anonymous
  • anonymous
Find all the zeros of the equation: -3x^4+27x^2+1200=0 PLEASE HELP!! I don't know how to do this and it would be very helpful if someone could explain
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
their are 3 0's
Michele_Laino
  • Michele_Laino
first step: I introduce a new variable \(z\) such that \(z=x^2\) so I can reqrite the equation like below: \[ - 3{z^2} + 27z + 1200 = 0\] which can be solved easily with respect to \(z\)
Michele_Laino
  • Michele_Laino
oops.. I can rewrite*...

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anonymous
  • anonymous
Okay, so then what is the next step?
Michele_Laino
  • Michele_Laino
we have to solve such equation for \(z\). In order to do that, I factor out \(-3\) so I can write this: \[ - 3\left( {{z^2} - 9z - 400} \right) = 0\] tefore we have to solve this equation: \[{z^2} - 9z - 400 = 0\] using the standard formula: \[z = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] where \(a=1,\;b=-9,\;c=-400\)
Michele_Laino
  • Michele_Laino
oops.. therefore*
anonymous
  • anonymous
Okay, give me a min. I'll do that.
anonymous
  • anonymous
Z=25, or Z=-16?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Okay, now what?
Michele_Laino
  • Michele_Laino
next I use the definition of the variable \(z\), so I get the two subsequent quadratic equations: \[\begin{gathered} {x^2} = 25 \hfill \\ {x^2} = - 16 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
since the square of any real number is always a positive number, then we can see that the second quadratic equation has not solution
Michele_Laino
  • Michele_Laino
namely there is not a number \(n\) such that \(n^2=-16\), am i right?
anonymous
  • anonymous
Yes.
Michele_Laino
  • Michele_Laino
ok! Therefore we have to solve this equation: \(\huge x^2=25\) please what are the solution of such equation?
anonymous
  • anonymous
x=5
Michele_Laino
  • Michele_Laino
hint: what is \((-5)^2=...?\)
anonymous
  • anonymous
25
Michele_Laino
  • Michele_Laino
thta's right! So the real solution of your starting equation, are: \(\huge x_1=5,\;x_2=-5\)
Michele_Laino
  • Michele_Laino
oops..solutions*
anonymous
  • anonymous
Okay! Now what?
Michele_Laino
  • Michele_Laino
we have finished. More precisely, your equation is a fourth grade equation, so we have two real solutions, and two imaginary solutions
anonymous
  • anonymous
So it's 5, -5, and then what?
Michele_Laino
  • Michele_Laino
the other imaginary solutions, are given solving this equation: \(x^2=-16\) such solutions are: \(x_3=4i,\;x_4=-4i\) where \(i\) is such that \(i^2=-1\)
anonymous
  • anonymous
Okay! That actually makes sense. Thanks so much for the help! You're a lifesaver.
Michele_Laino
  • Michele_Laino
:)

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