Solve 2/x=5-8x

$$\frac{ 2 }{ x } =5-8x$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{39}i}{16}\approx 0.3125+0.390312375i$$

$$x=\frac{-\sqrt{39}i+5}{16}\approx 0.3125-0.390312375i$$

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.

$$2=x\times 5-8xx$$

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

$$2=x\times 5-8x^{2}$$

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

$$x\times 5-8x^{2}=2$$

Subtract 2 from both sides.

$$x\times 5-8x^{2}-2=0$$

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.

$$-8x^{2}+5x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-8\right)\left(-2\right)}}{2\left(-8\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-8\right)\left(-2\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-5±\sqrt{25+32\left(-2\right)}}{2\left(-8\right)}$$

Multiply 32 times -2.

$$x=\frac{-5±\sqrt{25-64}}{2\left(-8\right)}$$

Add 25 to -64.

$$x=\frac{-5±\sqrt{-39}}{2\left(-8\right)}$$

Take the square root of -39.

$$x=\frac{-5±\sqrt{39}i}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-5±\sqrt{39}i}{-16}$$

Now solve the equation x=\frac{-5±\sqrt{39}i}{-16} when ± is plus. Add -5 to i\sqrt{39}.

$$x=\frac{-5+\sqrt{39}i}{-16}$$

Divide -5+i\sqrt{39} by -16.

$$x=\frac{-\sqrt{39}i+5}{16}$$

Now solve the equation x=\frac{-5±\sqrt{39}i}{-16} when ± is minus. Subtract i\sqrt{39} from -5.

$$x=\frac{-\sqrt{39}i-5}{-16}$$

Divide -5-i\sqrt{39} by -16.

$$x=\frac{5+\sqrt{39}i}{16}$$

The equation is now solved.

$$x=\frac{-\sqrt{39}i+5}{16} x=\frac{5+\sqrt{39}i}{16}$$

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.

$$2=x\times 5-8xx$$

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

$$2=x\times 5-8x^{2}$$

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

$$x\times 5-8x^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-8x^{2}+5x=2$$

Divide both sides by -8.

$$\frac{-8x^{2}+5x}{-8}=\frac{2}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{5}{-8}x=\frac{2}{-8}$$

Divide 5 by -8.

$$x^{2}-\frac{5}{8}x=\frac{2}{-8}$$

Reduce the fraction \frac{2}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{8}x=-\frac{1}{4}$$

Divide -\frac{5}{8}, the coefficient of the x term, by 2 to get -\frac{5}{16}. Then add the square of -\frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{8}x+\left(-\frac{5}{16}\right)^{2}=-\frac{1}{4}+\left(-\frac{5}{16}\right)^{2}$$

Square -\frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{1}{4}+\frac{25}{256}$$

Add -\frac{1}{4} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{39}{256}$$

Factor x^{2}-\frac{5}{8}x+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{16}\right)^{2}=-\frac{39}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{16}\right)^{2}}=\sqrt{-\frac{39}{256}}$$

Simplify.

$$x-\frac{5}{16}=\frac{\sqrt{39}i}{16} x-\frac{5}{16}=-\frac{\sqrt{39}i}{16}$$

Add \frac{5}{16} to both sides of the equation.

$$x=\frac{5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i+5}{16}$$