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Alison123

  • 5 years ago

how do you solve 2/3 x squared + x= 4 + 2/3

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  1. anonymous
    • 5 years ago
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    This? \[\frac{2}{3}x^{2}+x= 4+\frac{2}{3}\]

  2. Alison123
    • 5 years ago
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    yeah that

  3. anonymous
    • 5 years ago
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    When you have a polynomial like this, you want to get everything on one side so it equals 0:\[\frac{2}{3}x^{2} + x = \frac{14}{3}\]\[\frac{2}{3}x^{2} + \frac{3}{3}x - \frac{14}{3} = 0\] Then to make it easier, you can divide the whole thing by 3:\[2x^{2} + 3x -14 = 0\] Then you can factor it:\[(2x+7)(x-2)=0\] And now it can be solved, because for this to be zero, at least one of the parentheses needs to equal 0. So solve the first parentheses: \[(2x+7) = 0\]\[2x = -7\]\[x = -\frac{7}{2}\] Then the 2nd:\[(x-2) = 0\]\[x=2\] So you have 2 values that would work for x: -7/2 and 2

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