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abbie1

  • 3 years ago

What are the restrictions for 4x^2-x+4x/5(3x-1)^2?? Please help!!!

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  1. abbie1
    • 3 years ago
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    @hartnn ?? Can you help me! I have been working on this problem for forever.

  2. hartnn
    • 3 years ago
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    restrictions on values of 'x' right ?

  3. abbie1
    • 3 years ago
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    Yes I already simplified it and multiplied it together, and this is the end product. But I don't know the restrictions bc it's a complex denominator.

  4. hartnn
    • 3 years ago
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    complex denominator ? denominator = 5 (3x-1)^2 right ? restrictions : denominator = 0 so, ''x can't take values for which denominator = 5 (3x-1)^2 = 0 so, 3x-1 = 0 can you solve this for x ?

  5. abbie1
    • 3 years ago
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    Oh, so it's 1/3, but what about the 5? And would you write the final answer-numerator like 4x^2-x+4x-1, or (x+1)(4x-1)?

  6. hartnn
    • 3 years ago
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    about the 5 : \(5 (3x-1)^2=0 \\ \) divide both sides by 5 , 0/5 = 0 \((3x-1)^2=0\) take square root, 3x-1 = 0 x=1/3 is the restriction. numerator is a polynomial , so in this case it does not play any role.

  7. abbie1
    • 3 years ago
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    THANK YOU!!!! :D

  8. hartnn
    • 3 years ago
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    welcome ^_^

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