Solve x^4-3x^2+2/1-x=1+x

$$\frac{ { x }^{ 4 } -3 { x }^{ 2 } +2 }{ 1-x } = 1+x$$

Solve for x

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$x^{4}-3x^{2}+2=-x+1+\left(-x+1\right)x$$

Use the distributive property to multiply -x+1 by x.

$$x^{4}-3x^{2}+2=-x+1-x^{2}+x$$

Combine -x and x to get 0.

$$x^{4}-3x^{2}+2=1-x^{2}$$

Subtract 1 from both sides.

$$x^{4}-3x^{2}+2-1=-x^{2}$$

Subtract 1 from 2 to get 1.

$$x^{4}-3x^{2}+1=-x^{2}$$

Add x^{2} to both sides.

$$x^{4}-3x^{2}+1+x^{2}=0$$

Combine -3x^{2} and x^{2} to get -2x^{2}.

$$x^{4}-2x^{2}+1=0$$

Substitute t for x^{2}.

$$t^{2}-2t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{2±0}{2}$$

Solutions are the same.

$$t=1$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=-1 x=1$$

Variable x cannot be equal to 1.

$$x=-1$$