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.
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<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.
Answer:
Step-by-step explanation:
You are stupid
Answer:
2
Step-by-step explanation:
Answer:
Step-by-step explanation:
The volume of a rectanguiar shape like this one is V = L * W * H, where the letters represent Length, Width and Height. Here L is the longest dimension and is 28 - 2x; W is the width and is 22-2x; and finally, x is the height. Thus, the volume of this box must be
V(x) = (28 - 2x)*(22 - 2x)*x
and we want to maximize V(x).
One way of doing that is to graph V(x) and look for any local maximum of the graph. We'd want to determine the value of x for which V(x) is a maximum.
Another way, for those who know some calculus, is to use the first and second derivatives to identify the value of x at which V is at a maximum.
I have provided the function that you requested. If you'd like for us to go all the way to a solution, please repost your question.
i would probably use an iv or a blood simple