Exact roots!!!

[Synergy]

Hey guys,

Was wondering how I could do this:

the question asks to find the exact roots of each equation and then for equation (g) I have this:

1 - 1/x = 1/x^3 - 1/x^2

I got this far:

0 = 1/x^3 - 1/x^2 + 1/x - 1
Then I looked for common factors of (-1) which were either +/- 1, when I used the factor theorem to find if (x-1) was a factor I proved it correct! So now that is where I am stuck:

(1/x^3 - 1/x^2 + 1/x -1) / (x-1) = ???

How would I do this? I am thinking maybe you use like the negative intervals as in: 1/x^3 becomes x^-3 but then when I divide I would get x^-4 or 1/x^4 which I do not want and can not have I believe!

Can someone help out please?

[VBAhack]

Welcome to the forums!

Suggest the following:

1. Plot the function. If you have Excel, it's real easy. This gives you a qualitative understanding of the function and can (perhaps) give you an idea of how many real roots it has and where they are. This isn't essential, but it sure helps.

2. Get a common denominator for the terms. As an illustration, if I had x/2 + 1/3, that would be (3x + 2)/(2*3) = (3x + 2)/6.

3. After step 2, factor the numerator. That should tell you what the roots are.

[Glaysher]

Hey guys,

Was wondering how I could do this:

the question asks to find the exact roots of each equation and then for equation (g) I have this:

1 - 1/x = 1/x^3 - 1/x^2

I got this far:

0 = 1/x^3 - 1/x^2 + 1/x - 1
Then I looked for common factors of (-1) which were either +/- 1, when I used the factor theorem to find if (x-1) was a factor I proved it correct! So now that is where I am stuck:

(1/x^3 - 1/x^2 + 1/x -1) / (x-1) = ???

How would I do this? I am thinking maybe you use like the negative intervals as in: 1/x^3 becomes x^-3 but then when I divide I would get x^-4 or 1/x^4 which I do not want and can not have I believe!

Can someone help out please?

Start with

0 = 1/x³ - 1/x² + 1/x - 1

Multiply both sides by x³

0 = 1 - x + x² - x³

x³ - x² + x - 1 = 0

Cubic equation, use factor theorem to factorise

x = 1

1 - 1 + 1 - 1 = 0

So x - 1 is a factor and x = 1 a root

x³ - x² + x - 1 = (x - 1)(x² + bx + c)

Work out what b and c are by multiplying out the right hand side and comparing with the left hand side and you will end up with a quadratic

I suspect from the way your question is phrased that you will need the quadratic formula to find the two other roots exactly but I don't have time to finish the solution now.

[zaza]

You just need to imagine the solution and it will be obvious...




Alternatively, use your eyes.

[Glaysher]

Back now with a bit more time

Multiply out right hand side

= x³ + bx² + cx - x² - bx - c

Equate coefficients

- c = -1
So c = 1

bx² - x² = -x²
So b = 0

S factorisation is now

(x - 1)(x² + 1)

x² + 1 has no real roots

Has complex roots i and -i

So roots are 1, i and -i

[eranga262154]

Back now with a bit more time

Multiply out right hand side

= x³ + bx² + cx - x² - bx - c

Equate coefficients

- c = -1
So c = 1

bx² - x² = -x²
So b = 0

S factorisation is now

(x - 1)(x² + 1)

x² + 1 has no real roots

Has complex roots i and -i

So roots are 1, i and -i

This is the exact answer which I got. As here start with simple things like simplification of expression. Then carry-on your work. It is very easy most of the time.