Question

Determine the sign of c, if equality ax^2 + bx + c = 0 doesn't have any solutions and inequality a + b + c < 0 is right.

Answer 1

Is it ax saquare?

Answer 2

square

Answer 3

Yes.
\[ax^2 + bx + c = 0; a+ b + c < 0\]

Answer 4

you want sign of c right?

Answer 5

I understand that from ax^2 + bx + c = 0 not having any solutions, I can write that:
b^2 - 4ac < 0
But I don't see how I would obtain the sign of c in this way.

Answer 6

@Hawk, yes.

Answer 7

ax^2 + bx + c = 0
Doesn't have solutions only when a and c have same signs.

Answer 8

Yep when we take \[b^{2}-4ac\]

Answer 9

Sign of c can be positive i think

Answer 10

Just noticed that, if a and c are negative, then inequality b < a + c holds true.

Answer 11

Yep

Answer 12

So ax

Answer 13

\[a^{x}2 +b+c=0\]

Answer 14

Sorry, but why?

Answer 15

.

Answer 16

@ganeshie8 Hello! Do you have any ideas?

Answer 17

I might have some idea, but I'm busy in kitchen atm. Will give this a good try after dinner :)

Answer 18

@ganeshie8 thank you very much! :)

Answer 19

well `c` must be less than zero

Answer 20

4ac > b^2
so either a and c are both negative or they are both positive

Answer 21

for this equation,
a<0
x co-ordinate of vertex <0
\[\large \bf \frac{-b}{2a}<0\]
If a<0,then b should be less than zero to satisfy it i.e b<0

To prove this
a + b + c < 0
For good solution, c must be less than zero

Answer 22

and you can see that y-intercept of the graph is negative !
therefore, c<0

Answer 23

what do you think? @welshfella

Answer 24

looks good

Answer 25

As an alternative, observe that \(a+b+c= p(1)\). Now since your curve must maintain the same sign (otherwise, if it doesn't, then it will have a root by IVT) we can say that \(p(0)\) has the same sign as \(p(1)\). But \(p(0)= c\)...

Answer 26

IVT worked here like a charm !