Question

1. solve for all values of x.

\\(\frac{9x}{x - 2}-\frac{8}{x^{2}-2x}=\frac{1}{x - 2}\\)
erin and christa were working on cubing binomials for math homework. erin believed they could time with a shortcut. she wrote down the rule below for christa to follow.
\\((a + b)^{3}=a^{3}+b^{3}\\)
is this shortcut always work? justify your result algebraically.

Explanation:

Step1: Find the least - common denominator (LCD)

The denominators are \(x - 2\) and \(x\). The LCD is \(x(x - 2)\).

Step2: Multiply each term by the LCD

Multiply \(\frac{9x}{x - 2}-\frac{8}{x}=\frac{1}{x - 2}\) by \(x(x - 2)\):
\[x(x - 2)\times\frac{9x}{x - 2}-x(x - 2)\times\frac{8}{x}=x(x - 2)\times\frac{1}{x - 2}\]
\[9x\times x-8(x - 2)=x\]

Step3: Expand and simplify

\[9x^{2}-8x + 16=x\]
\[9x^{2}-8x - x+16 = 0\]
\[9x^{2}-9x + 16=0\]
The discriminant of the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 9\), \(b=-9\), \(c = 16\)) is \(\Delta=b^{2}-4ac=(-9)^{2}-4\times9\times16=81 - 576=-495<0\)
So there are no real - valued solutions for \(x\).

Answer:

There are no real - valued solutions for \(x\).