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Mathematics 13 Online
OpenStudy (anonymous):

Solve x^2-5x>=0 can I factor out x and get x(x-5)..? or no

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

but that doesn't solve the problem, that tells you it is 0 at \(x=0\) and at \(x=5\) if you want to know where \(x^2-5x\geq 0\) you need to know that this is a parabola that opens up, and so it positive outside the zeros and negative between them

OpenStudy (anonymous):

I'm trying to find the domain of 1/4:sqrt(x^2-5x)

OpenStudy (anonymous):

ok then you are on the right path you need to solve \(x^2-5x\geq 0\) just like you wrote

OpenStudy (anonymous):

you know the zeros by factoring, as you did they are at 0 and 5 so this will be non-negative if \(x\leq 0\) or \(x\geq 5\)

OpenStudy (anonymous):

huh? I'm confused by what you just wrote.

OpenStudy (anonymous):

ok lets go slow

OpenStudy (anonymous):

you are trying to solve an inequality, not and equality if you wanted to solve \(x^2-5x=0\) it would be relatively easy you would factor and get \(x(x-5)=0\) so \(x=0\) or \(x=5\) so far so good?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

but you do not want to know where \(x^2-5x=0\) but rather where \(x^2-5x\geq 0\) i.e. the intervals for which \(x^2-5x\) is positive (ok actually non-negative)

OpenStudy (anonymous):

what does the graph of \(y=x^2-5x\) look like? it is a parabola that opens up, and has zeros at \(x=0\) and at \(x=5\)|dw:1346294288194:dw|

OpenStudy (anonymous):

so it is negative (below the \(x\) axis) between the zeros, and positive outside them is that clear or no?

OpenStudy (anonymous):

okay that makes a little sense.

OpenStudy (anonymous):

don't forget that when you solve an inequality, you solve both for greater than and less than, because those are the only options. you might have been taught a different way, namely to test points and see which are positive an negative, but is rather silly if you already know what the graph looks like also don't forget that positive is a synonym for above the \(x\) axis, and negative means below so you can see from the picture, which you can carry mentally (you don't have to graph) that a parabola that opens up is positive outside the zeros, by which i mean in this case it will be positive for all values of \(x<0\) and for all values of \(x>5\)

OpenStudy (anonymous):

|dw:1346294667381:dw|

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