# Root equation help

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

Homework Helper
Gold Member
Dearly Missed
Eeeh?

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

thomas49th
yeh, that's the one

Homework Helper
Gold Member
Dearly Missed
So, what is a common factor between the two terms on the right-hand side?

thomas49th
r² is the common factor

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

thomas49th
cross multiply?

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

Homework Helper
Gold Member
Dearly Missed
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?

Homework Helper
You want to get rid of the square on r.

What is the opposite of squaring?

thomas49th
square root the whole RHS

Homework Helper
Gold Member
Dearly Missed
Do you know what an equation is, and what is allowed to do with one?

Homework Helper
Gold Member
Dearly Missed
square root the whole RHS

Wrong.

WHAT must you take the square root of?

thomas49th

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

Homework Helper
Gold Member
Dearly Missed
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)

thomas49th
cheers, thanks for the help

So here's another one:

v² = u² + av² find v

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

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?

Homework Helper
Gold Member
Dearly Missed
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

Homework Helper
Gold Member
Dearly Missed
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?

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

Homework Helper
Gold Member
Dearly Missed
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?

Homework Helper
Gold Member
Dearly Missed
would changing it to

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

thomas49th
sory when changing from latex source code i pressed enter on window and it submitted:

Would this be considered simplyfying?

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

Homework Helper
Gold Member
Dearly Missed
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

Homework Helper
Gold Member
Dearly Missed
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

Last edited:
$$v^{2} = u^{2} + av^{\frac{x}5} 1. Rearrange to x 2. Rearrange to get v Science Advisor Homework Helper Gold Member Dearly Missed 1. Easy 2. Forget it. thomas49th okay. If you say it's easy ill believe you. Just, somthing about surds. Simplify http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd25.gif" [Broken] can you show me the steps you did it in Last edited by a moderator: Science Advisor Homework Helper Gold Member Dearly Missed I don't see that any rewriting of that expression is any simpler than the one given. thomas49th youre not suppost to leave surds on the bottom. Apparently the answer should be http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd28.gif" [Broken] But I looked at their method and it looked dogdy. Is the answer [tex]\frac{3(\sqrt{6} - \sqrt{2})}4$$