anonymous
  • anonymous
Factor x2 - 2x + 3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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aric200
  • aric200
Do you mean Factor x^2 - 2x + 3
UsukiDoll
  • UsukiDoll
@aric200 yes that's got to be the equation
UsukiDoll
  • UsukiDoll
anyway , to factor we focus on the middle term and the last term first then the sign patterns of the equation

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anonymous
  • anonymous
@aric200 yes sir
UsukiDoll
  • UsukiDoll
so the last term we have is a 3 and there is only one combination which is 1 x 3 = 3 so now let's grab the middle term. We notice that there is a -2 so our 1 must be positive and our 3 must be negative 1-3 = -2
aric200
  • aric200
It bothers me when people do not put the ^Carrots.
UsukiDoll
  • UsukiDoll
or am I reading this too fast? wait... I'm using discriminant formula
UsukiDoll
  • UsukiDoll
(-2)^2-4(1)(3) 4-12 = -8 oh nice.. we can't factor this -_- quadratic formula time
UsukiDoll
  • UsukiDoll
\[\frac{-b \pm \sqrt{b^2-4ac} }{2a} \]
UsukiDoll
  • UsukiDoll
so with the discriminant formula we already figured out the square root portion which was -8, so we let a = 1 b = -2 and c = 3 because our equation \[x^2-2x+3\] in the form of the standard quadratic equation which is \[ax^2+bx+c\]
UsukiDoll
  • UsukiDoll
now we just plug in our data \[\frac{2 \pm \sqrt{-8} }{2}\]
UsukiDoll
  • UsukiDoll
but we are not done... we can split that 8 up and negatives aren't allowed in the radicals which means we are going to get an imaginary or i out of this
UsukiDoll
  • UsukiDoll
\[\frac{2 \pm 2\sqrt{2}i }{2}\]
UsukiDoll
  • UsukiDoll
now we can get split this up and get rid of the 2's
UsukiDoll
  • UsukiDoll
\[\frac{2+2 \sqrt{2}i}{2}, \frac{2-2 \sqrt{2}i}{2}\]
UsukiDoll
  • UsukiDoll
\[1+\sqrt{2}i, 1-\sqrt{2}i\]
UsukiDoll
  • UsukiDoll
which is our roots..
anonymous
  • anonymous
@UsukiDoll so thats it?!
UsukiDoll
  • UsukiDoll
I think it should look like this? not sure \[(x-(1+\sqrt{2}i))(x-(1-\sqrt{2}i) )\]
UsukiDoll
  • UsukiDoll
@ganeshie8 I'm missing something in here...aren't I?
UsukiDoll
  • UsukiDoll
it's just that when I use to do those problems, all of my professors care about are the roots and I didn't have to put it back into factored form like what I just typed.
ganeshie8
  • ganeshie8
yeah the given quadratic cannot be factored over integers, it is prime

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