So this has stumped me for over an hour...

[irishScott]

It's one part of a multi-part problem, so no you're not doing my homework for me.

How do you simplify sqrt(v^4 + 4(u^2)(v^2) + 4u^4) such that the square root goes away? According to the hints provided with the textbook, there's a way. That or I've worked the determinant that leads to this wrong several times in a row.

Edit: Problem solved, thanks guys.

 

[bignateyk]

whats on the other side of the equation?

 

[blackdogdeek]

sqrt(v^4 + 4(u^2)(v^2) + 4u^4) = sqrt((v^2 + 2u^2)^2) = +-(v^2 + 2u^2)

 

[Macamus Prime]

Epoxy grout is the top of the line and best choice for any tile application. It can be substituted for sanded or unsanded grout. It is more sturdy than both as well as being waterproof and stain resistant.

 

[Possessed Freak]

sqrt(v^4 + 4(u^2)(v^2) + 4u^4) = sqrt((v^2 + 2u^2)^2) = +-(v^2 + 2u^2)

Ninja edit, was about to correct you!

[edit]What's worse is I did the math to verify and man whoever put in u's and v's in one problem are bastards, mine all look the same!

 

[irishScott]

Nothing. It's the magnitude of two crossed vectors.

r1 = <2u, v, 0>
r2 = <0, u, v>

r1 x r2 = <v^2, -2uv, 2u^2>

 

[guyver01]

2u^2+v^2

WolframAlpha is your friend

 

[blackdogdeek]

Ninja edit, was about to correct you!

only ninja can detect ninja edit. i bow to you, fellow ninja.

 

[irishScott]

sqrt(v^4 + 4(u^2)(v^2) + 4u^4) = sqrt((v^2 + 2u^2)^2) = +-(v^2 + 2u^2)

Holy #$%^ I'm an idiot. I'll blame the all-nighter. Thanks!

 

[Texashiker]

Can you at least give us an example with a practical application?

 

[TecHNooB]

wolframalpha reduced that shit in a fraction of a second.

 

[HN]

I will perform the dance of my people

 

[irishScott]

wolframalpha reduced that shit in a fraction of a second.

Yeah, shoulda thought of that too. Ironically enough I've been using it to check my plots all night, morning and fucking afternoon.

 

[Ornery Platypus]

It's one part of a multi-part problem, so no you're not doing my homework for me.

How do you simplify sqrt(v^4 + 4(u^2)(v^2) + 4u^4) such that the square root goes away? According to the hints provided with the textbook, there's a way. That or I've worked the determinant that leads to this wrong several times in a row.

Your expression simplifies to v^2 + 2u^2.

In general,

(a+b)^2 = a^2 + 2ab + b^2

alternatively,

sqrt(a^2 + 2ab + b^2) = a + b

If we make the assumption that this is indeed a square of the sum of two real numbers, solve following for a and b:
a^2 = v^4
2*a*b = 4*u^2*v^2

you get

a = v^2 and b = 2*u^2

Since solution satisfies b^2 = 4*u^4, the square assumption is correct.

Probably not the "formal" way to do it...