Complex Numbers And Quadratic Equations question 203
Question: If $ a<0 $ then the inequality $ ax^{2}-2x+4>0 $ has the solution represented by [AMU 2001]
Options:
A) $ \frac{1+\sqrt{1-4a}}{a}>x>\frac{1-\sqrt{1-4a}}{a} $
B) $ x<\frac{1-\sqrt{1-4a}}{a} $
C) x < 2
D) $ 2>x>\frac{1+\sqrt{1-4a}}{a} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ ax^{2}-2x+4>0 $
Þ $ x=\frac{2\pm \sqrt{4-16a}}{2a} $
Þ $ x=\frac{1\pm \sqrt{1-4a}}{a} $ \ $ \frac{1-\sqrt{1-4a}}{a}<x<\frac{1+\sqrt{1-4a}}{a} $ .
BETA
