Solve f(x)=1/2x+3=x

$$f ( x ) = \frac { 1 } { 2 x + 3 } = x$$

Solve for x

$$x=\frac{\sqrt{17}-3}{4}\approx 0.280776406$$

$$x=\frac{-\sqrt{17}-3}{4}\approx -1.780776406$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+3.

$$1=x\left(2x+3\right)$$

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

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

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

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

Subtract 1 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-3±\sqrt{9+8}}{2\times 2}$$

Add 9 to 8.

$$x=\frac{-3±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±\sqrt{17}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{4} when ± is plus. Add -3 to \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{4}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from -3.

$$x=\frac{-\sqrt{17}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-3}{4} x=\frac{-\sqrt{17}-3}{4}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+3.

$$1=x\left(2x+3\right)$$

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

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

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

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Add \frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{17}}{4} x+\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{17}-3}{4} x=\frac{-\sqrt{17}-3}{4}$$