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gjhfdfgBest ResponseYou've already chosen the best response.0
,dw:1360809292870:dw
 one year ago

ZeldaBest ResponseYou've already chosen the best response.0
So do this: y = 12/x Now plug this in the second equation.
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
How do I plug this in? Like, x^2 + 12/x = 40?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[x^2 + y^2=40\] \[x^2+(12/x)^2 = 40\] \[x^2 + \frac{144}{x^2} = 40\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Then multiply everything by \(x^2\) to get rid of the fraction and solve.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
When you multiply through by \(x^2\) you get \[x^4 + 144 = 40x^2\]or\[x^4 40x^2+144 =0\]How to solve that? Make a substitution \(u = x^2\) to turn it into a quadratic: \[u^240u+144 = 0\] Find the solutions for \(u\), then replace \(u\) with \(x^2\).
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Then solve those for \(x\). You'll end up with 4 solutions; use \(xy=12\) to get the accompanying values of \(y\).
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Why don't you tell me what part of that makes sense to you, and what part doesn't?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
You can also solve \[x^440x^2+144=0\]by factoring. You'll have \[(x^2a)(x^2b)=0\] where the values of a and b add to 40 and multiply to 144. 4 and 36 fit that description, so \[x^440x^2+144 = (x^236)(x^24) = 0\]Now find the values of \(x\) so that \[x^2=36\]\[x^2=4\](and don't forget about the negative values that solve those equations!)
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
First confusion is multiplying by x^2, you've shown the steps but I don't get how everything came out to be
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Everything is a big ball of confusion, I don't get why anything of this not coming through to me
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Okay, I was just trying to get rid of the fraction. You can multiply both sides of the equation by x^2 without damaging the equality, and it eliminates the fraction.
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
^ that doesnt make any sense. :/ I get (12/x)^2 = 144, except what happened to the x? Wouldnt it of made it 144x?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Look at it a different way: \[x^2 + \frac{144}{x^2} = 40\]We want to make a common denominator of \(x^2\) so we can add the two terms on the left. So, we can multiply the first term by \(\frac{x^2}{x^2}\) to get \[\frac{x^2}{1}*\frac{x^2}{x^2} + \frac{144}{x^2} = 40\]or\[\frac{x^4+144}{x^2} = 40\]But to solve that, we'll end up multiplying both sides by \(x^2\) giving \[x^4+144 = 40x^2\]and so I just saved a step multiplying by \(x^2\) instead of multiplying the first term by 1 (remember, \(x^2/x^2 = 1\)) and then the entire equation by \(x^2\). \[(\frac{12}{x})^2 = \frac{12}{x}*\frac{12}{x} = \frac{12*12}{x*x} = \frac{144}{x^2}\]
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Ok, I think I'm with you so far...
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Did you understand how I solved the equation by factoring?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Kinda, except how the 36 & 4 came in
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Ah nevermind, 4 * 36 = 144, that was the factored out version of 144 right?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Well, what are the factors of 144? 1*144 2*72 3*48 4*36 6*24 8*18 12*12
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Got it but why 4 * 36?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Because 4 and 36 add to 40 (the coefficient of the middle term) and multiply to 144 (the coefficient of the final term). \[(x^24)(x^236) = x^4 36x^2 4x^2+144 = x^4  40x^2 +144\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Remember, when we multiply \[(xa)(xb) = x^2axbx+ab = x^2(a+b)x+ab\] Compare that with our equation (pretend it in x instead of x^2 for the moment) \[x^240x+144\] We can see that (a+b) = 40, and ab = 144. 36 and 4 are the two numbers that satisfy those equations.
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Ok, so a=36, b= 4, a*b= 144, is what your saying right?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Yes, that is correct. So that leaves us with solving \[(x^236)(x^24) = 0\]Clearly, for that product to be 0, either component equals 0, so \[x^236=0\]\[x^24=0\]will give us the answers. How would you solve them?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
If its zero then wouldnt it make all the answers 0?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
No, because the value of x that makes x^236 = 0 isn't the same value as the value of x that makes x^24 = 0, is it?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
To find all the solutions, we have to find all the different values of x that can make that product = 0. There are 4 of them, because we have an x^4 in the equation.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Looking at it a different way, 0*0 is not the only way to get 0 in a multiplication. 1*0 = 0, 2*0 = 0, 13423523493459823487234.6734234*0 = 0, etc.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
I'm not sure I've ever actually multiplied 13423523493459823487234.6734234*0 before now, but I'm still pretty sure it equals 0 :)
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Please, no laughing, this is serious stuff ;)
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Oh no I'm not laughing lol. So I dont use the x^2 to find the answers?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Yeah, you do use x^2 to find the answers. I'll do two, you can do two. \[x^236=0\]\[x^2=36\]Take square root of both sides \[x=\pm \sqrt{36} = \pm 6\] Remember that \(6*6 =36\) so we have to include it as one of the solutions. Now we'll find the values of \(y\) that go with those values of \(x\):\[xy=12\]\[y=\frac{12}x\]\[y=\frac{12}6 = 2\]\[y=\frac{12}{6}=2\]My two solutions are \[(6,2),(6,2)\] What are your solutions for the other half? (\(x^24=0\))
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Ah, sorry, my solutions are (6,2) and (6,2)
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Would it be (2,3) & (2, 3) ?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
I agree with the x values, but I'm not so sure about the y values. How did you get them?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Why 4? 4 isn't the value you found for x...
