Complex Numbers And Quadratic Equations question 628
Question: If $ \alpha ,\beta $ are roots of $ ax^{2}+bx+b=0, $ then $ \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}+\sqrt{\frac{b}{a}} $ is ( $ b^{2}\ge 4ab, $ a and b are of same sign)
Options:
A) 0
B) 1
C) 2
D) $ 2\sqrt{\frac{b}{a}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{( \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }} )}^{2}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }+2=\frac{{{(\alpha +\beta )}^{2}}}{\alpha \beta } $ $ =\frac{{{( -\frac{b}{a} )}^{2}}}{( \frac{b}{a} )}=\frac{b}{a} $
$ \therefore \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\sqrt{\frac{b}{a}} $ [ $ \because \alpha ,\beta $ are real] $ \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}+\sqrt{\frac{b}{a}}=2\sqrt{\frac{b}{a}} $
BETA
