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 = [tex]u²\1-a[/text]<br /> v = $$\sqrt{u²/1-a}$$<br /> <br /> am I right?[/tex]

 

[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??