If f(x) = Ax^2+Bx+C, is f(0)*f(5)0?

Question

If  f(x) = Ax^2+Bx+C  , is f(0)*f(5)>0 ?

(1)  f(0)>0

(2) C:B:A is in the ratio 25:10:1

Question made by me.Feedback welcome. Kudos to the first correct solution. OA after some initial discussion.

Zarrolou · Answer

If f(x) =  Ax^2+Bx+C  , is f(0)*f(5)>0 ?

f(0)=C 
 f(5)=A25+B5+C

is  (C)(A25+B5+C)>0  ?

I. f(0)>0

So  C>0 , the first term is +ve, no info about the second. 
No sufficient

II. C:B:A is in the ratio 25:10:1 
They can be written in the form 25x:10x:1x, and x does not equal 0.
If are all positive (x>0)
 (C)(A25+B5+C)=(+)(+)>0 
If are all negative (x<0)
 (C)(A25+B5+C)=(-)(-)>0 
Sufficient
 B

mau5 · Answer

+1 Zarrolou! However, would like to show a more graphical approach: Consider a generalised quadratic expression : f(x) = Ax^2+Bx+C Any quadratic equation with real roots(i.e. f(x)=0) will intersect the X-axis atleast once and at-most twice. The former case will happen only when the roots are real and equal; the latter when the roots are real and unequal.For any quadratic where f(x)=0 and the roots are not real, for example, f(x) = x^2+x+1 , the curve will NEVER intersect the X-axis. Consider the attached diagram for an example : The quadratic which intersects the X-axis at (1,0) and (3,0) will have f(x)<0 for all values of x belonging to {1,3}-the shaded region; f(1) = f(3) = 0 and f(x)>0 for all other real values of x. Thus, the sign of f(x) changes between the two roots. If the curve has only one root, as shown in the diagram in the 2nd quadrant, f(x) =0 at only one value of x and f(x)>0 for all other real values of x. Now back to the question: F.S 1 states that f(0)>0. Also, we are asked whether f(0)*f(5)>0. Thus we have to find whether f(5)>0 or f(5)<0. Now, f(0)>0 in both the curves shown in the first quadrant. However, in the first graph with roots at (1,0) and (3,0), f(5)>0. In the second curve, with roots (1,0) and (6,0), f(5)<0. Insufficient. From F.S 2, we have the ratio between A,B and C. Thus,on substituition, f(x) = \frac{C}{25}(x+5)^2 . Now for C as positive/negative, the curve would open upwards/downwards. In any case, as the curve has equal and real root (x=-5), thus the value of f(x) for all other real values of x would be of the same sign. Thus, f(0)*f(5) will always be positive. Sufficient. B. Let me know if anything remains unclear.

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If f(x) = Ax^2+Bx+C, is f(0)*f(5)>0?

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