anonymous
  • anonymous
can some one tell me how to get factors i mean method 3b^2-20b-12
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
x=−b±(√b2−4ac)/2a
anonymous
  • anonymous
aaaaa what's that ? i don't think that's the way
anonymous
  • anonymous
this will give u two value for x..........& say it x1,x2...... now factor is (x-x1)(x-x2)...ok

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anonymous
  • anonymous
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anonymous
  • anonymous
what??!! no no!
anonymous
  • anonymous
(b-6) (3b-2) am i right ?
ash2326
  • ash2326
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 \[(x-(\frac{10+ \sqrt{34}}{3}))(x-(\frac{10- \sqrt{34}}{3}))\]
ash2326
  • ash2326
Sorry it'd be \[(x-(\frac{10+ 2\sqrt{34}}{3}))(x-(\frac{10 -2\sqrt{34}}{3}))\] I missed the 2:(
anonymous
  • anonymous
ah that's too advance i guess but right factors (b-6)(3b-2)
ash2326
  • ash2326
No, these are not the factors \[ (b-6)(3b-2)= 3b^2-20b+12\] but in your question you have \[3b^2-20b-12\]
anonymous
  • anonymous
yah :) do u know how to get (b-6)(3b-2)

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