Solving Surd Equation: Step-by-Step Guide for 5√x = 40/x

[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

 

[f(x)]

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

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

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

 

[danago]

Could square both sides, and then cross multiply.

 

[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

 

[ice109]

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

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

 

[Kurdt]

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

Whats weird about substitution? Its pretty standard practise anywhere

 

[ice109]

Whats weird about substitution? Its pretty standard practise anywhere

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

 

[steven10137]

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

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

nah, seems simple enough for me.

 

[ice109]

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

nah, seems simple enough for me.

what? how is that incorrect

 

[steven10137]

nah I am 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.

 

[ice109]

nah I am 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.

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

 

[malawi_glenn]

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

eventually you will meet problems of that kind ;)

 

[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?

 

[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.

 

[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?

 

[malawi_glenn]

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?

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

 

[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

 

[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.