Solve 4x+7/x=3

$$4x+ \frac{ 7 }{ x } =3$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{103}i}{8}\approx 0.375+1.268611446i$$

$$x=\frac{-\sqrt{103}i+3}{8}\approx 0.375-1.268611446i$$

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.

$$4xx+7=3x$$

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

$$4x^{2}+7=3x$$

Subtract 3x from both sides.

$$4x^{2}+7-3x=0$$

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.

$$4x^{2}-3x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -3 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 4\times 7}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9-16\times 7}}{2\times 4}$$

Multiply -16 times 7.

$$x=\frac{-\left(-3\right)±\sqrt{9-112}}{2\times 4}$$

Add 9 to -112.

$$x=\frac{-\left(-3\right)±\sqrt{-103}}{2\times 4}$$

Take the square root of -103.

$$x=\frac{-\left(-3\right)±\sqrt{103}i}{2\times 4}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{103}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{3±\sqrt{103}i}{8}$$

Now solve the equation x=\frac{3±\sqrt{103}i}{8} when ± is plus. Add 3 to i\sqrt{103}.

$$x=\frac{3+\sqrt{103}i}{8}$$

Now solve the equation x=\frac{3±\sqrt{103}i}{8} when ± is minus. Subtract i\sqrt{103} from 3.

$$x=\frac{-\sqrt{103}i+3}{8}$$

The equation is now solved.

$$x=\frac{3+\sqrt{103}i}{8} x=\frac{-\sqrt{103}i+3}{8}$$

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.

$$4xx+7=3x$$

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

$$4x^{2}+7=3x$$

Subtract 3x from both sides.

$$4x^{2}+7-3x=0$$

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

$$4x^{2}-3x=-7$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=-\frac{7}{4}+\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{7}{4}+\frac{9}{64}$$

Add -\frac{7}{4} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{103}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=-\frac{103}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{-\frac{103}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{\sqrt{103}i}{8} x-\frac{3}{8}=-\frac{\sqrt{103}i}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{3+\sqrt{103}i}{8} x=\frac{-\sqrt{103}i+3}{8}$$