Complex Numbers And Quadratic Equations question 798
Question: $ x^{2}+x+1+2k(x^{2}-x-1)=0 $ is a perfect square for how many values of k [Orissa JEE 2004]
Options:
A) 2
B) 0
C) 1
D) 3
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation $ (1+2k)x^{2}+(1-2k)x+(1-2k)=0 $ If equation is a perfect square then root are equal i.e., $ {{(1-2k)}^{2}}-4(1+2k)(1-2k)=0 $ i.e., $ k=\frac{1}{2},\frac{-3}{10} $ . Hence total number of values = 2.
BETA
