## kitsune0724 3 years ago By factoring how do you solve 3x^2/3-11x^1/3-4=0? My teacher doesn't do substitution. please and thank you.

1. cwrw238

(3x^1/3 + 1)(x^1/3 - 4 ) = 0

2. cwrw238

can you continue?

3. cwrw238

3x^1/3 + 1 = 0 x^(1/3) = -1/3 x^1/3 - 4 = 0 x^1/3 = 4

4. kitsune0724

let me see...

5. cwrw238

hint: x = x(1/3)^3

6. kitsune0724

(-1/3)^3 or (1/3)^3?

7. cwrw238

sorry - i didn't write it corruptly (x^(1/3)^ 3 = x^(1/3 * 3) = x^1 = x or in other words you have the cube root of and you cube it - that gives you x so if you cube both sides of the equations you get the value of x e g (x^(1/3)^ 3 =( -1/3)^3 so x = - 1/3^3 = -1/27 thats one of the roots

8. cwrw238

* correctly

9. kitsune0724

Thank you so much.

10. cwrw238

note x^(1/3) is same as cube root x

11. cwrw238

so to find other root you do same thing with x^1/3 = 4

12. cwrw238

yw

13. kitsune0724

Thank you again. But do you divide x^1/3=4? Like 4/1=4.

14. cwrw238

just cube both sides of th equation as we did to the first one (x^(1/3)^3 = x 4^3 = 4*4*4 = 64 x = 64

15. cwrw238

i suggest you review the laws of indices x^a + x^b = x^(a+b) x^a / x^b = x^(a-b) (x^a)^b = x^(a*b) = x^ ab

16. cwrw238

also |dw:1342632977864:dw|

17. cwrw238

18. kitsune0724

Thank you so much.

19. cwrw238

yw

20. kitsune0724