Complex Numbers And Quadratic Equations question 566
Question: If the roots of the equation $ ax^{2}+bx+c=0 $ are real and of the form $ \frac{\alpha }{\alpha -1} $ and $ \frac{\alpha +1}{\alpha } $ , then the value of $ {{(a+b+c)}^{2}} $ is [AMU 2000]
Options:
A) $ b^{2}-4ac $
B) $ b^{2}-2ac $
C) $ 2b^{2}-ac $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Here $ \frac{\alpha +1}{\alpha }+\frac{\alpha }{\alpha -1}=-\frac{b}{a} $ and $ \frac{\alpha +1}{\alpha -1}=\frac{c}{a} $ \ $ \alpha =\frac{c+a}{c-a} $ and $ \frac{2{{\alpha }^{2}}-1}{\alpha (\alpha -1)}=-\frac{b}{a} $ Substituting $ \alpha $ , we get $ {{(a+b+c)}^{2}}=b^{2}-4ac $ . Note: Students should remember this fact.
BETA
