anonymous
  • anonymous
Find the zeros of the function by using the Quadratic Formula. f(x) = 2x^2 - 2x + 3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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angela210793
  • angela210793
|dw:1332532630999:dw|
anonymous
  • anonymous
teacher said it was wrong :(
AccessDenied
  • AccessDenied
Does your teacher want complex solutions?

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anonymous
  • anonymous
a square root is involved in the answer somehow
angela210793
  • angela210793
huh????????????? O.o see u have |dw:1332532972855:dw|
AccessDenied
  • AccessDenied
Like, she was checking this part of the quadratic formula: \[ x = \frac{-b \pm \sqrt{\color{green}{b^2 - 4ac}}}{2a} \] if that's a negative value, then there are no real zeroes to the function
angela210793
  • angela210793
@AccessDenied is the right formula the one i wrote when D=0???? i don't remember it very well :S
AccessDenied
  • AccessDenied
umm, as in, the solution when D=0? It'd be the vertex -b/2a, since if D=0 then the square root of 0 is 0, so -b +- 0 is -b either way
angela210793
  • angela210793
i thought that..and then erased -_- thanks xD
AccessDenied
  • AccessDenied
I guess if your teacher wants the complex zeroes... \[ \begin{split} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a};~~~a=2,~b=-2,~and~c = 3\\ &= \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(3)}}{2(2)}\\ &= \frac{2 \pm \sqrt{4 - 24}}{4}\\ &= \frac{2 \pm \sqrt{\color{blue}{-20}}}{4}\\ &= \frac{2 \pm \sqrt{4}\sqrt{5}\sqrt{\color{blue}{-1}}}{4}\\ &= \frac{\cancel{2} 1 \pm \cancel{2}1\color{blue}i\sqrt{5}}{\cancel{4}2}\\ &= \frac{1 \pm \color{blue}i\sqrt{5}}{2} \end{split} \]

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