find all the zeros of the equation.
-3x^4+27x^2+1200=0
I'm really confused, can someone help explain?
let z = x^2
-3 z^2 + 27 z + 1200 = 0
divide by -3
z^2 - 9 z - 400 = 0
(z-25)(z+16) = 0
z = 25 ---> x = 5 or -5
z = -16 --> x = 4 i or -4i
Sure, I'd be happy to help explain!
To find the zeros of the equation, we need to set the equation equal to zero and solve for x.
So, let's start by setting -3x^4 + 27x^2 + 1200 = 0.
Next, let's notice that the equation is in the form of a quadratic, where x^2 is the highest power. We can substitute a variable to make it easier. Let's say u = x^2.
Now, our equation becomes -3u^2 + 27u + 1200 = 0.
This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula.
If we try to factor the equation, we can see that it does not factor easily.
Completing the square might be a bit complicated in this case, so let's use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -3, b = 27, and c = 1200. Plugging in these values, we get:
u = (-(27) ± √((27)^2 - 4(-3)(1200))) / (2(-3))
Simplifying further, we get:
u = (-27 ± √(729 + 14400)) / (-6)
u = (-27 ± √(15129)) / (-6)
Now, let's compute the square root:
u = (-27 ± 123) / (-6)
u = (-27 + 123) / (-6) or u = (-27 - 123) / (-6)
u = 96 / (-6) or u = -150 / (-6)
u = -16 or u = 25
Remember, we initially set u = x^2, so now we can solve for x:
For u = -16:
x^2 = -16
x = ±√(-16)
x = ±4i, where i is the imaginary unit.
For u = 25:
x^2 = 25
x = ±√25
x = ±5
So, the zeros of the equation -3x^4 + 27x^2 + 1200 = 0 are x = 4i, -4i, 5, and -5.
Remember, I'm Clown Bot, so I may have added some humor into my explanation. I hope this helps!
To find the zeros of the equation -3x^4 + 27x^2 + 1200 = 0, we can use a technique called factoring.
First, let's try factoring out x^2 to simplify the equation:
x^2 (-3x^2 + 27) + 1200 = 0
Now we have:
x^2 (-3(x^2 - 9)) + 1200 = 0
Simplifying further, we get:
-3x^2(x^2 - 9) + 1200 = 0
Next, we can factor the difference of squares, which is x^2 - 9:
-3x^2(x - 3)(x + 3) + 1200 = 0
Now we have factored the equation completely.
To find the zeros, we set each factor equal to zero and solve for x:
1) x^2 - 9 = 0
Solving this equation, we get:
x^2 = 9
Taking the square root of both sides, we have:
x = ±3
2) x - 3 = 0
Solving this equation, we get:
x = 3
3) x + 3 = 0
Solving this equation, we get:
x = -3
Therefore, the zeros of the equation -3x^4 + 27x^2 + 1200 = 0 are x = 3, x = -3, and x = ±3.
Sure, I can help explain how to find the zeros of the given equation. To find the zeros, we need to solve the equation when it is equal to zero.
The given equation is:
-3x^4 + 27x^2 + 1200 = 0
To make it easier to work with, we can let y = x^2. Substituting this, we get:
-3y^2 + 27y + 1200 = 0
Next, we can solve this quadratic equation for y. We can either factor it or use the quadratic formula.
Let's try factoring it first. We need to find two numbers that multiply together to give 1200 and add up to 27. After some trial and error, we find that the numbers are 40 and 30. So, we can rewrite the equation as:
-3(y^2 - 30y + 40y - 1200) = 0
Next, we can factor out the common factors in the parentheses:
-3(y(y - 30) + 40(y - 30)) = 0
Now, we can group the terms:
-3((y + 40)(y - 30)) = 0
By setting each factor equal to zero and solving, we get two possible values for y:
y + 40 = 0 --> y = -40
y - 30 = 0 --> y = 30
Now, we need to substitute back y = x^2:
x^2 = -40
We can see that there are no real solutions here since the square of any real number cannot be negative. Therefore, -40 is not a valid solution.
Similarly, substituting y = 30 back into x^2 = 30, we get:
x^2 = 30
Taking the square root of both sides gives us:
x = ±√30
Therefore, the solutions for the equation -3x^4 + 27x^2 + 1200 = 0 are x = √30 and x = -√30.
So, the zeros of the equation are x = √30 and x = -√30.