Solve 1+1/x+1/x^2=0

Hello, silverdragon!

Solve: \(\displaystyle \L\:1\,+\,\frac{1}{x}\,+\,\frac{1}{x^2}\:=\:0\)

Multiply by \(\displaystyle x^2:\;\;\L x^2\,+\,x\,+\,1\:=\:0\)

Then use the Quadratic Formula . . .

 
ok, when I try to use the quadratic formula I get no answer. I used a quadratic equation calculator and it says it's not defined.That their is no answer. :?

Now your suppose to FOIL and equal 0 right?

Example:

(x-3)(x+4)=0
x-3=0 or x+4=0
x=3....or.x=-4
 
silverdragon316 said:
ok, when I try to use the quadratic formula I get no answer. I used a quadratic equation calculator and it says it's not defined.
YOU solve using the quadratic...NOT the calculator :shock:

x^2 + x + 1 = 0

x = [-1 +- sqrt(1^2 - 4(1)(1))] / 2

x = [-1 +- sqrt(-3)] / 2

So what does that tell YOU?
 
silverdragon316 said:
when I try to use the quadratic formula I get no answer.
What do you mean by "no answer"? (When you reply, along with showing your work, please specify whether you have yet studied complex numbers; that is, the numbers with "i" in them.)

silverdragon316 said:
Now your suppose to FOILand equal 0 right?
I'm sorry, but I don't know what this means. You have a quadratic that you need to solve (and which is already equal to zero), not two binomials that you need to multiply (informally called "FOILing"). What are you proposing be multiplied?

Thank you.

Eliz.

P.S. I see that you posted more right as I was posting the above reply. As has been requested of you a few times before, please post replies and follow-ups as replies (that is, as answers contained in posts following the posts with the questions), rather than going back and doing stealth-edits. Thank you for your consideration.
 
ok, when I try to use the quadratic formula I get no answer. I used a quadratic equation calculator and it says it's not defined.

The discriminant \(\displaystyle \L b^{2} - 4ac\) for the quadratic will tell you if there are any real solutions to your polynomial.

\(\displaystyle \L x^2 + x + 1 = 0\)

\(\displaystyle \L b^{2} - 4ac = [1^{2} - 4(1)(1)] = [1-4] = -3\)

Your discriminant is < 0: Thus your polynomial contains no real solutions.
 
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