Solve 7/x+2x=7

$$\frac { 7 } { x } + 2 x = 7$$

Solve for x (complex solution)

$$x=\frac{7+\sqrt{7}i}{4}\approx 1.75+0.661437828i$$

$$x=\frac{-\sqrt{7}i+7}{4}\approx 1.75-0.661437828i$$

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 x.

$$7+2xx=7x$$

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

$$7+2x^{2}=7x$$

Subtract 7x from both sides.

$$7+2x^{2}-7x=0$$

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.

$$2x^{2}-7x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 7}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 7}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 7}}{2\times 2}$$

Multiply -8 times 7.

$$x=\frac{-\left(-7\right)±\sqrt{49-56}}{2\times 2}$$

Add 49 to -56.

$$x=\frac{-\left(-7\right)±\sqrt{-7}}{2\times 2}$$

Take the square root of -7.

$$x=\frac{-\left(-7\right)±\sqrt{7}i}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{7}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{7}i}{4}$$

Now solve the equation x=\frac{7±\sqrt{7}i}{4} when ± is plus. Add 7 to i\sqrt{7}.

$$x=\frac{7+\sqrt{7}i}{4}$$

Now solve the equation x=\frac{7±\sqrt{7}i}{4} when ± is minus. Subtract i\sqrt{7} from 7.

$$x=\frac{-\sqrt{7}i+7}{4}$$

The equation is now solved.

$$x=\frac{7+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+7}{4}$$

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 x.

$$7+2xx=7x$$

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

$$7+2x^{2}=7x$$

Subtract 7x from both sides.

$$7+2x^{2}-7x=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-7x=-7$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{7}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{7}{2}+\frac{49}{16}$$

Add -\frac{7}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{7}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{7}i}{4} x-\frac{7}{4}=-\frac{\sqrt{7}i}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{7+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+7}{4}$$