible by $ x-a $ and $ x-b $ , then
Options:
A) $ a=-b=-c $
B) $ a=2b=2c $
C) $ a=b=c $ , $ a=-2b=-2c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
As $ f(x)=x^{3}-3b^{2}x+2c^{3} $ is divisible by $ x-a $ and $ x-b $ , therefore $ f(a)=0\Rightarrow a^{3}-3b^{2}a+2c^{3}=0 $ …..(i) and $ f(b)=0 $
Þ $ b^{3}-3b^{3}+2c^{3}=0 $ …..(ii) From (ii), $ b=c $ From (i), $ a^{3}-3ab^{2}+2b^{3}=0 $ (Putting $ b=c $ )
$ \Rightarrow (a-b)(a^{2}+ab-2b^{2})=0 $
$ \Rightarrow $ $ a=b $ or $ a^{2}+ab=2b $ Thus $ a=b=c $ or $ a^{2}+ab=2b^{2} $ and $ b=c $ $ a^{2}+ab=2b^{2} $ is satisfied by $ a=-2b $ . But $ b=c $ . \ $ a^{2}+ab-2b^{2} $ and $ b=c $ is equivalent to $ a=-2b=-2c $
BETA
