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# Solve x^2+ 4/(x^2)=12 for all real values of x watch

1. Please show me how to solve this without a calculator!
2. (Original post by Phoenix Everill)
Please show me how to solve this without a calculator!
Multiply both sides by and solve. Please post your working if you need any further hints.
3. (Original post by RogerOxon)
Multiply both sides by and solve. Please post your working if you need any further hints.
(Original post by Phoenix Everill)
Please show me how to solve this without a calculator!
just put the second term into terms of x^(-2), using the 4.
then solve
4. (Original post by ThatsAGoodOne349)
just put the second term into terms of x^(-2), using the 4.
then solve
so:

x^4 + 4 = 12x^2
=> x^4 + 4 - 12 = 0
=> x^2 + 4x^(-2) -12 = 0

Then I don't know how to solve, sorry.
5. (Original post by Phoenix Everill)
so:

x^4 + 4 = 12x^2
=> x^4 + 4 - 12 = 0
=> x^2 + 4x^(-2) -12 = 0

Then I don't know how to solve, sorry.
you got that right
6. (Original post by Phoenix Everill)
so:

x^4 + 4 = 12x^2
=> x^4 + 4 - 12 = 0
=> x^2 + 4x^(-2) -12 = 0

Then I don't know how to solve, sorry.

(we need to ignore x=0, if it were a solution)

Now solve it as a quadratic in .
7. (Original post by Phoenix Everill)
so:

x^4 + 4 = 12x^2
=> x^4 + 4 - 12 = 0
=> x^2 + 4x^(-2) -12 = 0

Then I don't know how to solve, sorry.
(Original post by ThatsAGoodOne349)
you got that right
Nope - the intermediate step is wrong. What happened to the in the second line?
8. (Original post by RogerOxon)
Nope. What happened to the ?
nono
you got rid of the fraction

try using the x^4 +4 = 12x^2 and using this:
y =x^2
9. (Original post by Phoenix Everill)
so:

x^4 + 4 = 12x^2
=> x^4 + 4 - 12 = 0
=> x^2 + 4x^(-2) -12 = 0

Then I don't know how to solve, sorry.
(Original post by ThatsAGoodOne349)
you got that right
They didn't - the bold line is wrong.

(Original post by RogerOxon)

(we need to ignore x=0, if it were a solution)

Now solve it as a quadratic in .
(Original post by ThatsAGoodOne349)
nono
you got rid of the fraction
Why do you think that's a problem?
10. (Original post by RogerOxon)
Why do you think that's a problem?
My last post...
11. (Original post by ThatsAGoodOne349)
My last post...
You must have misread what I posted. We're in violent agreement.
12. (Original post by RogerOxon)
You must have misread what I posted. We're in violent agreement.
LOL
okay
let's go with the substitution thing
13. RogerOxon
you didn't mention substitution inn that post
14. (Original post by ThatsAGoodOne349)
RogerOxon
you didn't mention substitution inn that post
What substitution?
15. (Original post by RogerOxon)
What substitution?
y=x^2

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