Find all the series of the equation.
-3x^4+27x^2+1200=0
Use quadratic formula and show all work.
4 years ago
4 years ago
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
4 years ago
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
4 years ago
oops, first factor is x^2 - 16
so x = ± 4i, see oobleck's above
11 months ago
To find the series of the given equation, we can use the quadratic formula. The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solutions can be obtained using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our given equation: -3x^4 + 27x^2 + 1200 = 0, we can rearrange it to match the quadratic form by dividing through by -3:
x^4 - 9x^2 - 400 = 0
Comparing this equation with the quadratic form ax^2 + bx + c = 0, we find that a = 1, b = 0, and c = -400.
Now, let's substitute these values into the quadratic formula:
x = (-0 ± √(0^2 - 4(1)(-400))) / (2(1))
x = (± √1600) / 2
x = ± 40 / 2
Simplifying further, we have:
x = ± 20
So, the series of the equation is: x = -20 and x = 20.
By substituting these values back into the original equation, we can confirm that they satisfy it.