## anonymous 3 years ago please help!; how do you find the solutions for x^4-x^2-12=0

1. anonymous

Factor it out!

2. anonymous

(x^2 - 4)(x^2 + 3) = 0

3. anonymous

i did and got 0,3,-4

4. anonymous

i just want to see if im right

5. anonymous

no!

6. anonymous

x^2 - 4 = 0 (x-2)(x+2) = 0 x=2,-2

7. cwrw238

let y = x^2 then y^2 - y -12 = 0 can you solve this?

8. anonymous

wait there are more solutions

9. cwrw238

yup there will be 4 solution because the degree of the equation is 4

10. anonymous

x^2 + 3 = 0 x^2 = -3 $x = i \sqrt{3}, -i \sqrt{3}$

11. anonymous

the solutions are: $x = 2, -2, i \sqrt{3}, -i \sqrt{3}$

12. anonymous

am I correct @cwrw238

13. anonymous

i understand why there 4 solutions but where did you get them from?

14. anonymous

First we factored them out, didnt we? then we set each factor equal to zero

15. jim_thompson5910

It might be easier to let z = x^2 so z^2 = x^4 which allows us to go from x^4-x^2-12=0 to z^2-z-12=0

16. jim_thompson5910

solve z^2-z-12=0 and keep in mind that z = x^2 so you can solve for x

17. anonymous

o because it x^2=3 its the square root plus i right

18. anonymous

ok i got it - thank you so much guys

19. anonymous

there you go @jim_thompson5910 :)

20. anonymous

thank you! :)

21. anonymous

this site is really helpful: mathway.com

22. cwrw238

@tegasree - u r correct

23. anonymous

thank you! :)