Solve 18/k^2=5-1/k

$$\frac { 18 } { k ^ { 2 } } = 5 - \frac { 1 } { k }$$

Solve for k

$$k = -\frac{9}{5} = -1\frac{4}{5} = -1.8$$

$$k=2$$

Steps Using the Quadratic Formula

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k^{2}, the least common multiple of k^{2},k.

$$18=k^{2}\times 5-k$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}\times 5-k=18$$

Subtract 18 from both sides.

$$k^{2}\times 5-k-18=0$$

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.

$$5k^{2}-k-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -1 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-18\right)}}{2\times 5}$$

Multiply -4 times 5.

$$k=\frac{-\left(-1\right)±\sqrt{1-20\left(-18\right)}}{2\times 5}$$

Multiply -20 times -18.

$$k=\frac{-\left(-1\right)±\sqrt{1+360}}{2\times 5}$$

Add 1 to 360.

$$k=\frac{-\left(-1\right)±\sqrt{361}}{2\times 5}$$

Take the square root of 361.

$$k=\frac{-\left(-1\right)±19}{2\times 5}$$

The opposite of -1 is 1.

$$k=\frac{1±19}{2\times 5}$$

Multiply 2 times 5.

$$k=\frac{1±19}{10}$$

Now solve the equation k=\frac{1±19}{10} when ± is plus. Add 1 to 19.

$$k=\frac{20}{10}$$

Divide 20 by 10.

$$k=2$$

Now solve the equation k=\frac{1±19}{10} when ± is minus. Subtract 19 from 1.

$$k=-\frac{18}{10}$$

Reduce the fraction \frac{-18}{10} to lowest terms by extracting and canceling out 2.

$$k=-\frac{9}{5}$$

The equation is now solved.

$$k=2 k=-\frac{9}{5}$$

Steps for Completing the Square

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k^{2}, the least common multiple of k^{2},k.

$$18=k^{2}\times 5-k$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}\times 5-k=18$$

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.

$$5k^{2}-k=18$$

Divide both sides by 5.

$$\frac{5k^{2}-k}{5}=\frac{18}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$k^{2}-\frac{1}{5}k=\frac{18}{5}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{1}{5}k+\left(-\frac{1}{10}\right)^{2}=\frac{18}{5}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{1}{5}k+\frac{1}{100}=\frac{18}{5}+\frac{1}{100}$$

Add \frac{18}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{1}{5}k+\frac{1}{100}=\frac{361}{100}$$

Factor k^{2}-\frac{1}{5}k+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{1}{10}\right)^{2}=\frac{361}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{1}{10}\right)^{2}}=\sqrt{\frac{361}{100}}$$

Simplify.

$$k-\frac{1}{10}=\frac{19}{10} k-\frac{1}{10}=-\frac{19}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$k=2 k=-\frac{9}{5}$$