## dddan Group Title please help!; how do you find the solutions for x^4-x^2-12=0 one year ago one year ago

1. tejasree Group Title

Factor it out!

2. tejasree Group Title

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

3. dddan Group Title

i did and got 0,3,-4

4. dddan Group Title

i just want to see if im right

5. tejasree Group Title

no!

6. tejasree Group Title

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

7. cwrw238 Group Title

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

8. tejasree Group Title

wait there are more solutions

9. cwrw238 Group Title

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

10. tejasree Group Title

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

11. tejasree Group Title

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

12. tejasree Group Title

am I correct @cwrw238

13. dddan Group Title

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

14. tejasree Group Title

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

15. jim_thompson5910 Group Title

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 Group Title

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

17. dddan Group Title

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

18. dddan Group Title

ok i got it - thank you so much guys

19. tejasree Group Title

there you go @jim_thompson5910 :)

20. tejasree Group Title

thank you! :)

21. tejasree Group Title

this site is really helpful: mathway.com

22. cwrw238 Group Title

@tegasree - u r correct

23. tejasree Group Title

thank you! :)