Pick highest exponent of placeholder
that is the degree
x^2 and y^5
2<5
5>2
5th degree
Answer:
x =15 + 30n
Step-by-step explanation:
8cos(12x)+4=-4
Subtract 4 from each side
8cos(12x)+4-4=-4-4
8cos(12x)=-8
Divide by 8
8/8cos(12x)=-8/8
cos(12x) = -1
Take the arccos on each side
arccos(cos(12x)) = arccos(-1)
12x = 180 degrees
But we can go around in a circle of 360 and get the same result
12 x = 180+360n where n is an integer
Divide each side by 12
x = 180/12 +360n/12
x =15 + 30n
Answer:
- x = ±√3, and they are actual solutions
- x = 3, but it is an extraneous solution
Step-by-step explanation:
The method often recommended for solving an equation of this sort is to multiply by the product of the denominators, then solve the resulting polynomial equation. When you do that, you get ...
... x^2(6x -18) = (2x -6)(9)
... 6x^2(x -3) -18(x -3) = 0
...6(x -3)(x^2 -3) = 0
... x = 3, x = ±√3
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Alternatively, you can subtract the right side of the equation and collect terms to get ...
... x^2/(2(x -3)) - 9/(6(x -3)) = 0
... (1/2)(x^2 -3)/(x -3) = 0
Here, the solution will be values of x that make the numerator zero:
... x = ±√3
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So, the actual solutions are x = ±3, and x = 3 is an extraneous solution. The value x=3 is actually excluded from the domain of the original equation, because the equation is undefined at that point.
_____
<em>Comment on the graph</em>
For the graph, we have rewritten the equation so it is of the form f(x)=0. The graphing program is able to highlight zero crossings, so this is a convenient form. When the equation is multiplied as described above, the resulting cubic has an extra zero-crossing at x=3 (blue curve). This is the extraneous solution.