True or False?

If a, b, and c denote real numbers and p(x)=ax^2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

Please show work

Results 1 to 12 of 12

- June 4th 2009, 07:40 PM #1

- Joined
- Apr 2009
- Posts
- 253

- June 4th 2009, 07:58 PM #2
if

then there exist two different solutions

if

there is no solutions

there are two same solutions

for more information see link below

Quadratic function - Wikipedia, the free encyclopedia

- June 4th 2009, 08:31 PM #3
False.

Putting this equation in standard form we have

Assume that , we can conclud that this is graph of a parabola that opens upwards and has vetex . Again, since the parabola opens upward,

And since there are enumerable instances where this statement does not hold, the answer is false

- June 13th 2009, 07:45 PM #4

- Joined
- Apr 2009
- Posts
- 253

- June 13th 2009, 09:23 PM #5
The part in red doesn't make sense. What I think you're trying to say is that . An ordered pair cannot be less that a number.

And since there are enumerable instances where this statement does not hold, the answer is false**all**cases. The statement is an existential statement; one that asks us to find**at least one**case where the statement is true.

True or False?

If and , , then**there exists**a real number such that .**true**statement under certain cases. Again, it all depends on the value of the discriminant . If

, then there are two solutions (so there exists more than one ).

, there there is only one solution (thus exists).

, there there is no real solution. This implies that a does not exist.

To summarize, the only time when an exists is when . Also, the value of is the real zero(s) of the quadratic equation.

Consider . Since , we search for an such that . So in essence, we solve . But or . So in this case, there exists two possible values for : and , where both values for are real numbers.

- June 13th 2009, 09:48 PM #6

- Joined
- Apr 2009
- Posts
- 253

Oh ok I get it now

can you help me with this one:

http://www.mathhelpforum.com/math-he...z-1-2-1-a.html

- June 13th 2009, 09:57 PM #7

- Joined
- Apr 2009
- Posts
- 253

- June 13th 2009, 10:07 PM #8
Yes, that is true. Note that you imposed certain conditions on what you let a,b, and c equal. As I said above, the statement is true for certain cases/conditions, and this would be one of them. You just illustrated the case where we only have one value such that (I illustrated a case where we had more than one value).

- June 14th 2009, 10:14 AM #9
*So, you maintain that there exists a zero for every parabola in the form of ?*

*You were right about the last statement of my argument, a small error, but the result is still sound. I cahnged it. But by reading the question again, and taking note of the phrase*"there exists" ,*I interpret this to mean that*"there must exist" or 'there always exists".*And you and I both know that this is a false statement.*

I will attempt to reword the statement. Note that a,b, and c can be any real numbers.

--For any given parabola, there exists at least one place where it touches the x-axis.--

Is this ridiculous, or is this ridiculous?

*"Exists is not a predicate."*Immanuel Kant.

- June 14th 2009, 11:25 AM #10

- Joined
- Apr 2005
- Posts
- 14,973
- Thanks
- 1121

- June 14th 2009, 12:48 PM #11

- Joined
- Apr 2009
- Posts
- 253

- June 14th 2009, 12:55 PM #12