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kanigayan

  • 2 years ago

can some one tell me how to get factors i mean method 3b^2-20b-12

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  1. fontez
    • 2 years ago
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    x=−b±(√b2−4ac)/2a

  2. kanigayan
    • 2 years ago
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    aaaaa what's that ? i don't think that's the way

  3. amitlpu91
    • 2 years ago
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    this will give u two value for x..........& say it x1,x2...... now factor is (x-x1)(x-x2)...ok

  4. kanigayan
    • 2 years ago
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    |dw:1330616930509:dw|

  5. fontez
    • 2 years ago
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    what??!! no no!

  6. kanigayan
    • 2 years ago
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    (b-6) (3b-2) am i right ?

  7. ash2326
    • 2 years ago
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    we have \[3b^2-20b-12\] standard quadratic equation \[ax^2+bx+c=0\] we know its solution is given as \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] here a=3 b=-20 and c=-12 so \[x=\frac{-(-20)\pm \sqrt{(-20)^2-4\times 3\times -12}}{2\times 3}\] we get \[x=\frac{20\pm \sqrt{400+144}}{6}\] now we get \[x=\frac{20\pm \sqrt{400+144}}{6}\] we get \[x=\frac{20\pm \sqrt{544}}{6}\] \[544= 17\times 16\times 2\] so \[x=\frac{20\pm \sqrt{16\times 34}}{6}\] we get \[x=\frac{20\pm 4\sqrt{ 34}}{6}\] let's divide by 2 \[x=\frac{10\pm \sqrt{34}}{3}\] so the two factors are \[(x-(\frac{10+ \sqrt{34}}{3}))(x-(\frac{10- \sqrt{34}}{3}))\]

  8. ash2326
    • 2 years ago
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    Sorry it'd be \[(x-(\frac{10+ 2\sqrt{34}}{3}))(x-(\frac{10 -2\sqrt{34}}{3}))\] I missed the 2:(

  9. kanigayan
    • 2 years ago
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    ah that's too advance i guess but right factors (b-6)(3b-2)

  10. ash2326
    • 2 years ago
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    No, these are not the factors \[ (b-6)(3b-2)= 3b^2-20b+12\] but in your question you have \[3b^2-20b-12\]

  11. kanigayan
    • 2 years ago
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    yah :) do u know how to get (b-6)(3b-2)

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