Solving for all values of x including imaginary numbers

[shemer77]

Homework Statement

Solve for all values of x both real and imaginary
1) x^3=-8
2)(3/x+1)+(x/x-1)=(x-5/(x+1)(x-1))
3)2x^3-8x^2+5x=0
4)x^4-7x^2+12=0
5)x - sqroot ( 6 - 5x ) = 0
6)(absvalue(3x)) + 6 = 10

The Attempt at a Solution

I tried all of them but came out with wonky answers or I got stuck,
Please help me, DONT GIVE ME THE ANSWERS, show me how to do them, or guide me :)

 

[mutton]

1) You want to get rid of the exponent 3. What can you do to both sides of the equation to make this happen?

2) Do you mean x/x - 1 = 1 - 1 = 0 or x/(x - 1)?

3) Every term on the left side has something in common: a factor.

4) Notice that $$(x^2)^2 = x^4$$, so it is a quadratic.

5) Remember that you can't take the square root of a negative number, so make sure all your answers agree with this fact.

6) |3x| is 3x if 3x is positive, and -3x if 3x is negative. Consider both cases.

 

[shemer77]

1) All i get is 2i, and -2, but there should be one more answer since its x^3
2)its 3 over x+1 plus x over x-1 equals x-5 over (x+1)(x-1)
3)I got x(x-1)(2x-5) so my answers were x,1,5/2
4)I did quad formula and i got 4, and 3 as my answers
5)i got -6 and 1
6)im not sure on how to do this one still

 

[mutton]

Sorry, I forgot you wanted imaginary values as well, so ignore some of what I said earlier.

3) x cannot be an answer, as you want to find x.

4) Those are roots for $$x^2$$, not x.

 

[mutton]

1) 2i is wrong because $$(2i)^3 = 8i^3 = -8i$$. Writing $$-8 = 8e^{i \pi}$$, the three roots are $$2e^{i \pi/3}, 2e^{i 3\pi/3}, 2e^{i 5\pi/3}$$. Are you familiar with this notation? If not, you can write $$x^3 + 8 = 0$$ and factor that as a difference of cubes.

2) The common denominator on the left side is the denominator on the right side, so combine the two terms on the left.

5) That seems right.

6) 4 = |3x| = 3|x|. |x| = 4/3 represents a circle in the complex plane.

 

[shemer77]

1) I did difference of cubes and got (x-2)(x^2+2x+4) then i did quadratic for that second part and got -2$$\pm2i\sqrt{3}$$ all over 2. so what would my answers be?
2)I made the denominators the same across the whole equation so i got 3x-3+x^2+x=x-5 and i condensed that to, x^2-3x+2=0 and i factored it to (x-2)(x-1) so my answers are 2 and 1 right?
3)so are my answers 1 and 5/2?
4)I don't know what to do, since i can't factor or use quad formula
5) yay
6)so would my answer be -4/3 and 4/3?

 

[mutton]

1) Expanding (x - 2)(x^2 + 2x + 4) gives x^3 - 8, which is not x^3 + 8. Check your answers by seeing that their cubes are -8.

2) 1 cannot be an answer because x - 1 is in the denominators, and you can't divide by 0.

3) The biggest power of x is x^3, so like you said before there are 3 roots. When you have a product of 3 factors = 0, consider the case where each factor = 0. In the question, should the 8 be a 7?

4) x^2 = 4 or x^2 = 3. These are like the first question.

6) There are infinitely more answers if you want imaginary values as well.