GoldenLeaf

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Nombre de usuarioGoldenLeaf
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+2
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0
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GoldenLeaf  29 abr 2015
 #2
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0
12 may 2014
 #4
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+5

http://www.piday.org/million/

 

That's nowhere even close

12 may 2014
 #1
avatar+1006 
+3

$${\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1\,600}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{40\,000}}$$

 

Solve for each individual variable set first. Since 'y' is first, I will do that.

 

25y^2 = 40000

 

Divide by 25 on both sides.

 

y^2 = 1600

y = 40

 

Now, solve for 'x'

 

-1600x^2 = 40000

 

Divide by -1600

 

x^2 = -25

x = +- 5i (positive or negative)

 

I am going to try something else, so hang on a second; I'm not necessarily sure if this is what you meant, but this is solving for each individual variable as itself.

 

NUMBER 2

 

So first divide the whole equation by 25.

 

y^2 - 64x^2 = 1600

 

Add 64x^2 to both sides.

 

y^2 = 64x^2 + 1600

 

Take the root.

 

y = 8x + 40

 

Now, plug in (8x + 40) for 'y'.

 

25(8x + 40)^2 - 1600x^2 = 40000

 

Even the whole thing out now. This means taking the '^2' of (8x + 40) and distributing it to said term.

 

25 * 64x^2 + 1600 - 1600x^2 = 40000

 

Add [-1600x^2] to both sides.

 

25 * (64x^2 + 1600) = 1600x^2 + 40000

 

Take the square root of the whole thing.

 

$${\sqrt{{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{64}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,600}} = {\mathtt{1\,600}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40\,000}}}}$$

 

This works out unbelieveably nicely.

 

5 * (8x + 40) = 40x + 200

 

Distribute the 5 to the (8x + 40)

 

40x + 200 = 40x + 200

 

Interestingly enough, the whole equations cancels itself out.

 

Infinite solutions.

12 may 2014