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Mathematics 18 Online
OpenStudy (anonymous):

What is the possible discriminant of the graph? -11 0 25 73 GRAPH BELOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

OpenStudy (anonymous):

OpenStudy (anonymous):

first two are out because this has two real zeros

OpenStudy (anonymous):

My answer was C

OpenStudy (anonymous):

Just by looking nat where theyre pinpointed

OpenStudy (anonymous):

no that is out also for the following reason if the discriminant is a perfect square, the the zeros will be rational number it looks like your zeros are not rational

OpenStudy (anonymous):

hmmmm

OpenStudy (anonymous):

at least that is what is suggested by the picture. that the zeros are some irrational numbers

OpenStudy (anonymous):

I see that as well now looking closely

OpenStudy (anonymous):

when you use the quadratic formula you take \(\sqrt{b^2-4ac}\) as part of it. if \(b^2-4ac=25\) then you would get \(5\) and so your zeros would be integers or fractions

OpenStudy (anonymous):

In this case, integers

OpenStudy (anonymous):

What would I plug into the formula?

OpenStudy (anonymous):

Hey, but any number with a finite number of decimals is a rational number.

OpenStudy (anonymous):

not sure what you mean

OpenStudy (anonymous):

Nothing

OpenStudy (anonymous):

@ivanmlerner totally agreed, but this problem looks like it was made up by some math teacher of course the discriminant could be 25 and the zeros could be some fractions the question is not a good one

OpenStudy (anonymous):

I mean that the numbers that are in the graph are rational.

OpenStudy (anonymous):

they are labelled as such i am assuming that they are rounded. of course i am making an assumption and again, the problem is not a good one a correct answer would be "not enough information"

OpenStudy (anonymous):

then again maybe i am totally wrong what do the points actually say?

OpenStudy (anonymous):

I'm going with 25, because that's the only one I can actually back up

OpenStudy (anonymous):

Actually you have enouth information, you have two simetric points and with that you can find the x of the vertex that do not depend on c (ax^2+bx+c). With that you are left with two points that must satisfy an equation with two unknown variables, so it is possible.

OpenStudy (anonymous):

Thank you very much for your time, you're incredibly gracious

OpenStudy (anonymous):

But you are probably right @satellite73

OpenStudy (anonymous):

@InsanelyChaotic can you write what those coordinates actually are? i can't read them

OpenStudy (anonymous):

discriminate = b62 - 4ac? and yes

OpenStudy (anonymous):

-5, -1

OpenStudy (anonymous):

discriminant is \(b^2-4ac\) yes i was trying to read the coordinates of the graph you posted, but i cannot

OpenStudy (anonymous):

if the zeros are \(-5\) and \(-1\) it must be from some other graph however, if those are the zeros then the discriminant must be a perfect square, so it could be 25 and could not be 73 however, that is not what it looked like from the graph

OpenStudy (anonymous):

I think I made a slight error, and it's actually 75

OpenStudy (anonymous):

@satellite73 I was thinking about earlier, and I realized that even if the zeros are irrational, the discriminant can still be a perfect square.

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