## anonymous 4 years ago i need help solving this: x^(2/3)-2x^(1/3)=15. And the intrsutions say explain and demonstrate how to use substitution to solve the equation. well i don't understand this at all. please help!

1. anonymous

treat x^(2/3) as x^(1/3)^2

2. anonymous

ok put $z=x^{\frac{1}{3}}$

3. anonymous

then factor as x^2-2x-15=0, and substitute x^1/3 back in for x

4. anonymous

then by the laws of exponents $z^2=(x^{\frac{1}{3}})^2=x^{\frac{2}{3}}$

5. anonymous

so by replacing $x^{\frac{1}{3}}$ by $z$ you get an equation that looks like $z^2-2z=15$

6. anonymous

so the answer should be 5?

7. anonymous

8. anonymous

and you can solve this for z by writing $z^2-2z-15=0$ $(z-5)(z+3)=0$ so $z=5$ or $z=-3$

9. anonymous

that is what i was getting 5 or -3 but then when i try to substitute back in i do not get anything that looks right. i absolutely think that i am doing something wrong still

10. anonymous

but now we have to remember that $z=x^{\frac{1}{3}}$so that means that $x^{\frac{1}{3}}=5$ or $x^{\frac{1}{3}}=-3$

11. anonymous

and so $x=(5)^2=125$ or $x=(-3)^3=-27$

12. anonymous

so i substitute back in -25 or 127?

13. anonymous

i mean -27 or 125

14. anonymous

you substitute x^(1/3) back in for z

15. anonymous

x^(1/3)=5 x^(1/3)=-3

16. anonymous

then cube both sides of each equation to find x

17. anonymous

(x^(1/3))^3=5^3 x=125 (x^(1/3))^3=-3^3 x=-27

18. anonymous

so both are good solutions? im trying to follow sorry, i am just very bad at this

19. anonymous

so both are good solutions? im trying to follow sorry, i am just very bad at this

20. anonymous

Yes both are good solutions because they are in the domain of the original function.