SOLUTION: Ok... the problem is (1/x)-(1/3)=(6/x^2). The instructions are to Solve the equation. I tried the problem and i got No Solution. I'm not really sure if this is the correct answer.

Question 78424This question is from textbook Advanced Algebra
: Ok... the problem is (1/x)-(1/3)=(6/x^2). The instructions are to Solve the equation. I tried the problem and i got No Solution. I'm not really sure if this is the correct answer. I don't really understand this area in the chapter very much at all. Your help would be very Appreciated! Thanks! This question is from textbook Advanced Algebra

Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Multiply both sides by the LCD which is

Distribute

Subtract 18 from both sides

Arrange in descending order
Now use the quadratic formula to solve for x:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -63 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -63 is + or - .

The solution is

Here's your graph:

Since we have a complex answer, this means we have no real solution. So you were right and there are no solutions. If you graphed the 2 equations, you would notice that they never intersect.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(1/x)-(1/3)=(6/x^2)
The lcm is 3x^2.
Multiply thru by 3x^2 to get:
3x-x^2=18
x^2-3x+18=0
(x-6)(x+3)=0
x=6 or x=-3
---------
Checking these answers:
If x=6: 1/6-2/6=6/36------false
If x=-3: -1/3-1/3=6/9-----false
Therefore No Solution.
===========Cheers,
Stan H.
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