Question

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

Answer 1

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

Answer 2

can you continue?

Answer 3

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

x^1/3 - 4 = 0
x^1/3 = 4

Answer 4

let me see...

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

* correctly

Answer 9

Thank you so much.

Answer 10

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

Answer 11

so to find other root you do same thing with

x^1/3 = 4

Answer 12

yw

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

also