Find the Lowest common denominator ,

[ingrid]

1+ 2x-¹ - 8x-² =0

and then i think it turns into
1 + 1/2x - 1/8x2 = 0

so on the bottom i am left with
1 + 2x - 8x²

i think 2x would be the lowest common denominator but the way we did it in class, we would cancel out everything on the bottom

 

[mmm4444bot]

2x^(-1) = 2(1/x) = 2/x

In other words, the exponent -1 applies only to x.

8x^(-2) = 8(1/x^2) = 8/x^2

In other words, the exponent -2 applies only to x.

So, those denominators are x and x^2.

1 + 2/x - 8/x^2 = 0

Multiply both sides of this equation by the LCD, and the denominators will cancel.

 

[ingrid]

then the LCD would be x(x²) ??

 

[mmm4444bot]

the LCD would be x(x²) Not quite.

The LCD is x^2.

(Look for the highest power of x; that's it. Don't multiply the powers of x together.)

-------------------------------------------------------------------

If there's more than one variable, it goes like this.

Denominators: x^2 y^5 and x^7 y^2 and x y^8

The highest power of x is x^7.

The highest power of y is y^8.

Therefore, the LCD of the three denominators is x^7 y^8.

 

[BigGlenntheHeavy]

\(\displaystyle 1+2x^{-1}-8x^{-2} \ = \ 0\)

\(\displaystyle 1+\frac{2}{x}-\frac{8}{x^2} \ = \ 0\)

\(\displaystyle \frac{x^2+2x-8}{x^2} \ = \ 0\)

\(\displaystyle x^2\bigg(\frac{x^2+2x-8}{x^2}\bigg) \ = \ (0)(x^2)\)

\(\displaystyle x^2+2x-8 \ = \ 0\)

\(\displaystyle (x+4)(x-2) \ = \ 0 \ \implies \ x \ = \ -4,2\)

\(\displaystyle It \ behooves \ you \ to \ check \ the \ two \ solutions \ in \ the \ original \ equation.\)

 

[Denis]

\(\displaystyle 1+\frac{2}{x}-\frac{8}{x^2} \ = \ 0\)

WHY all them steps, BigG?
Multiply by x^2:
x^2 + 2x - 8 = 0
(x - 2)(x + 4) = 0
x = 2 or -4
Done!

 

[mmm4444bot]

WHY all them steps, BigG?

It's all good. Glenn shows how to rewrite without negative exponents first and, even though it's not necessary, how x^2 works as a LCD to write a single ratio on the LHS.

These are good steps to understand. I hope the original poster learns something from them.