1. tejasree

Factor it out!

2. tejasree

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

3. dddan

i did and got 0,3,-4

4. dddan

i just want to see if im right

5. tejasree

no!

6. tejasree

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. tejasree

wait there are more solutions

9. cwrw238

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

10. tejasree

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

11. tejasree

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

12. tejasree

am I correct @cwrw238

13. dddan

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

14. tejasree

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. dddan

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

18. dddan

ok i got it - thank you so much guys

19. tejasree

there you go @jim_thompson5910 :)

20. tejasree

thank you! :)

21. tejasree

this site is really helpful: mathway.com

22. cwrw238

@tegasree - u r correct

23. tejasree

thank you! :)