# Root equation help

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

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arildno
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Eeeh?

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

yeh, that's the one

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

r² is the common factor

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

cross multiply?

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

arildno
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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?

do you square the whole RHS?

HallsofIvy
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You want to get rid of the square on r.

What is the opposite of squaring?

square root the whole RHS

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

arildno
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square root the whole RHS
Wrong.

WHAT must you take the square root of?

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

arildno
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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)

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?

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?

arildno
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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?

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

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

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

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

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

would changing it to

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

be simplfying?

arildno
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would changing it to

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