Solve 6-5/k-1/k^2=0

$$6 - \frac { 5 } { k } - \frac { 1 } { k ^ { 2 } } = 0$$

Solve for k

$$k=-\frac{1}{6}\approx -0.166666667$$

$$k=1$$

Steps Using Factoring By Grouping

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,k^{2}.

$$k^{2}\times 6-k\times 5-1=0$$

Multiply -1 and 5 to get -5.

$$k^{2}\times 6-5k-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6k^{2}+ak+bk-1. To find a and b, set up a system to be solved.

$$a+b=-5 ab=6\left(-1\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 6k^{2}-5k-1 as \left(6k^{2}-6k\right)+\left(k-1\right).

$$\left(6k^{2}-6k\right)+\left(k-1\right)$$

Factor out 6k in 6k^{2}-6k.

$$6k\left(k-1\right)+k-1$$

Factor out common term k-1 by using distributive property.

$$\left(k-1\right)\left(6k+1\right)$$

To find equation solutions, solve k-1=0 and 6k+1=0.

$$k=1 k=-\frac{1}{6}$$

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,k^{2}.

$$k^{2}\times 6-k\times 5-1=0$$

Multiply -1 and 5 to get -5.

$$k^{2}\times 6-5k-1=0$$

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.

$$6k^{2}-5k-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-1\right)}}{2\times 6}$$

Square -5.

$$k=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-1\right)}}{2\times 6}$$

Multiply -4 times 6.

$$k=\frac{-\left(-5\right)±\sqrt{25-24\left(-1\right)}}{2\times 6}$$

Multiply -24 times -1.

$$k=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 6}$$

Add 25 to 24.

$$k=\frac{-\left(-5\right)±\sqrt{49}}{2\times 6}$$

Take the square root of 49.

$$k=\frac{-\left(-5\right)±7}{2\times 6}$$

The opposite of -5 is 5.

$$k=\frac{5±7}{2\times 6}$$

Multiply 2 times 6.

$$k=\frac{5±7}{12}$$

Now solve the equation k=\frac{5±7}{12} when ± is plus. Add 5 to 7.

$$k=\frac{12}{12}$$

Divide 12 by 12.

$$k=1$$

Now solve the equation k=\frac{5±7}{12} when ± is minus. Subtract 7 from 5.

$$k=-\frac{2}{12}$$

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

$$k=-\frac{1}{6}$$

The equation is now solved.

$$k=1 k=-\frac{1}{6}$$

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,k^{2}.

$$k^{2}\times 6-k\times 5-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$k^{2}\times 6-k\times 5=1$$

Multiply -1 and 5 to get -5.

$$k^{2}\times 6-5k=1$$

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.

$$6k^{2}-5k=1$$

Divide both sides by 6.

$$\frac{6k^{2}-5k}{6}=\frac{1}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$k^{2}-\frac{5}{6}k=\frac{1}{6}$$

Divide -\frac{5}{6}, the coefficient of the x term, by 2 to get -\frac{5}{12}. Then add the square of -\frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{5}{6}k+\left(-\frac{5}{12}\right)^{2}=\frac{1}{6}+\left(-\frac{5}{12}\right)^{2}$$

Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{5}{6}k+\frac{25}{144}=\frac{1}{6}+\frac{25}{144}$$

Add \frac{1}{6} to \frac{25}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{5}{6}k+\frac{25}{144}=\frac{49}{144}$$

Factor k^{2}-\frac{5}{6}k+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{5}{12}\right)^{2}=\frac{49}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{5}{12}\right)^{2}}=\sqrt{\frac{49}{144}}$$

Simplify.

$$k-\frac{5}{12}=\frac{7}{12} k-\frac{5}{12}=-\frac{7}{12}$$

Add \frac{5}{12} to both sides of the equation.

$$k=1 k=-\frac{1}{6}$$