# Solving surd equation

1. Jun 4, 2007

### thomas49th

(urgent)Solving surd equation

Solve $$5\sqrt{x} = \frac{40}{x}$$

how would I start by solving this?
Cross multipling seems to be difficul here
Thx

Last edited: Jun 4, 2007
2. Jun 4, 2007

### f(x)

$$x\sqrt{x}=8$$
$$x^{3/2}=2^3$$
$$x^{3/2}=4^{3/2}$$

Last edited: Jun 4, 2007
3. Jun 4, 2007

### danago

Could square both sides, and then cross multiply.

4. Jun 4, 2007

### steven10137

substitution ...

let $$u=\sqrt{x}$$

$$5u = \frac{40}{u^2}$$

$$5u - \frac{40}{u^2} = 0$$
find common denominator and simplify:
$$u=2$$

then

$$\sqrt{x}=2$$
therefore $$x=4$$

** lol beat me to it

Last edited: Jun 4, 2007
5. Jun 4, 2007

### ice109

i've seen weird things like this before. where did you learn this? like what country

6. Jun 4, 2007

### Kurdt

Staff Emeritus
Whats weird about substitution? Its pretty standard practise anywhere

7. Jun 4, 2007

### ice109

i was never taught that and
i don't see the point? $$\sqrt{x} = x^\frac{1}{2}$$ and then just add exponents of polynomials with like bases when multiplying

8. Jun 5, 2007

### steven10137

well ... you weren't taught correctly, lol

nah, seems simple enough for me.

9. Jun 5, 2007

### ice109

what? how is that incorrect

10. Jun 5, 2007

### steven10137

nah im not saying your working is wrong, i'm just saying substitution is a better way, which can be applied; Particularly when doing harder and more complex problems.

11. Jun 5, 2007

### ice109

i seriously don't see how, please show me an example

12. Jun 5, 2007

### malawi_glenn

eventually you will meet problems of that kind ;)

13. Jun 5, 2007

### steven10137

ok, this is waaaayy off topic .... but, as per request i shall give multiple:

perhaps basic:
$$x - 3\sqrt{x} = -2$$
let $$u=\sqrt{x}$$
$$u^2 - 3u + 2 = 0$$
$$(u-1)(u-2) = 0$$
$$u=1$$, $$u=2$$
$$u=\sqrt{x}$$
then
$$1=\sqrt{x}$$ and $$x=1$$
or
$$2=\sqrt{x}$$ and $$x=4$$

would you like another ... perhaps trigonometric, involving quotients or exponentials?

14. Jun 5, 2007

### danago

Its basically the same as treating some expression (in this case, $$\sqrt{x}$$ as a variable or symbol itself, and not treating it as an actual expression until the end. So instead of treating $$\sqrt{x}$$ as the square root of x, its just a symbol, which we treat like any other symbol, such as a, x, y, $$\theta$$ etc.

15. Jun 5, 2007

### Gib Z

Perhaps a better example of where a substitution would help:

Solve : $$x^8+x^4+\pi =0$$

Personally I would let u=x^4, what would you do?

16. Jun 5, 2007

### malawi_glenn

I would solve it with my texas ti-83 =P

17. Mar 20, 2008

### binomialgurl

Surds HELP!!

im having trouble trying to figure out a problem. can someone help me.

(2/6 -root3)^2 - (2/6+root3)^2

if that makes sense to anyone help.

I think that the denominator of both has to be rationalised but do i expand the squared brackets first or later or what?

any ideas welcome thanks

18. Mar 20, 2008

### Tedjn

Do you mean

$$\left(\frac{2}{6-\sqrt{3}}\right)^2 - \left(\frac{2}{6+\sqrt{3}}\right)^2$$

What are you supposed to do with this? Simplify?

EDIT: Sorry, I didn't see that this question was moved into its separate thread. I'll re-post this there.