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

  • Thread starter thomas49th
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In summary: What are you supposed to do with this? Simplify?If you're solving for x, then you could simplify by grouping like terms together and then simplify the product.
  • #1
thomas49th
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(urgent)Solving surd equation

Solve [tex]5\sqrt{x} = \frac{40}{x}[/tex]

how would I start by solving this?
Cross multipling seems to be difficul here
Thx
 
Last edited:
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  • #2
thomas49th said:
Solve [tex]5\sqrt{x} = \frac{40}{x}[/tex]

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

[tex] x\sqrt{x}=8 [/tex]
[tex]x^{3/2}=2^3 [/tex]
[tex]x^{3/2}=4^{3/2}[/tex]
 
Last edited:
  • #3
Could square both sides, and then cross multiply.
 
  • #4
substitution ...

let [tex]u=\sqrt{x}[/tex]

[tex]5u = \frac{40}{u^2}[/tex]

[tex]5u - \frac{40}{u^2} = 0[/tex]
find common denominator and simplify:
[tex]u=2[/tex]

then

[tex]\sqrt{x}=2[/tex]
therefore [tex]x=4[/tex]

** lol beat me to it
 
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  • #5
steven10137 said:
substitution ...

let [tex]u=\sqrt{x}[/tex]

[tex]5u = \frac{40}{u^2}[/tex]

[tex]5u - \frac{40}{u^2} = 0[/tex]
find common denominator and simplify:
[tex]u=2[/tex]

then

[tex]\sqrt{x}=2[/tex]
therefore [tex]x=4[/tex]

** lol beat me to it

i've seen weird things like this before. where did you learn this? like what country
 
  • #6
ice109 said:
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 :confused:
 
  • #7
Kurdt said:
Whats weird about substitution? Its pretty standard practise anywhere :confused:

i was never taught that and
i don't see the point? [tex]\sqrt{x} = x^\frac{1}{2}[/tex] and then just add exponents of polynomials with like bases when multiplying
 
  • #8
ice109 said:
i was never taught that and
i don't see the point? [tex]\sqrt{x} = x^\frac{1}{2}[/tex] 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.
 
  • #9
steven10137 said:
well ... you weren't taught correctly, lol

nah, seems simple enough for me.

what? how is that incorrect
 
  • #10
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.
 
  • #11
steven10137 said:
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
 
  • #12
ice109 said:
i seriously don't see how, please show me an example

eventually you will meet problems of that kind ;)
 
  • #13
ok, this is waaaayy off topic ... but, as per request i shall give multiple:

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

would you like another ... perhaps trigonometric, involving quotients or exponentials?
 
  • #14
Its basically the same as treating some expression (in this case, [tex]\sqrt{x}[/tex] as a variable or symbol itself, and not treating it as an actual expression until the end. So instead of treating [tex]\sqrt{x}[/tex] as the square root of x, its just a symbol, which we treat like any other symbol, such as a, x, y, [tex]\theta[/tex] etc.
 
  • #15
Perhaps a better example of where a substitution would help:

Solve : [tex]x^8+x^4+\pi =0[/tex]

Personally I would let u=x^4, what would you do?
 
  • #16
Gib Z said:
Perhaps a better example of where a substitution would help:

Solve : [tex]x^8+x^4+\pi =0[/tex]

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

I would solve it with my texas ti-83 =P
 
  • #17
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
Do you mean

[tex]\left(\frac{2}{6-\sqrt{3}}\right)^2 - \left(\frac{2}{6+\sqrt{3}}\right)^2[/tex]

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.
 

1. What is a surd equation?

A surd equation is an equation that contains a square root or other root expressions, such as cube roots or fourth roots.

2. How do I solve a surd equation?

To solve a surd equation, you need to isolate the surd term on one side of the equation and square both sides of the equation to eliminate the radical. Then, solve for the variable using basic algebraic principles.

3. What is the first step in solving 5√x = 40/x?

The first step is to isolate the surd term, which in this case is 5√x, on one side of the equation. To do this, you can multiply both sides of the equation by x to get 5√x * x = 40. This will simplify to 5x = 40.

4. How do I eliminate the radical in the equation?

To eliminate the radical, you need to square both sides of the equation. This will result in 5x^2 = 1600. Then, you can solve for x by dividing both sides by 5 to get x = 320.

5. Are there any restrictions on the values of x in this equation?

Yes, there are restrictions on the values of x. Since we are dealing with a square root, the value of x must be greater than or equal to 0. Additionally, since there is a denominator of x in the equation, x cannot equal 0. So, the solution for x is x = 320, as long as x is greater than or equal to 0 and not equal to 0.

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