anonymous
  • anonymous
Can you simplify this further? *Using the quadratic formula to factor x^2-6x-4* The answer is supposed to be 6(+-)√53 / 2 ,but can't you simplify by 2? 6 and 2 go into 2. Then could it be 3(+-)√68/1 or no?
Mathematics
jamiebookeater
  • jamiebookeater
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Hero
  • Hero
Hi neverforgettovisitme
anonymous
  • anonymous
hi
anonymous
  • anonymous
lol

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anonymous
  • anonymous
6(+-)√53 / 2 is simplified
Hero
  • Hero
:D
anonymous
  • anonymous
answer is is \[3\pm\sqrt{13}\]
anonymous
  • anonymous
wait how is it simplified? in another problem there was a problem where you divided by 3 because it was a common factor
anonymous
  • anonymous
wait i'll go look for it
anonymous
  • anonymous
-9x^2-9x+6
anonymous
  • anonymous
no you cannot because the number inside the radical contains no perfect squares. this is the best you can do
anonymous
  • anonymous
later in the problem the guy got 9(+-)3√33 --------- -18 then he simplified by 3
anonymous
  • anonymous
khanacademy
anonymous
  • anonymous
for \[-9x^2-9x+6=0\] you can start with \[3x^2+3x-2=0\] and reduce before you start
anonymous
  • anonymous
oh i see
anonymous
  • anonymous
hold the phone solution to \[ x^2-6x-4=0\]is not what you wrote. it is what tomas A wrote
anonymous
  • anonymous
\[\frac{9\pm3\sqrt{33}}{-18}=\] you can simplify this since \[\frac{3(3\pm\sqrt{33})}{-6\cdot3}=\] \[\frac{1(3\pm\sqrt{33})}{-6\cdot1}=\]
anonymous
  • anonymous
\[x^2-6x-4=0\] \[x^2-6x=4\] \[(x-3)^2=4+9=13\] \[x-3=\pm\sqrt{13}\] \[x=3\pm\sqrt{13}\]

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