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Should I of done it by 2?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
So then it should be (2,6) & (2, 6)
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Right. Notice the symmetry with my solutions? There's a reason for that! Here's a picture:
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
So are you comfortable with your understanding of what we did? Would you like another problem to try?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
I think I'm still shaky with the substitution method, I could use another problem :P
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Just a minute, I want to make sure I don't give you one that is too awkward! Okay, how about this? You'll only have 2 solutions for this one: \[(x5)^2+4y^2=100, x+y=10\]
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Okay, I'll give this one a try,
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Ok this looks more complicated. o_o so, (x5) ^2 + 4y^2 = 100, x + y = 10, I take the square root of 5 which is 25, makes the problem (x25) + 4y^2 = 100, Am I going in the right direction so far?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Square root of 5 isn't 25, 5 is the square root of 25. I suggest that you first solve x+y=10 to give you x = 10y. Now plug that in to the other formula in place of x...
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
It might look more complicated, but I think it is a bit easier...
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
"now plug that in ..." is the substitution process we're practicing...
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
How would I plug that in the x's place? (105) ^2 + 4y^2 = 100?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[(x5)^2+4y^2=100\]\[x=10y\]\[((10y)5)^2+4y^2=100\]\[(5y)^2+4y^2=100\]Now expand \[(5y)^2\]and continue on to the solution
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[x+y= 10 \implies x = 10y\]Right?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
(the arrow means "implies")
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
So we just take our equation and write (10y) wherever we see x. \[(x5)^2+4y^2=100\]\[((10y)5)^2 + 4y^2=100\]\[(10y5)^2 + 4y^2=100\]\[(5y)^2+4y^2=100\]If we multiply out (expand) (5y)^2 \[(5y)^2=(5y)(5y) = 5*55y5y+(y)(y) = 2510y+y^2\]so our equation goes from \[(5y)^2+4y^2=100\]to\[(2510y+y^2)+4y^2=100\]and after rearranging\[5y^210y+25=100\]What would you do to solve that bad boy?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
yes, you would solve it? I'm happy to hear that :)
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
I have to go out for a while, so I'll post the rest of the steps in my next post, don't look ahead until you've tried to solve it, okay?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Yes as in reply to, " x+y=10⟹x=10−y Right?" lol
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Alrighty, take your time. It'll take a bit to get use to these problems
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Everything makes sense up to step 4 where the 10 was taken out, How/why was it taken out?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
10y5 = 105y = 5y, no?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
*face palm* Got it lol
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
And probably a little extra care needs to be taken when you multiply things like (5y)(5y) since the order is different than you are used to...
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
no...take another look.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
if you prefer, it's the same as \((1)(y5)(1)(y5)\) because \(5y = 1(y5)\) The \((1)(1) = 1\), and \((y5)(y5) = y^25y5y+25 = y^210y+25\)
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
5*5= 25, 2 's = + 2 y's = y^2 I thought?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[(ab)(ab) = a(ab)  b(ab) = a^2  ab  ba + b^2 = a^2 2ab + b^2\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
You are thinking of \[(a+b)(ab) = a^2ba + ba  b^2 = a^2b^2\]I think...
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Who knows what I was thinking...
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Today's unrelated, unsolicited life tip: when walking home in the dark, do not under any circumstances startle a skunk by nearly stepping on it. Guess what happened to me last night? :)
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Yikes & yuck aha. Hopefully you had some tomato juice on hand
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Hydrogen peroxide and baking soda is a better combo. Fortunately it was cold enough that I wasn't walking around in short sleeves, and just my clothing got zapped. Unfortunately, I seem to have brushed against many surfaces in the house on the way :( If I don't respond, I'm probably busy cleaning something!
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Here are the final steps, don't read on until you're ready to check your work or completely stuck: How to solve \[5y^210y+25=100\] Let's move the 100 onto the same side as everything else: \[5y^210y+25100=0\]\[5y^210y75=0\]Now let's divide out the common factor of 5:\[y^22y15=0\]Which two numbers multiply to 15 and add to 2? 5 and 3 do, so we can factor that as\[(y5)(y+3)=0\]Now solve each of those for 0\[y5=0, y=5\]\[y+3=0, y=3\]And finally use our substitution equation to find the corresponding values of x:\[x=10y\]\[x=105 = 5\]\[x=10(3)=13\]So our two solutions are (5,5) and (13,3).
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Hah no worries, its only a practice problem I'm in no rush. :) I've had my share of ' skunk perfume' when I was younger. Had a dog who loved to catch them & bring them to me. Unfortunately(the odor), it was still alive & wanted to share its odor. I'm off to go grab a bite to eat, I'll annoy you with my questions later. :P
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
^thats an easier way way to understand it, having it step by step by step
 one year ago
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