Question

ax^2 + bx + c=0 ..f(1)<0...f(-1)>-4...f(3)>5
Find the sign of a

Answer 1

dumb way of working it is by setting up 3 inequalities and then eliminating b and c

Answer 2

but i feel we can use IVT here as there is a sign change between 1 and 3, so there must be a zero in the interval (1, 3)

Answer 3

I think il work it the dumb way :

a + b + c < 0
-a - b - c -4 < 0
-9a - 3b -c +5 < 0
----------------------
eliminate

Answer 4

Seems like we have 3 unknowns and 3 conditions. Since we're looking at all these limiting cases, I don't think it will affect the general shape of the parabola will it? It'll either be up or down.\[\left[\begin{matrix}1 & 1 &1\\ 1 & 1 &1 \\ 9&3&1\end{matrix}\right] \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)=\left(\begin{matrix}0 \\ -4 \\ 5\end{matrix}\right)\]

Invert the matrix and left multiply to get a, b, and c of the limiting case.

Answer 5

That middle 1 should be -1.

Answer 6

Hmm sorry I don't think my way will really work very well since at some point the values might end up flattening out into a straight line. However I think if we take the derivative of f(x)=ax^2+bx+c with respect to a and solve for when a=0 we should get the point where the slope changes from up to down or down to up.

Answer 7

@dan815 wanna try this ?

Answer 8

well the location of roots ...doesnt help either .!

Answer 9

class 11 ...i know differentiation but can this be solely solved using graphical quadratic approach

Answer 10

there are several ways to solve this

Answer 11

did you try elimination ?

Answer 12

ax^2 + bx + c=0 ..f(1)<0...f(-1)>-4...f(3)>5


f(1) < 0 gives below inequality
`a + b + c < 0`

Answer 13

yeah i know how the 3 inequalities emerged ..

Answer 14

though its a boring method, eliminating b and c should be easy right ?

Answer 15

cuz we don't have a second method yet... so... lets try elimination and see what we get...

Answer 16

alright then ..:) i posted this to know approach.

Answer 17

yeah ganeshie8 your way seems pretty straight forward

Answer 18

i think the question is created just to strengthen the algebra of elimination rather then ..lol

Answer 19

shayad....taking the cases first a>0 then a<0 and seeing where do we get the graphical contradiction

Answer 20

@ganeshie8 could we think of something like this

Answer 21

oh yes seems legit

Answer 22

have to prove though