I just dont get it. Can someone please help.

Problems are attached

Chester

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- Aug 1st 2006, 12:00 PM #1

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- Aug 1st 2006, 12:06 PM #2

- Aug 1st 2006, 12:11 PM #3
These should be explained to you better, so here I go:

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

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

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

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

- Aug 1st 2006, 01:53 PM #4

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- Aug 1st 2006, 02:02 PM #5

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- Aug 1st 2006, 02:07 PM #6

- Aug 1st 2006, 03:56 PM #7
btw. the roots of $\displaystyle \sqrt{m^6}$ are $\displaystyle \pm m^3$

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

Do you know how to use foil?

$\displaystyle F:\;3\times3=9$

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

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

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

now you add:

$\displaystyle F+O+I+L$

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

$\displaystyle 9-28=-19$ voila!

- Aug 1st 2006, 04:08 PM #8

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- Aug 1st 2006, 04:15 PM #9Originally Posted by
**Chester**

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

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

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

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

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

name the other problems you need help on

- Aug 1st 2006, 05:03 PM #10
#2 I'm having trouble understanding what the problem is with #2, since $\displaystyle \frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6-1}}$ than therefore $\displaystyle \frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6-1}}=0$

but I assume you've done a typo somewhere...

#3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- Aug 1st 2006, 05:37 PM #11
alright you have the expression $\displaystyle \frac{\sqrt6\times\sqrt{14}}{\sqrt7\times\sqrt3}$

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

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

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

- Aug 1st 2006, 05:49 PM #12

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- Aug 1st 2006, 05:59 PM #13

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- Aug 1st 2006, 06:04 PM #14Originally Posted by
**Chester**

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

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

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

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

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

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

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

let's test it out:

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

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

you see? it works!

- Aug 1st 2006, 06:19 PM #15

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## Thanks for the explanation

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

$\displaystyle

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

$

Not

$\displaystyle

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

Do I solve this the same way?