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anonymous
 one year ago
Factor x2  2x + 3
anonymous
 one year ago
Factor x2  2x + 3

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aric200
 one year ago
Best ResponseYou've already chosen the best response.1Do you mean Factor x^2  2x + 3

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2@aric200 yes that's got to be the equation

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2anyway , to factor we focus on the middle term and the last term first then the sign patterns of the equation

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so 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 13 = 2

aric200
 one year ago
Best ResponseYou've already chosen the best response.1It bothers me when people do not put the ^Carrots.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2or am I reading this too fast? wait... I'm using discriminant formula

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2(2)^24(1)(3) 412 = 8 oh nice.. we can't factor this _ quadratic formula time

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{b \pm \sqrt{b^24ac} }{2a} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so 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 \[x^22x+3\] in the form of the standard quadratic equation which is \[ax^2+bx+c\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now we just plug in our data \[\frac{2 \pm \sqrt{8} }{2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{2 \pm 2\sqrt{2}i }{2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now we can get split this up and get rid of the 2's

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{2+2 \sqrt{2}i}{2}, \frac{22 \sqrt{2}i}{2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[1+\sqrt{2}i, 1\sqrt{2}i\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2which is our roots..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll so thats it?!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I think it should look like this? not sure \[(x(1+\sqrt{2}i))(x(1\sqrt{2}i) )\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 I'm missing something in here...aren't I?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2it'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
 one year ago
Best ResponseYou've already chosen the best response.0yeah the given quadratic cannot be factored over integers, it is prime
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