## anonymous one year ago Will give medal and become fan (accepting only correct answers)! Solve for x. (x^4 - 1) / (x^3) = 0

1. anonymous

2. zpupster

multply both sides by x^3 factor (x^2-1)(x^2+1) and again (x^2-1) = (x+1)(x-1) set to zero x+1=0 x-1=0 What are your roots??? (x^2+1) can never be 0 when y=0 x is?? see graph

3. Nnesha

okay so move x^3 to the right side as ~zpupster said multiply both sides by x^3

4. anonymous

Okay so then it would be $x ^{3} \times (x ^{4} - 1) \div x ^{3} = 0$

5. anonymous

But the x^3 can be cancelled so... the answer would be:$x ^{4} - 1 = 0$

6. anonymous

Right?

7. Nnesha

yep you need to multiply *both sides* so 0 times x^3 = 0 now apply difference of square rule

8. Nnesha

$\huge\rm a^2-b^2 =(a+b)(a-b)$

9. anonymous

$(x ^{2} + 1)(x ^{2} - 1)$

10. anonymous

$(x ^{2}+1)(x+1)(x-1)$

11. anonymous

Is that it?

12. Nnesha

yep right you would apply difference when there is negative sign so x^2-1 apply difference of square

13. Nnesha

14. Nnesha

yes right set all 3 parentheses equal to zero solve for x

15. anonymous

How do you find the square root of -1?

16. Nnesha

well do you need for this question ?

17. Nnesha

ohh gotcha so you will get an imaginary solution

18. anonymous

Is it going to be $\pm 1$

19. Nnesha

$\huge rm \sqrt{-1} = i$

20. anonymous

Oh like that! So the answer is x = 1, -1, and i

21. Nnesha

$\sqrt{-1}=i$

22. Nnesha

yep

23. anonymous

Alright thanks again!

24. Nnesha

my pleasure :=)