NotTim
Solve the following equation, xe[2,0pi]. tan^3x-9tan^2x+17tanx-6=0
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NotTim
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Well I tried putting x=1, I got an error.
ricnus
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\[x \approx3.14159 n+1.10715\]
\[x \approx3.14159 n+0.429998\]
\[x \approx3.14159 n+1.4191\]
myininaya
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first see if you can solve this
\[u^3-9u^2+17u-6=0\]
JamesJ
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Thanks myin; that's the way.
myininaya
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if you try u=2, you will see you have one solution
LagrangeSon678
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thats what i wasn gonna say replace tan with a variable
myininaya
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then using synthetic division you can find the other 2 solutions
NotTim
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I dunno what that is.
myininaya
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synethic division+quadratic formula (or perhaps factoring if we have other rational zeros)
nubeer
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u can also urs long division to get factors
NotTim
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Wait a sec guys, I'm trying to understand this all...
JamesJ
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A fancy way of saying that as u = 2 is a solution, (u-2) must be a factor of the polynomial. Hence you can write
\[ u^3−9u^2+17^u−6 = (u-2)q(u) \]
for some second order polynomial
JamesJ
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make that 17u, not 17^u
myininaya
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lol
NotTim
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What happened to tan?
JamesJ
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i'll let the other chefs take over
myininaya
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we will get back to the tan business after we solve that polynomial=0 for u
NotTim
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Ok, so I'm doing the whole F(#) thing at this point right?
myininaya
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6
what?
myininaya
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what does F(#) mean?
NotTim
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Looking for the value of x that makes everything 0.
ricnus
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u = 2
u = 1/2 (7-sqrt(37))
u = 1/2 (7+sqrt(37))
FoolForMath
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If you think how myininaya magically found that \( u=2 \) there is a analytical way of doing it which is known as Rational root test.
myininaya
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right fool
NotTim
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What if I got an error on my calculator?
NotTim
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Useless its tanx (as an entirety) that I using....
LagrangeSon678
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Your trying to plug number right into the equation?
myininaya
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i went through the possible rational zeros and found a rational zero
i didn't go through all possible rational zeros as i only needed one so i can just mess with a quadratic
NotTim
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Lagrange, yes...
NotTim
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Not the right idea huh?
FoolForMath
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You need to NotThink a bit.
NotTim
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...ok? I do that a lot.
LagrangeSon678
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Follow those steps that myin showed, youll end with the right answers
NotTim
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I'm just looking at this and my notes, and I have not a clue.
NotTim
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Is u=tanx?
myininaya
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yes
NotTim
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Ok.
FoolForMath
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it's more like Let u = tan x then form a polynomial in u, factor it and then substitute back the u =tan x and then solve it ..
NotTim
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ERkk.
NotTim
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Imma do this slowly...
If yo ugotta do something while you help me, do it now as I work this thru...
NotTim
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IS factor Thereom involved?
PaxPolaris
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yes...if u-2 is a factor of P(u) ...P(2)=0
PaxPolaris
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Let tan (x) = u \[u ^{3}-9u ^{2}+17u-6=0\] \[\left( u-2 \right)\left( u^2 - 7u -3\right) =0\]
NotTim
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wait wait waitt....
And that's also not how I was taught (or shown in the notes).
I'll try it out after I'm done my thing tho/.
NotTim
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Sorry about that. My internet went wonky.
NotTim
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And now I divide the thing by u-2 right?
myininaya
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u^2-7u+3
____________________
u-2| u^3-9u^2+17u-6
-(u^3-2u^2)
-------------------
-7u^2+17u-6
-(-7u^2+14u)
---------------
3u-6
-(3u-6)
--------
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NotTim
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ok
myininaya
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pax did i make a mistake?
myininaya
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our factorizations are a lttle different
i have
\[\left( u-2 \right)\left( u^2 - 7u +3\right) =0\]
FoolForMath
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myininaya, I wish If I can give you more medals for your patience.
NotTim
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Lol.
myininaya
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are you talking about me and my division above? lol
NotTim
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I think just waiting for me to finish up my stuff here. Thanks by the way.
myininaya
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oh i had a thing i was doing lol
PaxPolaris
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sorry u r right. it's +3
NotTim
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ITs +3?
myininaya
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just making sure
when it gets late i tend to screw up on little things
myininaya
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like sometimes i don't know how to add or multiply sometimes
myininaya
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we were talking about this:
\[\left( u-2 \right)\left( u^2 - 7u +3\right) =0 \]
NotTim
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ok. that scared me for a sec.
FoolForMath
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yep myininaya I was talking about that :D
myininaya
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lol
myininaya
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so we know we have u=2
we also have
\[u=\frac{7 \pm \sqrt{49-4(3)}}{2}\]
don't forget to replace u with tan(x)
and solve for x
NotTim
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Finally, I'm done the division.
myininaya
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lol
myininaya
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good
myininaya
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you have
\[u-2=0 \text{ or } u^2-7u+3=0\]
myininaya
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we need to use the quadratic formula to solve the second equation
NotTim
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I hope I remember this for the test.
myininaya
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so this is trig class?
myininaya
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i could be wrong but this is too long to put on a test
myininaya
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there might be a similar problem he/she puts there
myininaya
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but shorter
NotTim
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Whoops. I didn't type that...Double post or something.
myininaya
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i could be wrong
LagrangeSon678
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sometimes teachers get problems from special books
myininaya
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I'm sleepy guys
I'm sure if you need more help NotTim (who we decided was not myininaya earlier ;)lol) , fool and pax will continue to help unless they are are going.
NotTim
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K. Thanks man.
myininaya
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Do I look like a man to you?
myininaya
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Please say no
myininaya
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lol
NotTim
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Are you?
NotTim
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I actually have bad vision and human identification problems.
NotTim
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When I'm walking around I will misidentify everyone as the same person.
myininaya
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Ok. Its cool. No worries.
NotTim
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any1 know what I should do at (m-2)(m^2-7m+3)?
NotTim
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For m^2-7m+3, do I have to do the quadratic formula?
NotTim
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I guess I already know what to do with m-2, so I'm fine there.
NotTim
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ERm, are you people just gunna lurk there? It's fine, just....
ktklown
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find all places that u can be 0, then, recalling that u = tanx, solve for x.
ktklown
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you already found one solution is u=2, you can find the others by using the quadratic formula on the quadratic part of your factoring.
NotTim
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Ok.
NotTim
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Hey, if one of my answers is 6.54, and I wanna get the other angle with it, what do i Do?
NotTim
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Hi?
NotTim
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Um, sorry if I sound abrupt, but could anyone please answer my most recent question?