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dddan

  • 3 years ago

please help!! how do u solve for b^3+20b=9b^2?

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  1. campbell_st
    • 3 years ago
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    rewrite the equation so that you have \[b^3 - 9b^2 + 20b = 0\] now take b as a common factor \[b(b^2 - 9b + 20) = 0\] factor the quadratic in the brackets... you need the factors of 20 than add to -9.... they are both negatives... then you should be able to find the solutions.

  2. dddan
    • 3 years ago
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    so is it 5 and 4?

  3. campbell_st
    • 3 years ago
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    well you'll have \[b(b -5)(b-4) = 0\] so the solutions are b = 0, 4, 5

  4. dddan
    • 3 years ago
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    thank you

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