SOLUTION: Dealing with contradictions: If a, b, and c are nonzero real numbers such that a>0 and the equation ax^2+bx+c has no real solutions, then show that ax^2+bx+c<0 cannot be true for

Question 32757: Dealing with contradictions: If a, b, and c are nonzero real numbers such that a>0 and the equation ax^2+bx+c has no real solutions, then show that ax^2+bx+c<0 cannot be true for all values of x.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
AX^2+BX+C=0...HAS NO REAL SOLUTION...SO DISCRIMINANT = B^2-4AC ..IS
NEGATIVE....OR....4AC-B^2 IS POSITIVE
LET F(X)=AX^2+BX+C=A{X^2+BX/A+C/A}=A[{X^2+2(X)(B/2A)+(B/2A)^2}-(B/2A)^2+(C/A)]
=A[{X+(B/A)}^2}+{(C/A)-B^2/4A^2}]=A[{X+(B/A)}^2+{(4AC-B^2)/4A^2}]
SO F(x)=A..SINCE {X+(B/A)}^2 IS
ALWAYS POSITIVE BEING A PERFECT SQUAREAND {(4AC-B^2)} IS POSITIVE AS
SHOWN ABOVE AND 4A^2 IS ALSO POSITIVE BEING A PERFECT SQUARE...
HENCE F(X) IS ONLY DEPENDENT ON SIGN OF A...
F(X) IS POSITIVE IF A IS POSITIVE ..AND
F(X) IS NEGATIVE IF A IS NEGATIVE..
SO IF WE ARE GIVEN THAT A >0..OR POSITIVE THEN
F(X)=AX^2 + BX + C CAN NOT BE NEGATIVE
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