Complex Numbers And Quadratic Equations question 672
Question: If $ \frac{1}{2-\sqrt{-2}} $ is one of the roots of $ ax^{2}+bx+c=0, $ where a, b, c are real, then what are the values of a, b, c respectively?
Options:
A) $ 6,-4,1 $
B) $ 4,6,-1 $
C) $ 3,-2,1 $
D) $ 6,4,1 $
Show Answer
Answer:
Correct Answer: A
Solution:
Given quadratic equation is $ {ax^{2}} + bx + c = 0 $ whose one root is $ \frac{1}{2-\sqrt{-2}} $ Consider $ \frac{1}{2-\sqrt{-2}}=\frac{1}{2-\sqrt{2i}}\times \frac{2+\sqrt{2i}}{2+\sqrt{2}i} $ $ =\frac{2+\sqrt{2}i}{4+2}=\frac{2+\sqrt{2}i}{6} $
$ \therefore $ Another root will be $ \frac{2-\sqrt{2}i}{6} $ ( $ \because $ complex roots always occurs in pairs) Thus, sum of roots $ =( \frac{2+\sqrt{2}i}{6} )( \frac{2-\sqrt{2}i}{6} ) $ $ =\frac{4+2}{36}=\frac{1}{6} $
$ \therefore $ Required equation is $ {x^{2}}- ( sum of roots )x + ( product of roots ) = 0 $ $ x^{2}-\frac{4}{6}x+\frac{1}{6}=0 $
$ \Rightarrow 6x^{2}-4x+1=0 $ Thus, the values of a, b, c are 6, - 4, 1 respectively
BETA
