Destinyyyy
  • Destinyyyy
Simplify the equation. Then solve by the quadratic formula..
Mathematics
jamiebookeater
  • jamiebookeater
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Destinyyyy
  • Destinyyyy
3x^2-4x+3=0
Destinyyyy
  • Destinyyyy
x= -(-4)+- square root (-4)^2 -4(3)(3)/2(3)
Destinyyyy
  • Destinyyyy
4+square root 16-36/6 4+- square root -20/6 6(4+-4i square root 5)/6 cancel out the 6 My final answer is 4+-2i square root 5 but its wrong

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anonymous
  • anonymous
why don't you just use (4 +/- sqrt(20)i)/6
Destinyyyy
  • Destinyyyy
huh?? O.o
zzr0ck3r
  • zzr0ck3r
\(3x^2-4x+3=0\implies \dfrac{-(-4)\pm \sqrt{(-4)^2-4(3)(3)}}{2(3)}=\dfrac{4\pm\sqrt{16-36}}{6}\\=\dfrac{4\pm2\sqrt{5}i}{6}=\dfrac{2\pm\sqrt{5}i}{3}\)
Destinyyyy
  • Destinyyyy
Sorry.. Im assuming the 2 is from 4... But im a bit confused by the rest
Destinyyyy
  • Destinyyyy
And shouldn't the i be before square root 5???
zzr0ck3r
  • zzr0ck3r
the order of i does not matter, and I will show some more steps
zzr0ck3r
  • zzr0ck3r
\(\dfrac{4\pm2\sqrt{5}i}{6}=\dfrac{2(2+\sqrt{5}i)}{2*3}=(\dfrac{2}{2})\dfrac{2+\sqrt{5}i}{3}=(\dfrac{\cancel{2}}{\cancel{2}})\dfrac{2+\sqrt{5}i}{3}=\dfrac{2\pm\sqrt{5}i}{3}=\dfrac{2\pm i\sqrt{5}}{3}\)
Destinyyyy
  • Destinyyyy
What is the last one??
zzr0ck3r
  • zzr0ck3r
\(\dfrac{4\pm2\sqrt{5}i}{6}=\dfrac{2(2+\sqrt{5}i)}{2*3}=(\dfrac{2}{2})\dfrac{2+\sqrt{5}i}{3}=(\dfrac{\cancel{2}}{\cancel{2}})\dfrac{2+\sqrt{5}i}{3}=\\\dfrac{2\pm\sqrt{5}i}{3}=\dfrac{2\pm i\sqrt{5}}{3}\)

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