Find all the zeroes of the equation.
–3x4+ 27x2 + 1200 = 0
To solve the equation -3x^4 + 27x^2 + 1200 = 0, we can substitute y = x^2, which gives us the equation -3y^2 + 27y + 1200 = 0.
Next, we can solve this quadratic equation for y. Using the quadratic formula, y = (-b ± √(b^2 - 4ac))/(2a), where a = -3, b = 27, and c = 1200.
y = (-(27) ± √((27)^2 - 4(-3)(1200)))/(2(-3))
= (-27 ± √(729 + 14400))/(-6)
= (-27 ± √(15129))/(-6)
= (-27 ± 123)/(-6)
Therefore, y = (-27 + 123)/(-6) or y = (-27 - 123)/(-6).
For y = (-27 + 123)/(-6), we get y = 96/(-6) = -16.
For y = (-27 - 123)/(-6), we get y = -150/(-6) = 25.
Now, we substitute y back into the equation y = x^2, so x^2 = -16 or x^2 = 25.
For x^2 = -16, there are no real solutions because the square of a real number cannot be negative.
For x^2 = 25, we take the square root of both sides, giving x = ±√(25).
Therefore, the zeroes of the equation -3x^4 + 27x^2 + 1200 = 0 are x = -5 and x = 5.
To find the zeroes of the equation –3x^4 + 27x^2 + 1200 = 0, we can use the quadratic formula since we have a quadratic equation in terms of x^2.
First, let's substitute x^2 with a variable for simplicity. Let u = x^2. Therefore, our equation becomes:
-3u^2 + 27u + 1200 = 0
Now, we can solve for u using the quadratic formula, which states that for any quadratic equation of the form au^2 + bu + c = 0, the solutions (or zeroes) can be found using the following formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
Applying the quadratic formula to our equation, we have:
u = (-27 ± √(27^2 - 4(-3)(1200))) / (2(-3))
u = (-27 ± √(729 + 14400)) / (-6)
u = (-27 ± √15129) / (-6)
u = (-27 ± 123) / (-6)
This gives us two potential solutions for u:
u1 = (-27 + 123) / (-6) = 96/(-6) = -16
u2 = (-27 - 123) / (-6) = -150/(-6) = 25
Now, we substitute back u with x^2:
For u1: x^2 = -16
To find the square root of a negative number, we need to introduce complex numbers. Taking the square root of -16 gives us two complex solutions: x1 = √16i and x2 = -√16i.
For u2: x^2 = 25
We can take the square root of 25 to find two real solutions: x3 = 5 and x4 = -5.
Therefore, the zeroes of the equation –3x^4 + 27x^2 + 1200 = 0 are:
x1 = √16i
x2 = -√16i
x3 = 5
x4 = -5
To find the zeroes of the equation –3x^4 + 27x^2 + 1200 = 0, we can solve it by factoring or using the quadratic formula.
Let's use factoring.
First, let's factor out a common factor: 3.
3(-x^4 + 9x^2 + 400) = 0
Next, let's substitute y = x^2:
3(-y^2 + 9y + 400) = 0
Next, let's factor the quadratic in terms of y:
3(y - 25)(y + 16) = 0
Now, let's substitute back y = x^2:
(y - 25)(y + 16) = 0
Set each factor equal to zero and solve for x:
y - 25 = 0 or y + 16 = 0
If y - 25 = 0:
x^2 - 25 = 0
Taking the square root of both sides:
x = ±√25
So, x = ±5.
If y + 16 = 0:
x^2 + 16 = 0
Taking the square root of both sides:
x = ±√(-16)
Since there is no real square root of a negative number, x has no real solutions for this part.
Therefore, the zeroes of the equation –3x^4 + 27x^2 + 1200 = 0 are x = ±5.