Solve x+1divx=5

$$x+1 \div x=5$$

Solve for x

$$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$$

$$x=\frac{5-\sqrt{21}}{2}\approx 0.208712153$$

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.

$$xx+1=5x$$

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

$$x^{2}+1=5x$$

Subtract 5x from both sides.

$$x^{2}+1-5x=0$$

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.

$$x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

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

Add 25 to -4.

$$x=\frac{-\left(-5\right)±\sqrt{21}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{21}}{2}$$

Now solve the equation x=\frac{5±\sqrt{21}}{2} when ± is plus. Add 5 to \sqrt{21}.

$$x=\frac{\sqrt{21}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{21}}{2} when ± is minus. Subtract \sqrt{21} from 5.

$$x=\frac{5-\sqrt{21}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$

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.

$$xx+1=5x$$

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

$$x^{2}+1=5x$$

Subtract 5x from both sides.

$$x^{2}+1-5x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-5x=-1$$

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -1 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2} x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$