Answer:
Equation 3
Step-by-step explanation:
Lets see which of the functions has -2 as a zero root. We will go in order:
(1) (-2)^4 - 3(-2)^3 + 3(-2)^2 -3(-2) + 2 = 16 - 3(-8) + 3(4) + 6 +2 = 16 +24 +12 + 6 +2 =60 >0
So, (1) is wrong!
(2) (-2)^4 + 3(-2)^3 + 3(-2)^2 - 3(-2) - 2 = 16 - 24 + 12 + 6 - 2 =34 - 26 = 8 > 0
(2) is also wrong!
(3) (-2)^4 + 3(-2)^3 + 3(-2)^2 +3(-2) + 2 = 16 - 24 + 12 - 6 + 2 = 30 -30 = 0
The zero root x=-2 fits, what about x=-1?
(-1)^4 + 3(-1)^3 + 3(-1)^2 +3(-1) + 2 = 1 - 3 + 3 - 3 + 2 = 6 - 6 = 0
So, equation (3) fits both!
Finally, lets see (4):
(-2)^4 - 3(-2)^3 - 3(-2)^2 + 3(-2) + 2 = 16 + 24 - 12 - 6 + 2 = 42 - 18 = 24 > 0
So, (4) is also wrong.
Only equation 3 fits both zero roots!
The quartic function with x=-1 and x=-2 real roots is x^4+6x^3 +12x^2+12x+4. Quartic functions are polynomial functions of degree 4; quadratic equations resources also help understand the concept. In essence, finding roots of quartic functions follow the same logic as that of quadratic functions.
The subject matter pertains to quartic functions in mathematics. Quartic functions are polynomial functions with a degree of 4. From the question, the given zeros are x=-1 and x=-2, having multiplicity of 2 each (since there are only two real zeros). Thus, the quartic function with these zeros will be (x+1)^2*(x+2)^2. This can be expanded to x^4+6x^3 +12x^2+12x+4.
Exemplifying the relevance of The Solution of Quadratic Equations, normally known as second-order polynomials or quadratic functions, such equations can also be used to find zeros of the functions when set to equal zero. In this scenario, quartic functions are a degree higher, but the same principle applies in finding the roots when the equation is set equal to zero.
#SPJ3
Answer:
n=1
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.
Answer:
30n²+54+24n
Step-by-step explanation:
5nX6n=30n²
5nX6n=30n²5nX6=30
5nX6n=30n²5nX6=304X6n=24n
5nX6n=30n²5nX6=304X6n=24n4X6=24
=30n²+30+24n+24
=30n²+54+24n