Question

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!

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

so the answer should be 5?

Answer 7

there are two answers

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

and so
\[x=(5)^2=125\] or
\[x=(-3)^3=-27\]

Answer 12

so i substitute back in -25 or 127?

Answer 13

i mean -27 or 125

Answer 14

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

Answer 15

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

Answer 16

then cube both sides of each equation to find x

Answer 17

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

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

Answer 18

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

Answer 19

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

Answer 20

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