Complex Numbers And Quadratic Equations question 607
Question: Difference between the corresponding roots of $ x^{2}+ax+b=0 $ and $ x^{2}+bx+a=0 $ is same and $ a\ne b $ , then [AIEEE 2002]
Options:
A) $ a+b+4=0 $
B) $ a+b-4=0 $
C) $ a-b-4=0 $
D) $ a-b+4=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {\alpha_1},{\beta_1} $ are the roots of the eqn $ x^{2}+ax+b=0 $
$ \Rightarrow x=\frac{-a\pm \sqrt{a^{2}-4b}}{2} $
Þ $ {\alpha_1}=\frac{-a+\sqrt{a^{2}-4b}}{2},{\beta_1}=\frac{-a-\sqrt{a^{2}-4b}}{2} $ and $ {\alpha_2},{\beta_2} $ are the roots of the equation $ x^{2}+bx+a=0 $ So, $ {\alpha_2}=\frac{-b+\sqrt{b^{2}-4a}}{2}, $ $ {\beta_2}=\frac{-b-\sqrt{b^{2}-4a}}{2} $ Now $ {\alpha_1}-{\beta_1}=\sqrt{a^{2}-4b} $ ; $ {\alpha_2}-{\beta_2}=\sqrt{b^{2}-4a} $ Given, $ {\alpha_1}-{\beta_1}={\alpha_2}-{\beta_2} $
$ \Rightarrow \sqrt{a^{2}-4b}=\sqrt{b^{2}-4a} $
$ \Rightarrow a^{2}-b^{2}=-4(a-b) $
$ \Rightarrow a+b+4=0 $ .
BETA
