anonymous
  • anonymous
solve (y+1)(y-3) = -6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Luigi0210
  • Luigi0210
Welcome to openstudy :) First what you do is FOIL
anonymous
  • anonymous
foil what part though? I have only had problems that looks like 3(y+1)(y-3)=whatever
anonymous
  • anonymous
You must expand it out like so: \[(y+1)(y-3)=y(y-3)+1(y-3)=y^2-3y+y-3=y^2-2y+3\] So you have \[y^2-2y+3=-6\] And you can move everything over to one side like so: \[y^2-2y+3+6=0\phantom{spc}\rightarrow\phantom{spc}y^2-2y+9=0\] So now that you have a quadratic equation, you can use the quadratic formula

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anonymous
  • anonymous
I can demonstrate the quadratic equation if you want!
Luigi0210
  • Luigi0210
|dw:1378247768513:dw|
anonymous
  • anonymous
should it be -3? how did it turn into +3?
anonymous
  • anonymous
Im sorry your right anna
Luigi0210
  • Luigi0210
Nice catch, I was about to say.
anonymous
  • anonymous
So you would actually obtain the equation \[y^2-2y+3\] Better? :)
anonymous
  • anonymous
ok!! just making sure!! thank you so much!! I have so many math problems to do..and I have not been in algebra since high school!!
Luigi0210
  • Luigi0210
It's okay, you will learn it with time.
anonymous
  • anonymous
Anytime :) So, in general, for an equation in the form of \(ax^2+bx+c=0\), the solutions are the equation: \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Luigi0210
  • Luigi0210
..Or you could just try factoring.
anonymous
  • anonymous
can we finish the quadratic equation? I think i have it..i just want to make sure!
anonymous
  • anonymous
@Luigi0210 That would be a quicker method but im trying to give a more general answer
anonymous
  • anonymous
And so for the equation of \(y^2-2y+3=0\), we have: \(a=1\) \(b=-2\) \(c=3\) And we can sub these values into the quadratic equation: \[x=\frac{2\pm\sqrt{4-4(1)(3)}}{2}\] \[x=\frac{2\pm\sqrt{-8}}{2}\] \[x=\frac{2\pm2\sqrt{-2}}{2}\] \[x=1\pm\sqrt{-2}\] So therefore since the sign under the root sign is negative, this equation has no "real" roots. Only imaginary ones. So in math we made a unit, \(i\) called the imaginary number satisfying \(i^2=-1\). So we define it as \(i=\sqrt{-1}\) Which of course is not a comprehendable number but one which we use nonetheless. So we just sid that the roots are: \[x=1\pm\sqrt{-2}=1\pm\sqrt{-1*2}=1\pm\sqrt{-1}\sqrt{2}=1\pm i\sqrt{2}\] Those are the two imaginary rootsto the equation.
anonymous
  • anonymous
ok i started with a 9 as c..
anonymous
  • anonymous
Haha noo remember I made a mistake and that it actually was supposed to be \(y^2-2y+3\) ?
anonymous
  • anonymous
ok got it!! thanks!!
anonymous
  • anonymous
Anytime Anna :-)

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