anonymous
  • anonymous
if a=2 ; b=-5; c=5 how would you solve the quadratic equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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hartnn
  • hartnn
then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
anonymous
  • anonymous
Use the quadratic foumula \[x_1 =\frac{ -b + \sqrt{b^2-4ac}} {2a}\] \[x_2 =\frac{ -b - \sqrt{b^2-4ac}} {2a}\]
anonymous
  • anonymous
ya but i got 4+- the square root of -24 / 4 does this make sense?

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hartnn
  • hartnn
your b=-5, how did u get 4+/- ... ?
anonymous
  • anonymous
sry the b=-4 and the c=5
anonymous
  • anonymous
You have imaginary roots. Is that what you're worried about?
anonymous
  • anonymous
yup
anonymous
  • anonymous
Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients
hartnn
  • hartnn
u actually have imaginary roots for this..
anonymous
  • anonymous
alright can u tell me where to take it from the imaginary roots?
hartnn
  • hartnn
how did u get square root of -24 ?
anonymous
  • anonymous
because with the coefficients, its (-4)^2-4(2)(5) , in the square root right?
hartnn
  • hartnn
lol! b=-5 not -4
anonymous
  • anonymous
no i i mixed up the coefficents in the original question. i was given a=2;b=-4;c=5
hartnn
  • hartnn
ok, \(\huge{x=\frac{4 \pm \sqrt{-24}}{4}}=\frac{4 \pm 2\sqrt{-6}}{4}=\frac{2\pm i\sqrt{6}}{2}\)
anonymous
  • anonymous
alright thank you very much!

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