Solve +4/4-x=x/4-x+x+4

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

Solve for x

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by -x+4.

$$4=x+\left(-x+4\right)x+\left(-x+4\right)\times 4$$

Use the distributive property to multiply -x+4 by x.

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

Combine x and 4x to get 5x.

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

Use the distributive property to multiply -x+4 by 4.

$$4=5x-x^{2}-4x+16$$

Combine 5x and -4x to get x.

$$4=x-x^{2}+16$$

Swap sides so that all variable terms are on the left hand side.

$$x-x^{2}+16=4$$

Subtract 4 from both sides.

$$x-x^{2}+16-4=0$$

Subtract 4 from 16 to get 12.

$$x-x^{2}+12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-1±\sqrt{1+48}}{2\left(-1\right)}$$

Add 1 to 48.

$$x=\frac{-1±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-1±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±7}{-2}$$

Now solve the equation x=\frac{-1±7}{-2} when ± is plus. Add -1 to 7.

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

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{-1±7}{-2} when ± is minus. Subtract 7 from -1.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-3 x=4$$

Variable x cannot be equal to 4.

$$x=-3$$

Steps for Completing the Square

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by -x+4.

$$4=x+\left(-x+4\right)x+\left(-x+4\right)\times 4$$

Use the distributive property to multiply -x+4 by x.

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

Combine x and 4x to get 5x.

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

Use the distributive property to multiply -x+4 by 4.

$$4=5x-x^{2}-4x+16$$

Combine 5x and -4x to get x.

$$4=x-x^{2}+16$$

Swap sides so that all variable terms are on the left hand side.

$$x-x^{2}+16=4$$

Subtract 16 from both sides.

$$x-x^{2}=4-16$$

Subtract 16 from 4 to get -12.

$$x-x^{2}=-12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+x=-12$$

Divide both sides by -1.

$$\frac{-x^{2}+x}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{1}{-1}x=-\frac{12}{-1}$$

Divide 1 by -1.

$$x^{2}-x=-\frac{12}{-1}$$

Divide -12 by -1.

$$x^{2}-x=12$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{49}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{7}{2} x-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=4 x=-3$$

Variable x cannot be equal to 4.

$$x=-3$$