help please :)

mattoram

New member
Joined
Sep 12, 2005
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19
1) square root 2x(square root 10-square root x) *I don't even know how to attempt this.

2) (3 square root x+1)^2
*my figuring: 3*3=9......square root of 9 is 3
answer ? 3x+1?????? Is this correct

3) square root 3xy^3 * square root 18x^3y
*my figuring: 3 * 18=54x^4y^4....square root is 9*6
square root of 9 is 3.....my answer is 3x^2y^2 square root 6? is that correct?

any help, I would appreciate, thanks!
 
What are the instructions? We don’t know what to do either!’

What are we to do with \(\displaystyle \sqrt {2x}\left( {\sqrt {10 - \sqrt x } } \right)}\) or is it \(\displaystyle \sqrt {2x} \left( {\sqrt {10} - \sqrt x } \right)\)?
 
In response,

it is the second one you wrote. Also, all it says is to simplify.
 
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\(\displaystyle \sqrt {2x} \left( {\sqrt {10} - \sqrt x } \right) = \sqrt {20x} - \sqrt {2x^2 }\).
 
1) square root 2x(square root 10-square root x)

If you can't use "CODE", then above is ok this way:

Simplify: sqrt(2x)[sqrt(10) - sqrt(x)]

Then just multiply the damn thing:
sqrt(2x) * sqrt(10) = sqrt(20x)
sqrt(2x) * sqrt(x) = sqrt(2x^2)
SO:
sqrt(20x) - sqrt(2x^2)
can be simplified further to:
2sqrt(5x) - xsqrt(2)
 
Hello, mattoram!

You should use parentheses to indicate "how far" the square root goes
. . otherwise, you're making us guess the problems.

1) square root 2x(square root 10-square root x)
If (big IF) the problem is: .\(\displaystyle \sqrt{2x}\cdot(\sqrt{10}\,-\,\sqrt{x})\)

. . we have: .\(\displaystyle \sqrt{2x}\cdot\sqrt{10}\,-\,\sqrt{2x}\cdot\sqrt{x}\;=\;\sqrt{20x}\,-\,\sqrt{2x^2}\)

Simplify: .\(\displaystyle \sqrt{4\cdot5x}\,-\,\sqrt{x^2\cdot2}\;=\;\sqrt{4}\cdot\sqrt{5x}\,-\,\sqrt{x^2}\cdot\sqrt{2}\;=\;2\sqrt{5x}\,-\,x\sqrt{2}\)

2) (3 square root x+1)^2
We have: .\(\displaystyle (3\sqrt{x+1})^2\;=\;3^2\cdot(\sqrt{x+1})^2\;=\;9(x+1)\)

3) square root 3xy^3 * square root 18x^3y
We have: . \(\displaystyle \sqrt{3xy^3}\cdot\sqrt{18x^3y}\)

Then: . \(\displaystyle \sqrt{54x^4y^4}\;=\;\sqrt{9\cdot6\cdot x^4\cdot y^4}\;=\;\sqrt{9}\cdot\sqrt{6}\cdot\sqrt{x^4}\cdot\sqrt{y^4}\;=\;3\cdot\sqrt{6}\cdot x^2\cdot y^2\)

Your reasoning and your answer are correct!
 
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