Solve 6/x+x=17

$$\frac { 6 } { x } + x = 17$$

Solve for x

$$x = \frac{\sqrt{265} + 17}{2} \approx 16.639410298$$

$$x=\frac{17-\sqrt{265}}{2}\approx 0.360589702$$

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.

$$6+xx=17x$$

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

$$6+x^{2}=17x$$

Subtract 17x from both sides.

$$6+x^{2}-17x=0$$

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.

$$x^{2}-17x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -17 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 6}}{2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-\left(-17\right)±\sqrt{289-24}}{2}$$

Add 289 to -24.

$$x=\frac{-\left(-17\right)±\sqrt{265}}{2}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{265}}{2}$$

Now solve the equation x=\frac{17±\sqrt{265}}{2} when ± is plus. Add 17 to \sqrt{265}.

$$x=\frac{\sqrt{265}+17}{2}$$

Now solve the equation x=\frac{17±\sqrt{265}}{2} when ± is minus. Subtract \sqrt{265} from 17.

$$x=\frac{17-\sqrt{265}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{265}+17}{2} x=\frac{17-\sqrt{265}}{2}$$

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.

$$6+xx=17x$$

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

$$6+x^{2}=17x$$

Subtract 17x from both sides.

$$6+x^{2}-17x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-17x=-6$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=-6+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=-6+\frac{289}{4}$$

Add -6 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{265}{4}$$

Factor x^{2}-17x+\frac{289}{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}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{265}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{265}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{\sqrt{265}}{2} x-\frac{17}{2}=-\frac{\sqrt{265}}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=\frac{\sqrt{265}+17}{2} x=\frac{17-\sqrt{265}}{2}$$