What is the answer to (2x^2)/(x+1)-x+1=0 ?

The answer to (2x^2)/(x+1)-x+1=0 is x=i,x=-i

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(2x^2)/(x+1)-x+1=0

\frac{2 x ^{2}}{x + 1} - x + 1 = 0

x = i, x = - i

Solution steps

\frac{2 x ^{2}}{x + 1} - x + 1 = 0

Multiply both sides by

x + 1

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

Solve

2 x^{2} - x (x + 1) + x + 1 = 0 : x = i, x = - i

x = i, x = - i

\frac{1}{5 ( k - 3 )} = \frac{3}{4}

\frac{r}{3 r} = 12

\frac{1}{2 x} = \frac{4}{3}

\frac{y + 2}{y} = \frac{5}{3}

\frac{6}{x - 8} = \frac{2}{x + 6}

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