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Cant get my head around this...

Name: Anonymous 2007-08-29 21:08 ID:xqJYApfe

It's probably something simple that I'm missing, but anyway...

sqrt(u - 3) = u - 5

the answer is supposedly u = 7
however, it's important to know why, so... well, why?

Name: Anonymous 2007-08-29 21:12 ID:FuBP5vb7

This is an even stupider troll than your one about quadratics. In fact, getting the solution if you don't see it straight away, involved a quadratic.

Name: Anonymous 2007-08-29 21:19 ID:xqJYApfe

>>2

Sigh... Look, flame me for being stupid all you want but I still need an answer. Obviously, you do not wish to help so I'll just have to hope someone else does.

Name: Anonymous 2007-08-29 21:21 ID:FuBP5vb7

Do you know how to solve a quadratic equation?

Name: Anonymous 2007-08-29 21:26 ID:syyzMCKW

sqrt(u - 3) = u - 5
u - 3 = u^2 -10u + 25
0 = u^2 - 11u + 28
0 = (u -4)(u - 7)
u = 4 and u = 7

Plug them em into the original equation to check for extraneous roots.  With u = 4 you get 1 = -1, so you can throw that one out.  With u = 7 you get 2 = 2.  So u = 7.

Name: Anonymous 2007-08-29 21:26 ID:xqJYApfe

I'll say no, since I'm not sure what a quadratic equation can manifest itself as.

Name: Anonymous 2007-08-29 21:28 ID:FuBP5vb7

>>6
How old are you? You're ACTUALLY not a troll?

Name: Anonymous 2007-08-29 21:28 ID:xqJYApfe

>>5

Where does that -10u come from?

Name: Anonymous 2007-08-29 21:31 ID:xqJYApfe

>>7

No, I am not trolling. I am trying to learn, but the material the school supplied me with is little more than numbers. No real explanation to why the formulas work the way they do.

Name: Anonymous 2007-08-29 21:42 ID:xqJYApfe

>>5

Thank you.

Name: 4tran 2007-08-30 1:17 ID:o61XZG2M

>>8
FOIL - It comes from the cross terms (-5*u and -5*u).  Study the distributive and associative properties of addition and multiplication.

Essentially, getting to the step with the -10u term requires squaring both sides to eliminate the square root.

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