Complex Numbers And Quadratic Equations question 558
Question: The equation whose roots are $ \frac{1}{3+\sqrt{2}} $ and $ \frac{1}{3-\sqrt{2}} $ is [MP PET 1994]
Options:
A) $ 7x^{2}-6x+1=0 $
B) $ 6x^{2}-7x+1=0 $
C) $ x^{2}-6x+7=0 $
D) $ x^{2}-7x+6=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
The required equation is $ x^{2}-( \frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}} )x+\frac{1}{3+\sqrt{2}}\times \frac{1}{3-\sqrt{2}}=0 $
$ \Rightarrow x^{2}-( \frac{6}{7} )x+\frac{1}{7}=0\Rightarrow 7x^{2}-6x+1=0 $
BETA
