Certainly! Let's solve the equation step by step:
Given:
[tex]\[
\left(\frac{1}{3}\right) \frac{x^2-2 x}{16-2 x}=4 x \sqrt{9}
\][/tex]
Firstly, note that [tex]\(\sqrt{9} = 3\)[/tex]. Substitute this value into the equation:
[tex]\[
\left(\frac{1}{3}\right) \frac{x^2-2 x}{16-2 x}=4 x \cdot 3
\][/tex]
Simplify the right side:
[tex]\[
\left(\frac{1}{3}\right) \frac{x^2-2 x}{16-2 x} = 12 x
\][/tex]
Next, multiply both sides of the equation by 3 to eliminate the fraction on the left:
[tex]\[
\frac{x^2-2 x}{16-2 x} = 36 x
\][/tex]
Now let's get rid of the fraction by multiplying both sides of the equation by [tex]\(16-2x\)[/tex]:
[tex]\[
x^2 - 2x = 36x(16 - 2x)
\][/tex]
Expand the right side:
[tex]\[
x^2 - 2x = 36x \cdot 16 - 36x \cdot 2x
\][/tex]
[tex]\[
x^2 - 2x = 576x - 72x^2
\][/tex]
Bring all the terms to one side of the equation to set it equal to zero:
[tex]\[
x^2 - 2x - 576x + 72x^2 = 0
\][/tex]
[tex]\[
73x^2 - 578x = 0
\][/tex]
Factor out the common term [tex]\(x\)[/tex]:
[tex]\[
x(73x - 578) = 0
\][/tex]
This equation will hold true if either:
[tex]\[
x = 0
\][/tex]
or
[tex]\[
73x - 578 = 0
\][/tex]
Solve the second equation for [tex]\(x\)[/tex]:
[tex]\[
73x - 578 = 0
\][/tex]
[tex]\[
73x = 578
\][/tex]
[tex]\[
x = \frac{578}{73}
\][/tex]
[tex]\[
x \approx 7.91780821917808
\][/tex]
So the solutions to the equation [tex]\(\left(\frac{1}{3}\right) \frac{x^2-2 x}{16-2 x}=4 x \sqrt{9}\)[/tex] are:
[tex]\[
x = 0 \quad \text{and} \quad x \approx 7.91780821917808
\][/tex]
