IIT JEE , AIEEE Forum - Mathematics , Algebra

eklavya2009

if a>0 , f(x)=ax + bx² + c and b^{2 _}4ac < 0 then f(x) >0

x

R. (T/F)

RyuAmakusa

false if b>0 then f(x)>0 for all real x. if b<0 then f(x)<0 for all real x.

netkid07

if u mean f(x)=ax^2 + bx + c

yup it's true.....

as

b^2 - 4ac < 0.....it follows that the eqn has no real roots.....as a>0...the parabola opens upwards....and it never intersects x-axis.....sooo....for any value of x....f(x) will always be +ve

sboosy

I think it is true ....infact this is the condition u ll usually apply in problems

spideyunlimited

i think u meant ax^2 + bx + c = 0 ............??

then
True...

as if b^2 - ac < 0 means imaginary roots. means on cartesian plane, the parabola of ax^2 + bx + c = 0 will remain always above 0 (because a is also positive)

RyuAmakusa

ha i think we got into the habit of having the eq as ax^2 + bx +c
but look at the q carefully.

sboosy

sorry ..didnt read the question properly ...i took it to be
ax^2+bx+c ...sorry for the miss ..mayb the time is a factor ...

harshthakar017

tru try plotting graph

spideyunlimited

yea well thts y i mentioned :)
otherwise your answer is perfect :)

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