Complex Numbers And Quadratic Equations question 768
If $ a>0,b>0,c>0 $ then both the roots of the equation $ ax^{2}+bx+c=0 $ are negative
Options:
A) Are real and negative numbers
B) Have negative real parts
C) Are rational numbers?
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The roots of the equations are given by $ x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} $ (i) Let $ b^{2}-4ac>0,b>0 $ Now if $ a>0,c>0,b^{2}-4ac<b^{2} $
Þ the roots are negative. (ii) Let $ b^{2}-4ac<0, $ then the roots are given by $ x=\frac{-b\pm i\sqrt{(4ac-b^{2})}}{2a},(i=\sqrt{-1}) $ Which are imaginary and have negative real part $ (\because b>0) $ \ In each case, the roots have negative real part.
BETA
