Question

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

Answer 1

their are 3 0's

Answer 2

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 \(z\)

Answer 3

oops.. I can rewrite*...

Answer 4

Okay, so then what is the next step?

Answer 5

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 \(a=1,\;b=-9,\;c=-400\)

Answer 6

oops.. therefore*

Answer 7

Okay, give me a min. I'll do that.

Answer 8

Z=25, or Z=-16?

Answer 9

that's right!

Answer 10

Okay, now what?

Answer 11

next I use the definition of the variable \(z\), so I get the two subsequent quadratic equations:
\[\begin{gathered}
{x^2} = 25 \hfill \\
{x^2} = - 16 \hfill \\
\end{gathered} \]

Answer 12

since the square of any real number is always a positive number, then we can see that the second quadratic equation has not solution

Answer 13

namely there is not a number \(n\) such that \(n^2=-16\), am i right?

Answer 14

Yes.

Answer 15

ok! Therefore we have to solve this equation:
\(\huge x^2=25\)
please what are the solution of such equation?

Answer 16

x=5

Answer 17

hint:
what is \((-5)^2=...?\)

Answer 18

25

Answer 19

thta's right! So the real solution of your starting equation, are:

\(\huge x_1=5,\;x_2=-5\)

Answer 20

oops..solutions*

Answer 21

Okay! Now what?

Answer 22

we have finished. More precisely, your equation is a fourth grade equation, so we have two real solutions, and two imaginary solutions

Answer 23

So it's 5, -5, and then what?

Answer 24

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 \(i^2=-1\)

Answer 25

Okay! That actually makes sense. Thanks so much for the help! You're a lifesaver.

Answer 26

:)