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By factoring how do you solve 3x^2/311x^1/34=0? My teacher doesn't do substitution. please and thank you.
 one year ago
 one year ago
By factoring how do you solve 3x^2/311x^1/34=0? My teacher doesn't do substitution. please and thank you.
 one year ago
 one year ago

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cwrw238Best ResponseYou've already chosen the best response.1
(3x^1/3 + 1)(x^1/3  4 ) = 0
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
3x^1/3 + 1 = 0 x^(1/3) = 1/3 x^1/3  4 = 0 x^1/3 = 4
 one year ago

kitsune0724Best ResponseYou've already chosen the best response.0
(1/3)^3 or (1/3)^3?
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
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
 one year ago

kitsune0724Best ResponseYou've already chosen the best response.0
Thank you so much.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
note x^(1/3) is same as cube root x
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
so to find other root you do same thing with x^1/3 = 4
 one year ago

kitsune0724Best ResponseYou've already chosen the best response.0
Thank you again. But do you divide x^1/3=4? Like 4/1=4.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
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
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
i suggest you review the laws of indices x^a + x^b = x^(a+b) x^a / x^b = x^(ab) (x^a)^b = x^(a*b) = x^ ab
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
also dw:1342632977864:dw
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
please ask if you need any more help
 one year ago

kitsune0724Best ResponseYou've already chosen the best response.0
Thank you so much.
 one year ago

kitsune0724Best ResponseYou've already chosen the best response.0
Could you please help with another problem?
 one year ago
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