Complex Numbers And Quadratic Equations question 786
Question: The expression $ y=ax^{2}+bx+c $ has always the same sign as c if
Options:
A) $ 4ac<b^{2} $
B) $ 4ac>b^{2} $
C) $ ac<b^{2} $
D) $ ac>b^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ f(x)=ax^{2}+bx+c $ . Then $ f(0)=c $ . Thus the graph of $ y=f(x) $ meets y-axis at (0, c). If $ c>0 $ , then by hypothesis $ f(x)>0 $ This means that the curve $ y=f(x) $ does not meet x-axis. If $ c<0 $ , then by hypothesis $ f(x)<0 $ , which means that the curve $ y=f(x) $ is always below x-axis and so it does not intersect with x-axis. Thus in both cases $ y=f(x) $ does not intersect with x-axis i.e. $ f(x)\ne 0 $ for any real x. Hence $ f(x)=0 $ i.e. $ ax^{2}+bx+c=0 $ has imaginary roots and so $ b^{2}<4ac $ .
BETA
