Question

What is true about the solution of StartFraction x squared Over 2 x minus 6 EndFraction = StartFraction 9 Over 6 x minus 18 EndFraction?
x = plus-or-minus StartRoot 3 EndRoot, and they are actual solutions.
x = plus-or-minus StartRoot 3 EndRoot, but they are extraneous solutions.
x = 3, and it is an actual solution.
x = 3, but it is an extraneous solution.

Answer 1

Answer:

$x=\pm\sqrt{3}$  and they are actual solutions

Step-by-step explanation:

we have

$\frac{x^2}{2x-6}=\frac{9}{6x-18}$

Factor the denominators both sides

$\frac{x^2}{2(x-3)}=\frac{9}{6(x-3)}$

Simplify

$\frac{x^2}{2}=\frac{9}{6}$

$x^2=\frac{18}{6}$

$x=\pm\sqrt{3}$

Verify

1) For $x=\sqrt{3}$

$\frac{\sqrt{3}^2}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}$

$\frac{3}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}$

$18=18$ ---> is true

therefore

$x=\sqrt{3}$ ----> is an actual solution

2) For $x=-\sqrt{3}$

$\frac{-\sqrt{3}^2}{2(-\sqrt{3}-3)}=\frac{9}{6(-\sqrt{3}-3)}$

$\frac{3}{2(-\sqrt{3}-3)}=\frac{9}{6(-\sqrt{3}-3)}$

$18=18$ ---> is true

therefore

$x=-\sqrt{3}$  ----> is an actual solution

therefore

Answer 2

Answer:

The answer is A!

Step-by-step explanation: