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

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

Solve for x

$$x=\frac{\sqrt{17}-3}{2}\approx 0.561552813$$

$$x=\frac{-\sqrt{17}-3}{2}\approx -3.561552813$$

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 4x^{2}, the least common multiple of 2x^{2},4x.

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

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

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

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

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

Subtract 2 from both sides.

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

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

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

Square 3.

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

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8}}{2}$$

Add 9 to 8.

$$x=\frac{-3±\sqrt{17}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{2} when ± is plus. Add -3 to \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from -3.

$$x=\frac{-\sqrt{17}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}$$

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 4x^{2}, the least common multiple of 2x^{2},4x.

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

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

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

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

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=2+\frac{9}{4}$$

Add 2 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{17}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{17}}{2} x+\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}$$