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

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x,2x.

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

Use the distributive property to multiply x by x+3.

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

Use the distributive property to multiply 2 by x-3.

$$x^{2}+3x-2=2x-6+4-x^{2}$$

Add -6 and 4 to get -2.

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

Subtract 2x from both sides.

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

Combine 3x and -2x to get x.

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

Subtract -2 from both sides.

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

The opposite of -2 is 2.

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

Add x^{2} to both sides.

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

Add -2 and 2 to get 0.

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

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

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

Factor out x.

$$x\left(2x+1\right)=0$$

To find equation solutions, solve x=0 and 2x+1=0.

$$x=0 x=-\frac{1}{2}$$

Variable x cannot be equal to 0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x,2x.

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

Use the distributive property to multiply x by x+3.

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

Use the distributive property to multiply 2 by x-3.

$$x^{2}+3x-2=2x-6+4-x^{2}$$

Add -6 and 4 to get -2.

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

Subtract 2x from both sides.

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

Combine 3x and -2x to get x.

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

Subtract -2 from both sides.

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

The opposite of -2 is 2.

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

Add x^{2} to both sides.

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

Add -2 and 2 to get 0.

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

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

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

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

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 2}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±1}{4}$$

Now solve the equation x=\frac{-1±1}{4} when ± is plus. Add -1 to 1.

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

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-1±1}{4} when ± is minus. Subtract 1 from -1.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=0 x=-\frac{1}{2}$$

Variable x cannot be equal to 0.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x,2x.

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

Use the distributive property to multiply x by x+3.

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

Use the distributive property to multiply 2 by x-3.

$$x^{2}+3x-2=2x-6+4-x^{2}$$

Add -6 and 4 to get -2.

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

Subtract 2x from both sides.

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

Combine 3x and -2x to get x.

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

Add x^{2} to both sides.

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

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

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

Add 2 to both sides.

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

Add -2 and 2 to get 0.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 0 by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{1}{4} x+\frac{1}{4}=-\frac{1}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=0 x=-\frac{1}{2}$$

Variable x cannot be equal to 0.

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