What is true about the solution of x^2/2x-6=9/6x-18? x=±√3, and they are actual x=±√3, but they are extraneous x = 3, and it is an actual x = 3, but it is an extraneous

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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Comment on the graph

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.

step-by-step explanation:

it will be a cylinder. with a diameter of 8 and 4 single pieces high

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

it alot will look like this 4.4