# Root equation help

1. Feb 11, 2007

### 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

2. Feb 11, 2007

### arildno

Eeeh?

$$A=\pi{r}^{2}+r^{2}\sqrt{k^{2}-1}$$
Agreed?

3. Feb 11, 2007

### thomas49th

yeh, that's the one

4. Feb 11, 2007

### arildno

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

5. Feb 11, 2007

### thomas49th

r² is the common factor

6. Feb 11, 2007

### arildno

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

7. Feb 11, 2007

### thomas49th

cross multiply?

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

8. Feb 11, 2007

### 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?

9. Feb 11, 2007

### thomas49th

do you square the whole RHS?

10. Feb 11, 2007

### HallsofIvy

Staff Emeritus
You want to get rid of the square on r.

What is the opposite of squaring?

11. Feb 11, 2007

### thomas49th

square root the whole RHS

12. Feb 11, 2007

### arildno

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

13. Feb 11, 2007

### arildno

Wrong.

WHAT must you take the square root of?

14. Feb 11, 2007

### thomas49th

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

15. Feb 11, 2007

### 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)

16. Feb 11, 2007

### thomas49th

cheers, thanks for the help

So heres another one:

v² = u² + av² find v

v² - av² = u²
v²(1 - a) = u²
v² = $$u²\1-a[/text] v = [tex]\sqrt{u²/1-a}$$

am I right?

17. Feb 11, 2007

### thomas49th

cheers, thanks for the help

So heres 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?

18. Feb 11, 2007

### arildno

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?

19. Feb 11, 2007

### thomas49th

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

20. Feb 11, 2007

### arildno

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