Solve x^2-5/x-2=1

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

Solve for x

$$x = \frac{\sqrt{13} + 1}{2} \approx 2.302775638$$

$$x=\frac{1-\sqrt{13}}{2}\approx -1.302775638$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

Subtract x from both sides.

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

Add 2 to both sides.

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

Add -5 and 2 to get -3.

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

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.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12}}{2}$$

Add 1 to 12.

$$x=\frac{-\left(-1\right)±\sqrt{13}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{13}}{2}$$

Now solve the equation x=\frac{1±\sqrt{13}}{2} when ± is plus. Add 1 to \sqrt{13}.

$$x=\frac{\sqrt{13}+1}{2}$$

Now solve the equation x=\frac{1±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from 1.

$$x=\frac{1-\sqrt{13}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+1}{2} x=\frac{1-\sqrt{13}}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

Subtract x from both sides.

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

Add 5 to both sides.

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

Add -2 and 5 to get 3.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=3+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 3 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{13}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{13}}{2} x-\frac{1}{2}=-\frac{\sqrt{13}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{13}+1}{2} x=\frac{1-\sqrt{13}}{2}$$