Solve (x-3)^2/1-x^2+1-x/1+x+10/x^2-1=0

$$\frac { ( x - 3 ) ^ { 2 } } { 1 - x ^ { 2 } } + \frac { 1 - x } { 1 + x } + \frac { 10 } { x ^ { 2 } - 1 } = 0$$

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of 1-x^{2},1+x,x^{2}-1.

$$-\left(x-3\right)^{2}+\left(x-1\right)\left(1-x\right)+10=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$-\left(x^{2}-6x+9\right)+\left(x-1\right)\left(1-x\right)+10=0$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$-x^{2}+6x-9+\left(x-1\right)\left(1-x\right)+10=0$$

Use the distributive property to multiply x-1 by 1-x and combine like terms.

$$-x^{2}+6x-9+2x-x^{2}-1+10=0$$

Combine 6x and 2x to get 8x.

$$-x^{2}+8x-9-x^{2}-1+10=0$$

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

$$-2x^{2}+8x-9-1+10=0$$

Subtract 1 from -9 to get -10.

$$-2x^{2}+8x-10+10=0$$

Add -10 and 10 to get 0.

$$-2x^{2}+8x=0$$

Factor out x.

$$x\left(-2x+8\right)=0$$

To find equation solutions, solve x=0 and -2x+8=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of 1-x^{2},1+x,x^{2}-1.

$$-\left(x-3\right)^{2}+\left(x-1\right)\left(1-x\right)+10=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$-\left(x^{2}-6x+9\right)+\left(x-1\right)\left(1-x\right)+10=0$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$-x^{2}+6x-9+\left(x-1\right)\left(1-x\right)+10=0$$

Use the distributive property to multiply x-1 by 1-x and combine like terms.

$$-x^{2}+6x-9+2x-x^{2}-1+10=0$$

Combine 6x and 2x to get 8x.

$$-x^{2}+8x-9-x^{2}-1+10=0$$

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

$$-2x^{2}+8x-9-1+10=0$$

Subtract 1 from -9 to get -10.

$$-2x^{2}+8x-10+10=0$$

Add -10 and 10 to get 0.

$$-2x^{2}+8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}}}{2\left(-2\right)}$$

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-8±8}{-4}$$

Now solve the equation x=\frac{-8±8}{-4} when ± is plus. Add -8 to 8.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-8±8}{-4} when ± is minus. Subtract 8 from -8.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=0 x=4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of 1-x^{2},1+x,x^{2}-1.

$$-\left(x-3\right)^{2}+\left(x-1\right)\left(1-x\right)+10=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$-\left(x^{2}-6x+9\right)+\left(x-1\right)\left(1-x\right)+10=0$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$-x^{2}+6x-9+\left(x-1\right)\left(1-x\right)+10=0$$

Use the distributive property to multiply x-1 by 1-x and combine like terms.

$$-x^{2}+6x-9+2x-x^{2}-1+10=0$$

Combine 6x and 2x to get 8x.

$$-x^{2}+8x-9-x^{2}-1+10=0$$

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

$$-2x^{2}+8x-9-1+10=0$$

Subtract 1 from -9 to get -10.

$$-2x^{2}+8x-10+10=0$$

Add -10 and 10 to get 0.

$$-2x^{2}+8x=0$$

Divide both sides by -2.

$$\frac{-2x^{2}+8x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{8}{-2}x=\frac{0}{-2}$$

Divide 8 by -2.

$$x^{2}-4x=\frac{0}{-2}$$

Divide 0 by -2.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$