radicals and roots: multiply, simplify; solve equation

[will7675]

OKAY! NOW! I have figured the squareroot symbol stuff out!
I hope this clears things up for everyone (and keeps you from getting
annoyed with my confusing typing!!)
If anyone could tell me if I am on the right track or totally messing this up
I would greatly appreciate it!
Also the last couple of problems I have no idea where to start, so if some one could tell me what to do to get started, or the first step or two(asap!) I would REALLY REALLY appreciate it!

Okay, here they are-

1.Multiply and simplify:
$\displaystyle \sqrt{8X^2} * \sqrt{8X^3}$
My attempt:
$\displaystyle \sqrt{64X^5}$
$\displaystyle \sqrt{8X^5}$
$\displaystyle \sqrt{8X^4X}$
Final answer=\(\displaystyle \8X^4 sqrt{X}\)


2. Multiply and simplify: $\displaystyle \sqrt{6y^2} * \sqrt{3y^3}$
I multiplied \(\displaystyle \6y^2\) and \(\displaystyle \3y^3\) and ended up with $\displaystyle \sqrt{18y^5}$ for final answer? Is this right?


3. Solve: X-1= $\displaystyle \sqrt{X+5}$
My notes said to square everything:
$\displaystyle (X-1)^2$=($\displaystyle \sqrt{X+5}$)^2
$\displaystyle X^2-1=X+5$
I subtracted X from each side and then:
$\displaystyle X^2-X-1=5$
I then subtracted 5 from each side and then:
$\displaystyle X^2-X-6=0$
I found the two numbers that would multiply to equal
-6, and add to equal X(or 1)
(X-3)(X+2)
Set to equal zero=
X=3 or X=-2



4. Solve:
$\displaystyle 4+ \sqrt{X-3}$=11

Started out trying to subtract 4 from both sides like my lecture notes
were showing me,but could not come up with a good answer, if anyone could give me a suggestion as of how to get started that would be great!



5. Solve: $\displaystyle \sqrt{2-X} = \sqrt{3X-7}$


I have no idea where to start on these last two(#3 and #4). From my instructor's lecture notes, I know I am to end up with 2 answers that I have to set to equal zero. I understand that part, just I am having a hard time getting started on the first part of the problems!, I have made attempts and keep getting stuck! help!

 

[galactus]

Re: ASAP! ROOTS AND RADICALS! I NEED HELP ASAP!

4. Solve:
$\displaystyle 4+ \sqrt{X-3}$=11

Started out trying to subtract 4 from both sides like my lecture notes
were showing me,but could not come up with a good answer, if anyone could give me a suggestion as of how to get started that would be great!

You know to square to eliminate the radical. This is no different.

Subtract 4 from both sides:

\(\displaystyle \L\\\sqrt{x-3}=7\)

Square both sides:

\(\displaystyle \L\\x-3=49\)

That's all there is to it.

 

[Mrspi]

Re: ASAP! ROOTS AND RADICALS! I NEED HELP ASAP!

OKAY! NOW! I have figured the squareroot symbol stuff out!
I hope this clears things up for everyone (and keeps you from getting
annoyed with my confusing typing!!)
If anyone could tell me if I am on the right track or totally messing this up
I would greatly appreciate it!
Also the last couple of problems I have no idea where to start, so if some one could tell me what to do to get started, or the first step or two(asap!) I would REALLY REALLY appreciate it!

Okay, here they are-

1.Multiply and simplify:
$\displaystyle \sqrt{8X^2} * \sqrt{8X^3}$
My attempt:
$\displaystyle \sqrt{64X^5}$ OK up to here
$\displaystyle \sqrt{8X^5}$
$\displaystyle \sqrt{8X^4X}$
Final answer=\(\displaystyle \8X^4 sqrt{X}\)

Continue from that point where you were "correct." Separate into the product of radicals. Note that I've put all "perfect squares" under separate radical signs.

$\displaystyle \sqrt{64} * \sqrt{x^4} * \sqrt{x}$

$\displaystyle 8 * x^2 * \sqrt{x}$

 

[stapel]

Multi-post.