Question

What is true about the solution above.

Answer 1

ANSWER

$x = \pm \sqrt{3}$

and they are actual solutions.

EXPLANATION

The given equation is:

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

Cross multiply

${x}^{2} (6x - 18) = 9(2x -6 )$

This implies;

${x}^{2} (6x - 18) - 9(2x - 6) = 0$

$3{x}^{2} (2x - 6) - 9(2x - 6) = 0$

Factor

$(3 {x}^{2} - 9)(2x - 6) = 0$

$3 {x}^{2} - 9 = 0 \: or \: 2x - 6= 0$

$3 {x}^{2} = 9 \: or \: 2x = 6$

${x}^{2} = 3\: or \: x = 3$

${x} = \pm \sqrt{3} \: or \: x = 3$

The domain of the given equation is

$x \ne3$

Therefore the actual solutions are

$x = \pm \sqrt{3}$

NB: x=3 is not in the domain of the given equation. It cannot be an extraneous solution.