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

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

Solve for x (complex solution)

$$x=\frac{3+\sqrt{7}i}{2}\approx 1.5+1.322875656i$$

$$x=\frac{-\sqrt{7}i+3}{2}\approx 1.5-1.322875656i$$

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

$$x^{2}+4=x\times 3$$

Subtract x\times 3 from both sides.

$$x^{2}+4-x\times 3=0$$

Multiply -1 and 3 to get -3.

$$x^{2}+4-3x=0$$

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.

$$x^{2}-3x+4=0$$

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

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

Square -3.

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

Multiply -4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9-16}}{2}$$

Add 9 to -16.

$$x=\frac{-\left(-3\right)±\sqrt{-7}}{2}$$

Take the square root of -7.

$$x=\frac{-\left(-3\right)±\sqrt{7}i}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{7}i}{2}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{2} when ± is plus. Add 3 to i\sqrt{7}.

$$x=\frac{3+\sqrt{7}i}{2}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{2} when ± is minus. Subtract i\sqrt{7} from 3.

$$x=\frac{-\sqrt{7}i+3}{2}$$

The equation is now solved.

$$x=\frac{3+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i+3}{2}$$

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

$$x^{2}+4=x\times 3$$

Subtract x\times 3 from both sides.

$$x^{2}+4-x\times 3=0$$

Multiply -1 and 3 to get -3.

$$x^{2}+4-3x=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-3x=-4$$

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.

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

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

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

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

$$x^{2}-3x+\frac{9}{4}=-\frac{7}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{7}i}{2} x-\frac{3}{2}=-\frac{\sqrt{7}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i+3}{2}$$