find all zeros of the equation
-3x^4+27x^2+1200=0
If someone could explain it would be helpful
To find the zeros of the equation -3x^4 + 27x^2 + 1200 = 0, we can use factoring and the quadratic formula. Here's how you can do it step by step:
Step 1: Let's start by factoring out a common factor if possible. In this case, we can factor out -3 from all terms:
-3(x^4 - 9x^2 - 400) = 0
Step 2: Now, let's focus on the expression in the parentheses: x^4 - 9x^2 - 400. To simplify this expression, we can substitute x^2 with a new variable, let's say y:
y^2 - 9y - 400 = 0
Step 3: Solve the quadratic equation y^2 - 9y - 400 = 0 by factoring or applying the quadratic formula. In this case, factoring doesn't work, so we'll use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -9, and c = -400. Plugging these values into the quadratic formula, we get:
y = [9 ± √((-9)^2 - 4(1)(-400))] / (2*1)
= [9 ± √(81 + 1600)] / 2
= (9 ± √(1681)) / 2
= (9 ± 41) / 2
Therefore, we have two possible values for y:
y1 = (9 + 41) / 2 = 50/2 = 25
y2 = (9 - 41) / 2 = -32/2 = -16
Step 4: Now, we have found the values of y. We need to substitute them back into the equation y = x^2:
For y1 = 25: x^2 = 25
Taking the square root of both sides, we get:
x1 = √25 = ±5
For y2 = -16: x^2 = -16
Since we can't take the square root of a negative number in real numbers, this value does not yield real solutions.
Therefore, the zeros or solutions to the original equation -3x^4 + 27x^2 + 1200 = 0 are x = ±5.
Please note that complex solutions may exist in other cases, but in this particular question, we don't have any.
To find the zeros of the equation -3x^4 + 27x^2 + 1200 = 0, we can use factoring and the zero product property.
Step 1: Start by factoring out any common factors. In this equation, there are no common factors.
Step 2: Rearrange the equation so that it is in quadratic form, where the highest power of x is 2. To do this, substitute y = x^2, which gives us the equation -3y^2 + 27y + 1200 = 0.
Step 3: Now, we can factor the quadratic equation. We need to find two numbers that multiply to give 1200 and add up to 27 (the coefficient of y). By inspection or trial and error, we can find these numbers to be 60 and 20. So, we can rewrite the equation as (-3y + 60)(y + 20) = 0.
Step 4: Set each factor equal to zero and solve for y:
-3y + 60 = 0 → y = 20
y + 20 = 0 → y = -20
Step 5: Substitute the values of y back into the equation y = x^2 to find the corresponding values of x:
For y = 20:
x^2 = 20 → x = ±√20 → x = ±4.47
For y = -20:
x^2 = -20 → No real solutions (the square root of a negative number is not a real number).
Therefore, the zeros of the equation -3x^4 + 27x^2 + 1200 = 0 are x = ±4.47.