Solve 2x+1/x=7-3/x

$$2 x + \frac { 1 } { x } = 7 - \frac { 3 } { x }$$

Solve for x

$$x = \frac{\sqrt{17} + 7}{4} \approx 2.780776406$$

$$x=\frac{7-\sqrt{17}}{4}\approx 0.719223594$$

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.

$$2xx+1=x\times 7-3$$

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

$$2x^{2}+1=x\times 7-3$$

Subtract x\times 7 from both sides.

$$2x^{2}+1-x\times 7=-3$$

Add 3 to both sides.

$$2x^{2}+1-x\times 7+3=0$$

Multiply -1 and 7 to get -7.

$$2x^{2}+1-7x+3=0$$

Add 1 and 3 to get 4.

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

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.

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

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

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

Square -7.

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

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-\left(-7\right)±\sqrt{49-32}}{2\times 2}$$

Add 49 to -32.

$$x=\frac{-\left(-7\right)±\sqrt{17}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{17}}{4}$$

Now solve the equation x=\frac{7±\sqrt{17}}{4} when ± is plus. Add 7 to \sqrt{17}.

$$x=\frac{\sqrt{17}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from 7.

$$x=\frac{7-\sqrt{17}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+7}{4} x=\frac{7-\sqrt{17}}{4}$$

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.

$$2xx+1=x\times 7-3$$

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

$$2x^{2}+1=x\times 7-3$$

Subtract x\times 7 from both sides.

$$2x^{2}+1-x\times 7=-3$$

Multiply -1 and 7 to get -7.

$$2x^{2}+1-7x=-3$$

Subtract 1 from both sides.

$$2x^{2}-7x=-3-1$$

Subtract 1 from -3 to get -4.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -4 by 2.

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

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.

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

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

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

Add -2 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{17}{16}$$

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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{17}}{4} x-\frac{7}{4}=-\frac{\sqrt{17}}{4}$$

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

$$x=\frac{\sqrt{17}+7}{4} x=\frac{7-\sqrt{17}}{4}$$