Question

What is true about the solution of x^2/2x-6=9/6x-18?

Answer 1

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

_____

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

_____

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.

_____

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.

Answer 2

Answer:

The answer is A, x = ±√3, and they are actual solutions!

Step-by-step explanation:

(x^2)/(2x-6) = (9)/(6x)-18 -

Multiply both sides of the equation by '6x-18':

3x^2 = 9

Divide both sides of the equation by '3':

x^2 = 3

Use the Pythagorean theorem:

x = ±√3

Both work because they are not ±3; ±3 would have been extraneous since they are factors of '2x-6' and '6x-18' and '2x-6' and '6x-18' are in the denominator.