What is the answer to 1+2x+3x^2=(1-x)^{-2} ?

The answer to 1+2x+3x^2=(1-x)^{-2} is x=0,x= 4/3

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1+2x+3x^2=(1-x)^{-2}

1 + 2 x + 3 x^{2} = (1 - x)^{- 2}

x = 0, x = \frac{4}{3}

Decimal Notation

x = 0, x = 1.33333 \dots

Solution steps

1 + 2 x + 3 x^{2} = (1 - x)^{- 2}

Refine

1 + 2 x + 3 x^{2} = \frac{1}{( 1 - x ) ^{2}}

Multiply both sides by

(1 - x)^{2}

(1 - x)^{2} + 2 x (1 - x)^{2} + 3 x^{2} (1 - x)^{2} = 1

Solve

(1 - x)^{2} + 2 x (1 - x)^{2} + 3 x^{2} (1 - x)^{2} = 1 : x = 0, x = \frac{4}{3}

x = 0, x = \frac{4}{3}

Verify Solutions

Find undefined (singularity) points:

x = 1

Combine undefined points with solutions:

x = 0, x = \frac{4}{3}

\frac{13}{x ^{2}} - \frac{2}{x} = \frac{1}{x}

\frac{( 2 x )}{( x - 2 )} - \frac{3}{( x - 3 )} = - \frac{3}{( x ^{2} - 5 x + 6 )}

\frac{13}{4 x - 5} = \frac{31}{3}

\frac{3}{x} = \frac{6}{2}

x^{- 4} - 10 x^{- 2} + 9 = 0

What is the answer to 1+2x+3x^2=(1-x)^{-2} ?
The answer to 1+2x+3x^2=(1-x)^{-2} is x=0,x= 4/3
Solve for x: 1+2x+3x^2=(1-x)^{-2}
The solution is x=0,x= 4/3