Yay! Another Math Problem!

[SkyHog]

Ok - so I have a math problem that I don't know if I have completely solved or not yet. Seems simple enough:


sqrt(27x3y5z) sqrt(243x5yz3)

I got an answer of

(3xy2 sqrt[3xyz])(9x2z sqrt[3xyz])

Seems to me I can multiply the two together, and if I do, I'll get

81x4y3z2

Is that right?

 

[kath]

sqrt(27x^3 y^5 z) sqrt(243x^5 y z^3)

I got an answer of

(3xy^2 sqrt[3xyz])(9x^2 sqrt[3xyz])

You lost a "z" in the second term...

81x^4 y^3 z^2

Is that right?

... but it must've just been a typo cuz this looks OK to me...
You can also do it by multiplying all the stuff under both square roots together.
sqrt(A) * sqrt(B ) = sqrt(A*B )

--Kath

 

[ScottM]

Since both terms has the same exponetial (1/2), you can mulitple all the terms togather and then take the root.

It would fist simplify to (6561x^8y^6z^4)^1/2

Take the root of each term and you would get 81x^4y^3z^2

So either all three of us right or all three of us do not know what we are doing ;-)

 

[Tim]

2+2=

 

[EdFred]

Since both terms has the same exponetial (1/2), you can mulitple all the terms togather and then take the root.

It would fist simplify to (6561x^8y^6z^4)^1/2

Take the root of each term and you would get 81x^4y^3z^2

So either all three of us right or all three of us do not know what we are doing ;-)

And I did it completely different. I multiplied and added the exponents.

(27x3y5 z)1/2 * (243x5 y z3)1/2
(3*31/2)(x3/2)(y5/2)(z1/2) * (9*31/2)(x5/2)(y1/2)(z3/2)

(81)(x8/2)(y6/2)(z4/2)

81x4y3z2


Edit: Chuckrocks