verify answers

[mattoram]

1) -4x sqrt xy-x sqrt xy

(-4x-1x) =-5x sqrt xy...is that correct?


2) 3 sqrt 12-2 sqrt 108

not sure how to even do this


3) sqrt x^2=9

sqrt 3...so sqrt x+ 3, is that correct???

Thank you for any assistance

 

[Denis]

Lookit Matt, you were told before to type sqrt x this way: sqrt(x)

Like, we can't tell with sqrt xy if you mean sqrt(xy) or sqrt(x)*y

Repost properly, else too bad, so sad :evil:

 

[mattoram]

in response

sorry,

1) sqrt (x^2+9)
I came up with sqrt (x+3), is that correct, and if not where did I go wrong.

2) 3 sqrt (12)-2 sqrt (108)
I came up with -6 sqrt (3), is that correct?

3) -4x sqrt (xy)-x sqrt (xy)
I came up with -5x sqrt (xy), is this correct?


Thank you again....BTW, I am Kristin, Matts wife!

 

[soroban]

Hello, Kristin!

I'm still guessing what the problems look like . . .

\(\displaystyle 1)\;\;-4x\sqrt{xy}\,-\,x\sqrt{xy}\)

\(\displaystyle \text{Answer: }-5x\sqrt{xy}\) . . . is that correct? . [ . . . yes!

\(\displaystyle 2)\;\;3\sqrt{12}\,-\,2\sqrt{108}\)

Remember how to "simplify" square roots?

. . Recall that: .\(\displaystyle \sqrt{12}\,=\,\sqrt{4\cdot3}\,=\,\sqrt{4}\cdot\sqrt{3}\,=\,2\sqrt{3}\)

. . and that: .\(\displaystyle \sqrt{108}\,=\,\sqrt{36\cdot3}\,=\,\sqrt{36}\cdot\sqrt{3}\,=\,6\sqrt{3}\)


We have: .\(\displaystyle 3\cdot\sqrt{12}\:-\:2\cdot\sqrt{108}\)
. . . . . . . . . . . \(\displaystyle \downarrow\;\;\;\;\;\;\downarrow\)
. . . . . . . . . \(\displaystyle 3\cdot\overbrace{2\sqrt{3}} \:-\:2\cdot\overbrace{6\sqrt{3}}\;=\;6\sqrt{3}\,-\,12\sqrt{3}\;=\;-6\sqrt{3}\)

\(\displaystyle 3)\;\;\sqrt{x^2}\,=\,9\)

Square both sides: .\(\displaystyle (\sqrt{x^2})^2\:=\:9^2\)

and we have: .\(\displaystyle x^2\:=\:81\)

Therefore: .\(\displaystyle x\:=\:\pm9\)

 

[mattoram]

Soroban

thanks for all your help tonight, I really appreciate everything!!!!

Thanks again

Kristin