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

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

Solve for x

$$x = \frac{\sqrt{21} + 1}{2} \approx 2.791287847$$

$$x=\frac{1-\sqrt{21}}{2}\approx -1.791287847$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by \left(x+5\right)x^{2}, the least common multiple of x^{2},x+5.

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

Subtract x^{2} from both sides.

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

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.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\times 5}}{2\left(-1\right)}$$

Multiply 4 times 5.

$$x=\frac{-1±\sqrt{1+20}}{2\left(-1\right)}$$

Add 1 to 20.

$$x=\frac{-1±\sqrt{21}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±\sqrt{21}}{-2}$$

Now solve the equation x=\frac{-1±\sqrt{21}}{-2} when ± is plus. Add -1 to \sqrt{21}.

$$x=\frac{\sqrt{21}-1}{-2}$$

Divide -1+\sqrt{21} by -2.

$$x=\frac{1-\sqrt{21}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{21}}{-2} when ± is minus. Subtract \sqrt{21} from -1.

$$x=\frac{-\sqrt{21}-1}{-2}$$

Divide -1-\sqrt{21} by -2.

$$x=\frac{\sqrt{21}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{21}}{2} x=\frac{\sqrt{21}+1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by \left(x+5\right)x^{2}, the least common multiple of x^{2},x+5.

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

Subtract x^{2} from both sides.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -5 by -1.

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

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.

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

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

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

Add 5 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{21}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{21}}{2} x-\frac{1}{2}=-\frac{\sqrt{21}}{2}$$

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

$$x=\frac{\sqrt{21}+1}{2} x=\frac{1-\sqrt{21}}{2}$$