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Envii
 2 years ago
Please help.
Envii
 2 years ago
Please help.

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Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0what I think you should do is first multiply the (x1) by (x^24)

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0then square both sides, and then collect the like terms

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0How do I multiply (x1) by (x^24)?

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0Yay. :D Then how do I combine like terms?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0you have to square both sides first, then collect the like terms, then isolate x

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0there might be an easier way, but i'm not sure

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0to get rid of the sqrt sign, you'll need to square both sides: \[(x^3x^24x+4)^2=x^42x^3\]

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0Wait, wouldn't it be (x^42x^3)^2 on that side?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0no, because a sqrt and a squared cancels out. \[(\sqrt{a})^2=a\]because\[\sqrt{a}=a^\frac{ 1 }{ 2 }\]and\[(\sqrt{a})^2=a^{\frac{ 1 }{ 2 }*2}\]which is\[a^1=a\]

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0Do I have to do the (x^3x^24x+4)^2 first?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0yes, now it's a very, very, very long process. Are you allowed to use a calculator for this part?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0so type \[(x^3 −x^2 −4x+4)^2\] into the calculator to get an answer

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0How do I do that if I'm using the google calculator?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0I've never used a google calc. let me see

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0The calculator I normally use is broken, sorry. :(

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0try use this site, http://www.wolframalpha.com/input/?i=%28x%5E3x%5E24x%2B4%29%5E2 I've already typed in the equation, look for the words expanded form

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0Alright I got it. What next?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0subtract x^4 from both sides and add 2x^3 to get one side equal to 0

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0Alright, now I can do x^2x^2 to get , correct?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0but, yes, use that idea to get all the terms on one side

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0x^6x^4=x^2 and 2x^5+2x^3=x^2, right?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0no, you can't do that. they're not like terms. like terms are: x and 4x x^2 and 7x^2 not x^6 and x^4

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0so what you should have gotten is: \[x^62x^57x^4+16x^3+8x^232x+16=x^4+2x^3\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Mind if I make a suggestion that will keep things simpler?

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0@whpalmer4 of course @Chelsea04 Wouldn't it be =x^42x^3?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Way back at the beginning, notice that \(x^24\) is the difference of two squares, and factors to \((x+2)(x2)\) We can write the problem as \[(x1) = \frac{\sqrt{x^42x^3}}{(x+2)(x2)} = \frac{\sqrt{x^3(x2)}}{(x+2)(x2)} = \frac{x\sqrt{x(x2)}}{(x+2)(x2)}\]

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0yea, but how does that help? I noticed it too, but didn't know what to do with it

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Now if we keep everything like that, and multiply both sides by the denominator, we get \[(x1)(x+2)(x2) = x\sqrt{x(x2)}\]and after we square both sides, we just tack on exponents and get rid of the radical sign: \[(x1)^2(x+2)^2(x2)^2 = x^2x(x2)\]and now we can cancel out a common term from both sides\[(x1)^2(x+2)^2(x2) = x^3\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2we marvel at how beautiful it is? :)

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0wouldn't you have to eventually expand it to solve for x? @whpalmer4 that's funny, lol :P

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Are we 100% certain this was the problem to be solved, and we're really supposed to find the roots? We've got a 5th order polynomial here...

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2If we expand it out, we get \[x^58x^3+2x^2+12x8=0\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Can't factor it, and there's no exact general solution for 5th order or higher polynomials, so Mathematica just finds numeric values for this one: \[\{\{x\to 2.53629\},\{x\to 1.59124\},\{x\to 0.8782\, 0.254473 i\},\]\[\{x\to 0.8782\, +0.254473 i\},\{x\to 2.37113\}\}\]

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0I am just so confused right now.

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0what exactly does the question ask?

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0I guess you could've just typed the whole thing into the calculator

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Now here's a thought, what if the problem is really \[x1 = \sqrt{\frac{x^42x^3}{x^24} }\]

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0that'd make more sense

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Then we have \[(x1)^2 = \frac{x^3(x2)}{(x2)(x+2)}\]

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0is that the question @Envii ?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Cancel out (x2) and multiply both sides by (x+2) gives us \[(x1)^2(x+2) = x^3\]Expand the left side \[(x^22x+1)(x+2) = x^3\]Expand again\[x^3+2x^22x^24x+x+2 = x^3\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2@Envil, why don't you try solving that?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2This version is MUCH easier :)

Chelsea04
 2 years ago
Best ResponseYou've already chosen the best response.0YUP!!! MUCH MUCH EASIER!!!

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2I did like the look of the other one in factored form, though :)

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2@Envil have you simplified and solved that equation yet?

Envii
 2 years ago
Best ResponseYou've already chosen the best response.0Working on it. Sorry, I suck at math. xD

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Here's the equation again: \[x^3+2x^22x^24x+x+2 = x^3\]What do you have left after you collect like terms?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Right. And what is 4x + x?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Come on, you're really close to the answer!

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2\[3x+2 = 0\]Add 3x to both sides, and then divide both sides by 3 to get x = ????

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2I'd keep it in exact form: x = 2/3

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2But now we need to check it, because we squared both sides and that sometimes introduces fictitious solutions. x1 = 2/3 1 = 1/3 Uh oh, that's trouble, because we've got a square root on the other side of the equals sign, which is going to give us a positive number. If we expand it out, \[\frac{1}3 = \sqrt{\frac{(\frac{2}{3})^42(\frac{2}{3})^3}{(\frac{2}{3})^2 4}} = \sqrt{\frac{(\frac{16}{81})2(\frac{8}{27})}{(\frac{4}{9}) 4}} = \sqrt{\frac{(\frac{16}{81})(\frac{48}{81})}{(\frac{4}{9}) (\frac{36}{9})}} = \sqrt{\frac{\frac{32}{81}}{\frac{32}{9}}}=\sqrt{\frac{1}9}\] So it isn't a solution after all.

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2Well, did I lose you in the painful fraction arithmetic, or in the decision that the solution wasn't valid?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.2After we plugged our "solution" back into the original equation and ground through the arithmetic, we ended up with this: \[\frac{1}{3} = \sqrt{\frac{1}9}\]Unfortunately, that isn't true. Therefore, our solution isn't valid. There's a similar problem that does have working solutions: \[1x = \sqrt{\frac{x^42x^3}{x^24}}\]We proceed the same way: \[(1x)^2(x+2)(x2) = x^3(x2)\]Cancel common factors\[(1x)^2(x+2)=x^3\]Expand\[(12x+x^2)(x+2)=x^3\]Expand again\[x+22x^24x+x^3+2x^2=x^3\]Collect like terms\[3x+2=0\]\[x=\frac{2}3\] If we plug that back into its original equation: \[1\frac{2}{3} = \sqrt{\frac{(\frac{2}{3})^42(\frac{2}{3})^3}{(\frac{2}{3})^24}} \]\[\frac{1}3 = \sqrt{\frac{1}{9}}\]and that is true...I wonder if that was the correct problem!
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