radicals and roots: multiply, simplify; solve equations

[will7675]

Am I on the right track or totally clueless!?!?

1. Multiply and simplify: (2sqrt3-4sqrt5)(sqrt3-2sqrt5)

My attempt:
I multiplied:

2(sqrt)3x3-2(sqrt)3-5 - 4(sqrt)5x3-4x negative 2(sqrt)5x5

I further multiplied and simplified to an answer of 4(sqrt)25 and then 4(sqrt)5 for a final answer?

2. Multiply and simplify: sqrt6y^2 times sqrt3y^3)

I multiplied 6y^2 and 3y^3 and ended up with (sqrt)18y^5 for final answer?

3. Solve a 4+ in front of sqrt and x-3 inside sqrt = 11
4+sqrt(x-3) =11

4. Solve (sqrt)with 2x-3 inside minus 2 = (sqrt)with x-5 inside

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, but I have made attempts and keep getting stuck! help!

 

[tkhunny]

Re: radicals and roots!?!?!?! help!

Am I on the right track or totally clueless!?!?

1. multiply and simplify:
(2sqrt3-4sqrt5)(sqrt3-2sqrt5)

my attempt-
i multiplied:
2(sqrt)3x3-2(sqrt)3-5 - 4(sqrt)5x3-4x negative 2(sqrt)5x5
i further multiplied and simplified to an answer of 4(sqrt)25
and then 4(sqrt)5 for a final answer?

I can agree to neither classification. You MUST practice better notation. In your work shown above, it is very difficult to see what is inside a square root and what is not.

\(\displaystyle (2\sqrt{3}-4\sqrt{5})*(\sqrt{3}-2\sqrt{5})\)

This is a binomial and another binomial. You should get four terms.

\(\displaystyle 2\sqrt{3}*sqrt{3} - 4\sqrt{3}\sqrt{5} - 4\sqrt{5}sqrt{3} + 8\sqrt{5}\sqrt{5}\)

You don't quite have that. Look at it a little harder.

\(\displaystyle 2*3 - 4\sqrt{15} - 4\sqrt{15} + 8*5\)

\(\displaystyle 6 - 8\sqrt{15} + 40\)

\(\displaystyle 46 - 8\sqrt{15}\)

 

[stapel]

Please clarify your post. For instance, what do you mean by "2(sqrt)3x3"? Is this "2sqrt[3]x³"? Or "2sqrt[3x]³"? Or "2sqrt[3x³"? Or something else? What do you mean by "4(sqrt)25"? Is this "4sqrt[25] = 4(5) = 20"? Or "the fourth root of 25"? Or something else?

Thank you.

Eliz.

 

[will7675]

ahhh

I just don't know how to write a square root symbol on here!
so when I put (sqrt) that stands for the symbol.
for example:
2(sqrt)16
would be a 2 right in front of the symbol, and the 16 inside...

 

[pka]

Re: ahhh

I just don't know how to write a square root symbol on here!
2(sqrt)16

Do you see the TeX tab above?
Click it. The type \2 sqrt {16} now click close tags. \(\displaystyle \ 2 sqrt {16}\)

 

[stapel]

2(sqrt)16 would be a 2 right in front of the symbol, and the 16 inside.

Unfortunately, that doesn't clarify your meaning when there is more than one character following the "(sqrt)". This is why parentheses after the "sqrt", surrounding the entire argument, would have been helpful.

Please use grouping symbols to clarify your meaning, or use the LaTeX to typeset. Thank you.

Eliz.

 

[tkhunny]

...and please don't use 'x' for multiplication.

 

[will7675]

REVISED-Roots and Radicals

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 6y^2 and 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:
(X-1)^2=(\(\displaystyle \sqrt{X+5}\))^2
X^2-1=X+5
I subtracted X from each side and then:
X^2-X-1=5
I then subtracted 5 from each side and then:
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!


_________________
Lauren Williams, the math idiot!

 

[tkhunny]

It helps if you keep the Domain in mind.

\(\displaystyle 4 + \sqrt{x-3} = 11\)

You are right.

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

Here's where the trouble starts. If we are to stay in Real Numbers, that square root must exist, right? So \(\displaystyle x - 3 \ge 0\) or \(\displaystyle x \ge 3\). Keeping this in mind:

x - 3 = 49

\(\displaystyle x = 52 \ge 3\) so we should be good.

 

[tkhunny]

It's really nice that you figured out how to use some LaTeX. What is not nice is that you posted the same problem a couple fo times, causing multiple volunteers to spend time on the same material. This is not very polite. Please don't do it again.

 

[stapel]

1) sqrt[64] does not equal sqrt[8]; sqrt[64] = 8.

sqrt[x⁴] = sqrt[(x²)²] = x²; sqrt[x⁴] does not equal x⁴.

2) Try factoring: sqrt[18y⁵] = sqrt[(3)(3)(2)(y)(y)(y)(y)(y)] = sqrt[(3)(3)] sqrt[2] sqrt[yy] sqrt[yy] sqrt[y]. Simplify.

3) (x - 1)² = (x - 1)(x - 1), not x² - 1². Expand properly, and then solve the resulting quadratic equation.

4) Subtracting 4 from either side is the best way to start. Please reply showing what you have done.

5) Square both sides. Then solve the resulting linear equation.

Eliz.