find all the zeros of the equation -3x^4+27x^2+1200=0
6 years ago
6 years ago
pre cal
6 years ago
I think they named it "PLZ" because they wanted someone to please help them. My understanding is that "PLZ" translates to please.
6 years ago
let z = x^2
then you have
3 z^2 -27 z -1200 = 0
or
z^2 - 9 z - 1200 = 0
z is about 40 or -30
(you can do quadratic accurately)
so
x = +sqrt 40 or - sqrt 40 or +i sqrt 30 or -i sqrt 30
11 months ago
To find the zeros of the equation -3x^4 + 27x^2 + 1200 = 0, we can use the quadratic formula or factor the equation if possible.
First, let's rewrite the equation by factoring out the common factor of -3:
-3(x^4 - 9x^2 - 400) = 0
Now, let's focus on solving the equation within the parentheses: x^4 - 9x^2 - 400 = 0.
To make this equation easier to work with, let's substitute a new variable. Let's say u = x^2. Now, we have:
u^2 - 9u - 400 = 0
We can now factor this quadratic equation:
(u - 25)(u + 16) = 0
Now, we need to substitute back u = x^2 into the factored equation:
(x^2 - 25)(x^2 + 16) = 0
Using the difference of squares formula, we can further factor:
(x - 5)(x + 5)(x^2 + 16) = 0
Setting each factor to zero, we get:
x - 5 = 0 -> x = 5
x + 5 = 0 -> x = -5
x^2 + 16 = 0 -> x^2 = -16 -> x = ±√(-16)
Since taking the square root of a negative number results in imaginary solutions, the only real solutions are x = 5 and x = -5.
Therefore, the zeros of the equation -3x^4 + 27x^2 + 1200 = 0 are x = 5 and x = -5.