Find all the series of the equation.
-3x^4+27x^2+1200=0
Use quadratic formula and show all work.
4 answers
I meant zeroes, not series
what ypu have is a quadratic in x^2:
-3(x^2)^2 + 27(x^2) + 1200 = 0
so, using the quadratic formula,
x^2 = (-27±√15129)/-6 = (-27±123)/-6
x^2 = -16 or 25
so, x = ±4i,±5
-3(x^2)^2 + 27(x^2) + 1200 = 0
so, using the quadratic formula,
x^2 = (-27±√15129)/-6 = (-27±123)/-6
x^2 = -16 or 25
so, x = ±4i,±5
First< I would divide each term by -3 to get
x^4 - 9x^2 - 400
can you think of 2 numbers which multiply to get -400, and add to -9 ?
how about -25 and 9 ?
(x^2 + 9)(x^2 - 25) = 0
x^2 = -9 or x^2 = 25
x = ± 3i or x = ± 5
x^4 - 9x^2 - 400
can you think of 2 numbers which multiply to get -400, and add to -9 ?
how about -25 and 9 ?
(x^2 + 9)(x^2 - 25) = 0
x^2 = -9 or x^2 = 25
x = ± 3i or x = ± 5
oops, first factor is x^2 - 16
so x = ± 4i, see oobleck's above
so x = ± 4i, see oobleck's above
1 answer