Solve 3x+1/2x=5

$$3x+ \frac{ 1 }{ 2x } =5$$

Solve for x

$$x = \frac{\sqrt{19} + 5}{6} \approx 1.559816491$$

$$x=\frac{5-\sqrt{19}}{6}\approx 0.106850176$$

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.

$$3x\times 2x+1=10x$$

Multiply 3 and 2 to get 6.

$$6xx+1=10x$$

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

$$6x^{2}+1=10x$$

Subtract 10x from both sides.

$$6x^{2}+1-10x=0$$

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.

$$6x^{2}-10x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -10 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 6}}{2\times 6}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-10\right)±\sqrt{100-24}}{2\times 6}$$

Add 100 to -24.

$$x=\frac{-\left(-10\right)±\sqrt{76}}{2\times 6}$$

Take the square root of 76.

$$x=\frac{-\left(-10\right)±2\sqrt{19}}{2\times 6}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{19}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{10±2\sqrt{19}}{12}$$

Now solve the equation x=\frac{10±2\sqrt{19}}{12} when ± is plus. Add 10 to 2\sqrt{19}.

$$x=\frac{2\sqrt{19}+10}{12}$$

Divide 10+2\sqrt{19} by 12.

$$x=\frac{\sqrt{19}+5}{6}$$

Now solve the equation x=\frac{10±2\sqrt{19}}{12} when ± is minus. Subtract 2\sqrt{19} from 10.

$$x=\frac{10-2\sqrt{19}}{12}$$

Divide 10-2\sqrt{19} by 12.

$$x=\frac{5-\sqrt{19}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{19}+5}{6} x=\frac{5-\sqrt{19}}{6}$$

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.

$$3x\times 2x+1=10x$$

Multiply 3 and 2 to get 6.

$$6xx+1=10x$$

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

$$6x^{2}+1=10x$$

Subtract 10x from both sides.

$$6x^{2}+1-10x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}-10x=-1$$

Divide both sides by 6.

$$\frac{6x^{2}-10x}{6}=-\frac{1}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{10}{6}\right)x=-\frac{1}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{3}x=-\frac{1}{6}$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{1}{6}+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{1}{6}+\frac{25}{36}$$

Add -\frac{1}{6} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{19}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{19}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{19}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{\sqrt{19}}{6} x-\frac{5}{6}=-\frac{\sqrt{19}}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{\sqrt{19}+5}{6} x=\frac{5-\sqrt{19}}{6}$$