kitsune0724
By factoring how do you solve 3x^2/3-11x^1/3-4=0? My teacher doesn't do substitution. please and thank you.
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cwrw238
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(3x^1/3 + 1)(x^1/3 - 4 ) = 0
cwrw238
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can you continue?
cwrw238
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3x^1/3 + 1 = 0
x^(1/3) = -1/3
x^1/3 - 4 = 0
x^1/3 = 4
kitsune0724
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let me see...
cwrw238
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hint: x = x(1/3)^3
kitsune0724
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(-1/3)^3 or (1/3)^3?
cwrw238
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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
cwrw238
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* correctly
kitsune0724
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Thank you so much.
cwrw238
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note x^(1/3) is same as cube root x
cwrw238
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so to find other root you do same thing with
x^1/3 = 4
cwrw238
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yw
kitsune0724
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Thank you again. But do you divide x^1/3=4? Like 4/1=4.
cwrw238
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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
cwrw238
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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
cwrw238
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also
|dw:1342632977864:dw|
cwrw238
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please ask if you need any more help
kitsune0724
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Thank you so much.
cwrw238
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yw
kitsune0724
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Could you please help with another problem?