Complex Numbers And Quadratic Equations question 794
Question: If $ k\in (-\infty ,-2)\cup (2,\infty ), $ then the roots of the equation $ x^{2}+2kx+4=0 $ are [DCE 2002]
Options:
A) Complex
B) Real and unequal
C) Real and equal
D) One real and one imaginary
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation is $ x^{2}+2kx+4=0 $ Put $ k=-3 $ , $ x^{2}-6x+4=0 $
$ \Rightarrow x=3+\sqrt{5},3-\sqrt{5} $ Put k = 3, $ x^{2}+6x+4=0 $
$ \Rightarrow x=-3+\sqrt{5},-3-\sqrt{5} $ i.e., Roots are real and unequal.
BETA
