Problems are attached

Chester

2. 1. $\frac{b}{3}=\frac{-3}{4}$ therefore $4b=-9$ then $b=-\frac{9}{4}$

3. These should be explained to you better, so here I go:

If you have an equation like: $\frac{a}{b}=\frac{x}{y}$ than you can use the cross-product rule to say: $ay=bx$

Radicals: Just something most people tend to not know, $\sqrt[x]{a}=(a)^{\frac{1}{x}}$

therefore when you see something like $\sqrt[3]{5b^9}$ you can say it equals $(5b^9)^{\frac{1}{3}}$ and then use the rule of exponents to say: $5^{\frac{1}{3}}\cdot b^{\frac{9}{3}}$ which simplifies to $b^3\sqrt[3]{5}$

for some reason I really don't feel like answer most of your questions now, so I'll let someone else do that.

4. ## Thanks

Quick,
Thank you very much for the explanation. Any help is very much appreciated.

Chester

5. ## Question

Would number one be

-9/4

Chester

6. Originally Posted by Chester
Would number one be

-9/4

Chester
Yes, thank you for pointing that out (I've changed my last post)

7. btw. the roots of $\sqrt{m^6}$ are $\pm m^3$

$\left(3-2\sqrt7\right)\left(3+2\sqrt7\right)$

Do you know how to use foil?

$F:\;3\times3=9$

$O:\;3\times2\sqrt7=6\sqrt{7}$

$I:\;-2\sqrt7\times3=-6\sqrt{7}$

$L:\;-2\sqrt7\times2\sqrt7=-4\sqrt{7^2}=-4\times7=-28$

$F+O+I+L$

$9\boxed{-6\sqrt{7}+6\sqrt{7}}-28$notice the things in the box cancel out therefore,

$9-28=-19$ voila!

8. ## Thanks again

I do know how to use FOIL. I have actually figured out a good deal of these sitting here for the last 12 hours.LOL I used the (a+b)(a-b)=(a^2-b^2). = 3^2-28= -19. I am still working on the last problems.

9. Originally Posted by Chester
I do know how to use FOIL. I have actually figured out a good deal of these sitting here for the last 12 hours.LOL I used the (a+b)(a-b)=(a^2-b^2). = 3^2-28= -19. I am still working on the last problems.
$x^{-2}=4$

simplify: $\frac{1}{x^2}=4$

reciprocate: $x^2=\frac{1}{4}$

find the square root of both sides: $x=\sqrt{\frac{1}{4}}$

extend: $x=\frac{\sqrt1}{\sqrt4}$

solve: $\boxed{x=\frac{1}{2}}$

name the other problems you need help on

10. #2 I'm having trouble understanding what the problem is with #2, since $\frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6-1}}$ than therefore $\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6-1}}=0$
but I assume you've done a typo somewhere...

#3
$3+\sqrt{x^2-8x}=0$

multiply both sides by $3-\sqrt{x^2-8x}$ and you'll get: $\left(3+\sqrt{x^2-8x}\right)\left(3-\sqrt{x^2-8x}\right)=0\left(3-\sqrt{x^2-8x}\right)$

now think back and you'll remember that (a+b)(a-b)=(a^2-b^2).

simplify: $3^2-\left(\sqrt{x^2-8x}\right)^2=0$

solve: $9-(x^2-8x)=0$

get rid of parentheses: $9-x^2+8x=0$

rearrange: $-x^2+8x+9=0$

now use the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

substitute: $x=\frac{-8\pm\sqrt{8^2-4(-1)(9)}}{2(-1)}$

simplify: $x=\frac{-8\pm\sqrt{64+36}}{-2}$

simplify $x=\frac{-8\pm\sqrt{100}}{-2}$

use the principle square root: $x=\frac{-8+10}{-2}$

subtract: $x=\frac{2}{-2}$

divide: $\boxed{x=-1}$

11. alright you have the expression $\frac{\sqrt6\times\sqrt{14}}{\sqrt7\times\sqrt3}$

for convenience, let's seperate the fractions: $\frac{\sqrt6}{\sqrt3}\times\frac{\sqrt{14}}{\sqrt7 }$

now simplify and you'll get: $\sqrt{\frac{6}{3}}\times\sqrt{\frac{14}{7}}$

divide and solve: $\sqrt2\times\sqrt2=2$

12. ## Unbelievable

This is not a hard one but my book is so bad that I could not even get that out of it. Just multiply across.

Thanks,
Chester

13. ## Oops

This one should be

$
\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6+1}}
$

Would it be difficult to explain this. Do I just divide and then subtract?

14. Originally Posted by Chester
This is not a hard one but my book is so bad that I could not even get that out of it. Just multiply across.

Thanks,
Chester
are you asking how to change an equation like this $\frac{a}{b}=\frac{x}{y}$ into this: $ax=by$

Well, I'll tell you even if this isn't what you're asking...

we have equation $\frac{a}{b}=\frac{x}{y}$ so to change this equation the first step would be by multiply both sides by $b$ look below:

multiply both sides by $b$: $\frac{a}{b}\times b=\frac{x}{y}\times b$

therefore: $\frac{ab}{b}=\frac{xb}{y}$

therefore: $\frac{a\!\not b}{\not b}=\frac{xb}{y}$

so then we get: $a=\frac{xb}{y}$

so multiply both sides by $y$ and you get: $ay=xb$

let's test it out:

$\frac{1}{2}=\frac{2}{4}$

change it around and you get: $1\times4=2\times2\quad\rightarrow\quad 4=4$

you see? it works!

15. ## Thanks for the explanation

I am a little confused (not that that’s hard to do at this point.) But the problem is

$
\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6+1}}
$

Not

$
\frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6+1}}$

Do I solve this the same way?

Page 1 of 2 12 Last