Solving a Root Equation for the Variable r

[thomas49th]

Hi, I have the equation

A = πr² + r²root(k²-1)

i need to rearrange it to find r

i go it to

r³ = (2A/π+root(k²-1)

to get just r (with no powers) what will the final equation look like and why.

Thanks

 

[arildno]

Eeeh?

Your original equation is:
A=\pi{r}^{2}+r^{2}\sqrt{k^{2}-1}
Agreed?

 

[thomas49th]

yeh, that's the one

 

[arildno]

So, what is a common factor between the two terms on the right-hand side?

 

[thomas49th]

r² is the common factor

 

[arildno]

So, you may rewrite your equation as:
r^{2}(\pi+\sqrt{k^{2}-1})=A
What would you do next?

 

[thomas49th]

cross multiply?

r^{2} = A/(\pi+\sqrt{k^{2}-1})

 

[arildno]

Which is to divide each side with the factor (\pi+\sqrt{k^{2}-1}), rather than cross-multiplication.

1. Now, does this equal what you posted before?

2. Since you now know the SQUARE of a number, how do we get what the number itself is?

 

[thomas49th]

do you square the whole RHS?

 

[HallsofIvy]

You want to get rid of the square on r.


What is the opposite of squaring?

 

[thomas49th]

sorry, my bad
square root the whole RHS

 

[arildno]

Do you know what an equation is, and what is allowed to do with one?

 

[arildno]

sorry, my bad
square root the whole RHS

Wrong.

WHAT must you take the square root of?

 

[thomas49th]

isn't the answer:

r= \sqrt{A/(\pi+\sqrt{k^{2}-1})}

 

[arildno]

1. You must take the square root of BOTH sides of the equation, not just of one of the sides as you said. (This you have done)

2. Put parentheses about the correct radicand.

 

[thomas49th]

cheers, thanks for the help

So here's another one:

v² = u² + av² find v

v² - av² = u²
v²(1 - a) = u²
v² = u²\1-a[/text]<br /> v = \sqrt{u²/1-a}<br /> <br /> am I right?

 

[thomas49th]

cheers, thanks for the help

So here's another one:

v² = u² + av² find v

v² - av² = u²
v²(1 - a) = u²
v² = u^{2}/1-a
v = \sqrt{u^{2}/1-a}

am I right?

 

[arildno]

Use PARENTHESES ABOUT YOUR DENOMINATOR!
Is it that hard to get?

Secondly, in the prior exercise I assumed that "r" was a radius, and hence necessarily a non-negative quantity (you didn't say).
Now, must "v" be a non-negative quantity?

 

[thomas49th]

v is meaningless, I'm just praticing rearranging the formula

 

[arildno]

Is it "meaningless"?
Is it not even a number?

 

[thomas49th]

well it must be a number...musn't it.
Did i get the question right?

v = \sqrt{u^{2}/(1-a)}

 

[arildno]

v can be either of the two numbers:
v=\pm\sqrt{\frac{u^{2}}{1-a}}

 

[thomas49th]

would changing it to

v=\frac{u}\sqrt{1-a}}

be simplfying?

 

[arildno]

would changing it to

v=\frac{u^{2}}\sqrt{{1-a}}

How would that come about?

 

[thomas49th]

sory when changing from latex source code i pressed enter on window and it submitted:

Would this be considered simplyfying?

v=\frac{u}\sqrt{(1-a)}

 

[arildno]

Almost; but you forget you have TWO solutions for v:
v=\pm\frac{u}{\sqrt{1-a}}

 

[thomas49th]

so that's simplifying, yet leaving a surd as a demoninator isn't. Howcome? What's so special with surds

 

[arildno]

so that's simplifying, yet leaving a surd as a demoninator isn't. Howcome? What's so special with surds

That's a matter of taste, mostly.
The first expression is about as simple; however, most would regard the square root of a square (i.e, your numerator) as a non-simplified expression.

 

[thomas49th]

How would this equation go then...

v^{2} = u^{2} + a^{2\5} find x

 

[arildno]

Eeh, what x?
The one under the table, or the one NOT appearing in your equation??

 

[thomas49th]

sorry, latex isn't working...
v^{2} = u^{2} + av^{\frac{x}5} <br /> <br /> 1. Rearrange to x<br /> 2. Rearrange to get v

 

[arildno]

1. Easy
2. Forget it.

 

[thomas49th]

okay. If you say it's easy ill believe you.

Just, somthing about surds.

Simplify

http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd25.gif"

can you show me the steps you did it in

 

[arildno]

I don't see that any rewriting of that expression is any simpler than the one given.

 

[thomas49th]

youre not suppost to leave surds on the bottom. Apparently the answer should be

http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd28.gif"

But I looked at their method and it looked dogdy. Is the answer

\frac{3(\sqrt{6} - \sqrt{2})}4

as the link above looks like it's 3/4 multiplied by root 6 - root 2

 

[arildno]

okay, so whenever you've got a surd in your denominator, you are to rationalize it. Fine by me; mind you, that is THEIR choice, not everybody's else's choice.

Remember that 1=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}
See if you can use that to derive their answer.

 

[thomas49th]

\frac{3}{(\sqrt{6} + \sqrt{2})}

\frac{3(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})}

use smilie face method on denominator

root 6 x root 6 = 6
root 2 x - root 2 = - 2

root 6 x - root 2 = - root 12
root 6 x root 2 = root 12

- root 12 and root 12 cancel each other out. 6 - 2 = 4



leaving you with
\frac{3(\sqrt{6} - \sqrt{2})}4

am I right or am I right?

 

[arildno]

Indeed you are right. Here is a true smilie for you:

 

[thomas49th]

Okay. I've got some more problems i wish to solve

Express
\frac{1}{x - 2} + \frac{2}{x+4}
as a single algergraic faction

 

[arildno]

So, what is the least common multiple of the denominators?

 

[thomas49th]

would it be 2?

 

[arildno]

Does x-2 divide 2?? Does x+4 divide 2??

 

[thomas49th]

is it (x-2)(x+4) = x² -8 + 2x

 

[thomas49th]

ahh me thinks you cross multiply?

\frac{x+4 + 2(x-2)}{(x-2)(x+4)}

 

[arildno]

I do not cross-multiply here either, I EXPAND both fractions in the manner you've just done.

 

[thomas49th]

To add the fractions I need to find a common denominator right?
What ever I do to the denominator I must do to the numerator right?
So how do I find the lost common mulitple of x-2 and x+4?


Thanks

 

[arildno]

You've done it just fine, as I've said.
Just don't call it cross-multiplication.

 

[thomas49th]

so I am at \frac{3x}{x^{2}+2x-8}
now where to? That's not the final answer is it?

 

[arildno]

Looks final to me, given your task to find a SINGLE algebraic fraction identical to the given sum.

 

[thomas49th]

Ive looked this up on an old test paper, and apparently the answer is 1/3

The final step before the answer I've written

3x / ((x-2)(x+4))

but I don't know how I got 1/3...do you?