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janessia
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\[x - 5\sqrt{x} +6=0\]
Wired
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Like the last one, get 5sqrt(x) by itself on one side of the equation, and then square both sides.
timo86m
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wired is right :)
janessia
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lol well so far i got \[x-6=5\sqrt{x}\]
janessia
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and idk where to go from there
timo86m
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divide both sides by 5 :)
janessia
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but i thought you had to square something or something like that
timo86m
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isolated for sqrt(x) you not there yet div by 5 first
Wired
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You can do it either way actually. I find it simpler to square both sides at that point so you don't have to deal with fractions. Those fractions would cancel out later anyway, so no need to introduce them in the first place.
timo86m
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that is also true :)
Wired
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\[(x-6)^{2} = (5\sqrt{x})^{2}\]
janessia
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how do i do \[(5\sqrt{x})^2\]
timo86m
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well
(a*b) ^n is a^n*b^n right?
janessia
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yes so the first is 25 and i still dont kno the second part
timo86m
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25*x :)
sqrt(x)^2 is x :)
janessia
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oh lol im dumb lol
timo86m
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no you not :)
janessia
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ha ha well i think so but thanks anyway :)
timo86m
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are not :P now finish the q :)
janessia
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so i got\[x-36=25x\]
janessia
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?????
timo86m
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that dont look right
janessia
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oh wait i did it wrong sorry lol
janessia
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x-36=25x???
timo86m
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(x−6)^2 is (x-6)*(x-6) use foil or whatever method you prefer.
janessia
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my head hurts
Wired
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Just as an FYI, here's another way to look at squaring a square root:
Note that:
\[\LARGE (X^{S})^{T} = X^{ST}\]
\[\LARGE \sqrt{x} = x^{1/2}\]
\[\LARGE (\sqrt{x})^{2} = (x^{1/2})^{2}=x^{(1/2)(2)} = x^{1}\]
janessia
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does this thing have no solution!!??? lol cuz thts wat it seems like!
janessia
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ohk thnk u
Wired
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\[x-6=5\sqrt{x}\]
\[(x-6)^{2}=(5\sqrt{x})^{2}\]
\[(x-6)(x-6)=25x\]
\[x(x-6)-6(x-6)=25x\]
Wired
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Do you see what to do now?
janessia
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distribute?
Wired
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Yep