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x^3-14x=0 How many rational solutions? How many irrational solutions?
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@Directrix @jim_thompson5910 @robtobey Help please1 :)
hey!!
this has to factor right? since there is a common factor of \(x\) in both terms !
Hi !
Your right
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once you write \[x(x^2-14)=0\] it should be easy to solve (easy enuf)
I got 3 answers
yes, you should \[\{0,\sqrt{14},-\sqrt{14}\}\]
There should be three though right because the leading coefficient had an exponent of 3
Yup thats what i got
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How many are rational and how many are irrational?
@misty1212
0 is rational for sure right?
and since \(14\) is NOT a perfect square, then \(\sqrt{14}\) is irrational
Okay so 1 rational and 2 irrational
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yup
Thank you so much !
\[\color\magenta\heartsuit\]
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