Solve -7x^2+4/x^3+1=5/x^2-x+1-1/x+1

$$\frac { - 7 x ^ { 2 } + 4 } { x ^ { 3 } + 1 } = \frac { 5 } { x ^ { 2 } - x + 1 } - \frac { 1 } { x + 1 }$$

Solve for x

$$x=0$$

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 \left(x+1\right)\left(x^{2}-x+1\right), the least common multiple of x^{3}+1,x^{2}-x+1,x+1.

$$-7x^{2}+4=\left(x+1\right)\times 5-\left(x^{2}-x+1\right)$$

Use the distributive property to multiply x+1 by 5.

$$-7x^{2}+4=5x+5-\left(x^{2}-x+1\right)$$

To find the opposite of x^{2}-x+1, find the opposite of each term.

$$-7x^{2}+4=5x+5-x^{2}+x-1$$

Combine 5x and x to get 6x.

$$-7x^{2}+4=6x+5-x^{2}-1$$

Subtract 1 from 5 to get 4.

$$-7x^{2}+4=6x+4-x^{2}$$

Subtract 6x from both sides.

$$-7x^{2}+4-6x=4-x^{2}$$

Subtract 4 from both sides.

$$-7x^{2}+4-6x-4=-x^{2}$$

Subtract 4 from 4 to get 0.

$$-7x^{2}-6x=-x^{2}$$

Add x^{2} to both sides.

$$-7x^{2}-6x+x^{2}=0$$

Combine -7x^{2} and x^{2} to get -6x^{2}.

$$-6x^{2}-6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}}}{2\left(-6\right)}$$

Take the square root of \left(-6\right)^{2}.

$$x=\frac{-\left(-6\right)±6}{2\left(-6\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±6}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{6±6}{-12}$$

Now solve the equation x=\frac{6±6}{-12} when ± is plus. Add 6 to 6.

$$x=\frac{12}{-12}$$

Divide 12 by -12.

$$x=-1$$

Now solve the equation x=\frac{6±6}{-12} when ± is minus. Subtract 6 from 6.

$$x=\frac{0}{-12}$$

Divide 0 by -12.

$$x=0$$

The equation is now solved.

$$x=-1 x=0$$

Variable x cannot be equal to -1.

$$x=0$$

Steps for Completing the Square

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x^{2}-x+1\right), the least common multiple of x^{3}+1,x^{2}-x+1,x+1.

$$-7x^{2}+4=\left(x+1\right)\times 5-\left(x^{2}-x+1\right)$$

Use the distributive property to multiply x+1 by 5.

$$-7x^{2}+4=5x+5-\left(x^{2}-x+1\right)$$

To find the opposite of x^{2}-x+1, find the opposite of each term.

$$-7x^{2}+4=5x+5-x^{2}+x-1$$

Combine 5x and x to get 6x.

$$-7x^{2}+4=6x+5-x^{2}-1$$

Subtract 1 from 5 to get 4.

$$-7x^{2}+4=6x+4-x^{2}$$

Subtract 6x from both sides.

$$-7x^{2}+4-6x=4-x^{2}$$

Add x^{2} to both sides.

$$-7x^{2}+4-6x+x^{2}=4$$

Combine -7x^{2} and x^{2} to get -6x^{2}.

$$-6x^{2}+4-6x=4$$

Subtract 4 from both sides.

$$-6x^{2}-6x=4-4$$

Subtract 4 from 4 to get 0.

$$-6x^{2}-6x=0$$

Divide both sides by -6.

$$\frac{-6x^{2}-6x}{-6}=\frac{0}{-6}$$

Dividing by -6 undoes the multiplication by -6.

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

Divide -6 by -6.

$$x^{2}+x=\frac{0}{-6}$$

Divide 0 by -6.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}+x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{1}{2} x+\frac{1}{2}=-\frac{1}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=0 x=-1$$

Variable x cannot be equal to -1.

$$x=0$$