How do I solve x/3 + 4/x = x+1/3?

Answer 1

Should this be:-
x / 3 + 4 / x = (x + 1) / 3
x² + 12 = x² + x
x = 12

Or is it:-

x/3 + 4/x = x + 1/3
x² + 12 = 3x² + x
2x² + x - 12 = 0
x = [- 1 ± √(1 + 96)] / 4
x = [- 1 ± √(97) ] / 4

You choose!

PS
Use brackets to avoid confusion.

Answer 2

x/3 + 4/x = x+1/3

Is it

x/3 + 4/x = x +(1/3) or (x+1)/3?

the solutions are different.

Answer 3

x/3 + 4/x = x+1/3
3x[x/3 + 4/x] = 3x[x + 1/3]
x^2 + 12 = 3x^2 + x
x^2 - 3x^2 - x + 12 = 0
- 2x^2 - x + 12 = 0
-1[2x^2 - x + 12] = -1(0)
2x^2 + x - 12 = 0
x = {-1 +/- sqrt[1^2 - 4(2)(-12)]}/2(2)
x = 2.212 and - 2.712 ANS

teddy boy

Answer 4

(x^2 + 12)/ 3x = (3x+1)/3

(x^2 + 12)/ x = 3x + 1

x^2 + 12 = 3x^2 + x

2x^2 + x - 12 = 0

then u solve the equation as i still cant think of the ways how to put it in to the box like belox..... so u use the formula which is b2- 4ac.... something like tat ...

Answer 5

x/3 + 4/x = x + 1/3
x + 12/x = 3x + 1
12/x - 1 = 2x
12 - x = 2x^2
-2x^2 -x + 12 = 0
x^2 + 1/2x - 6 = 0
now use quadratic eqn, FOIL, whatever...

Answer 6

x/3 + 4/x = x+1/3
x^2/3x+12/3x=x+1/3
x^2/3x+12/3x-x+1/3=0
x^2/3x+12/3x-3x/3x+x/3x=0
(x^2+12-3x+x)/3x=0
(x^2+-2x+12)/3x=0
x^2-2x+12=0

than solve for x, plug into quadratic formula

Answer 7

I'm afraid Dusty has mis-read 4/x to be x/4. Previous answers are worth looking at though! :-)

Answer 8

times it all by 12
then it goes: 4x +3x = 12x +4
7x = 12x +4
7x-12x = 4
-4x = 4
x = -1
simple isnt it?