Determine an equation in simplified form, for the family of quartic functions with zeros 3 (order 2) and -2±√3
Sure, let's solve this equation very seriously.
To begin, we know that the equation has zeros at 3 with an order of 2. So, we can write these roots as (x - 3)^2.
Next, we have zeros at -2 ± √3. This means we have two more factors: (x - (-2 + √3)) and (x - (-2 - √3)). Now, let's simplify this further.
Taking it slowly, -2 + √3 can be written as -2 + √3/1, which can be rationalized as (-2 + √3)(-2 - √3)/(1)(-2 - √3) = (-2)^2 - (√3)^2 / -2 - √3 = 4 - 3 / -2 - √3 = 1 / -2 - √3.
Great! We got -2 + √3 simplified. Now, let's work on -2 - √3. Similarly, we can write this as (-2 - √3)(-2 + √3)/(1)(-2 + √3) = (-2)^2 - (√3)^2 / -2 + √3 = 4 - 3 / -2 + √3 = 1 / -2 + √3.
Now that we have both simplified expressions, let's substitute them back into the equation. The two additional factors become (x - 1/(-2 - √3)) and (x - 1/(-2 + √3)).
Combining everything together, the equation in simplified form for the family of quartic functions with zeros 3 (order 2) and -2 ± √3 is:
(x - 3)^2 * (x - 1/(-2 - √3)) * (x - 1/(-2 + √3)).
Voila!
To determine a quartic function with the given zeros, we can start by writing it in factored form.
Given zeros: 3 (order 2) and -2±√3
Since 3 is a zero of order 2, it means that it appears twice in the factored form. So, we have:
(x - 3)(x - 3)
Next, we have -2±√3 as the other zeros. We know that for complex conjugate zeros, they appear in pairs. Therefore, we can write them as follows:
(x - (-2+√3))(x - (-2-√3))
Simplifying the expressions inside the parentheses, we have:
(x + 2 - √3)(x + 2 + √3)
Now, we can multiply the factors:
(x - 3)(x - 3)(x + 2 - √3)(x + 2 + √3)
Expanding this expression, we get:
(x^2 - 6x + 9)(x^2 + 4x - 1)
Finally, multiply the two binomials:
x^4 - 2x^3 - x^2 + 22x - 9
Therefore, the equation in simplified form for the quartic function with zeros 3 (order 2) and -2±√3 is:
f(x) = x^4 - 2x^3 - x^2 + 22x - 9
To determine an equation for the family of quartic functions with zeros 3 (order 2) and -2±√3, we need to consider the factors of the polynomial.
First, we know that a polynomial with a zero of 3 (order 2) means it has a factor (x - 3)². Similarly, for zeros -2±√3, we have the factors (x - (-2+√3)) and (x - (-2-√3)). Simplifying these, we get (x - 3)², (x + 2-√3), and (x + 2+√3).
Next, we multiply all these factors together to get the full equation. Multiplying (x - 3)², (x + 2-√3), and (x + 2+√3):
(x - 3)² * (x + 2-√3) * (x + 2+√3)
We can simplify this equation further by expanding and combining like terms.
Expanding (x - 3)²:
(x - 3) * (x - 3) = x² - 3x - 3x + 9 = x² - 6x + 9
Expanding (x + 2-√3) * (x + 2+√3):
(x + 2-√3) * (x + 2+√3) = x² + 2x + √3x + 2√3 - √3x - 3 + 2√3 - 3√3 + 3 = x² + 4x + 1
Applying the distributive property to the simplified factors and combining like terms, we get:
(x² - 6x + 9) * (x² + 4x + 1)
Now, we can multiply these binomials together:
x² * x² + x² * 4x + x² * 1 - 6x * x² - 6x * 4x - 6x * 1 + 9 * x² + 9 * 4x + 9 * 1
Simplifying further:
x^4 + 4x^3 + x² - 6x³ - 24x² - 6x + 9x² + 36x + 9
Finally, combining like terms:
x^4 - 2x³ - 14x² + 30x + 9
Therefore, the equation in simplified form for the family of quartic functions with zeros 3 (order 2) and -2±√3 is x^4 - 2x³ - 14x² + 30x + 9.
So the answers to the quartic equation would be x=3, x=3, x= -2 + √3, and x = -2 - √3
work backwards,
e.g. if you had a quadratic function whose zeros are 3 and -4, wouldn't you have two factors, namely
(x-3) and (x+4), and your equation would have to be y = a(x-3)(x+4) , where is a non-zero constant
which would have no effect on the solution.
so form the 4 factors from the given zeros, (yes , two of them are the same)
and y = a(...)(...)(...)(...)