Solve 1/a^2+1/a-6=0

$$\frac { 1 } { a ^ { 2 } } + \frac { 1 } { a } - 6 = 0$$

Solve for a

$$a=\frac{1}{2}=0.5$$

$$a=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring By Grouping

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

$$1+a+a^{2}\left(-6\right)=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-6a^{2}+a+1=0$$

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

$$a+b=1 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 1.

$$a=3 b=-2$$

Rewrite -6a^{2}+a+1 as \left(-6a^{2}+3a\right)+\left(-2a+1\right).

$$\left(-6a^{2}+3a\right)+\left(-2a+1\right)$$

Factor out -3a in the first and -1 in the second group.

$$-3a\left(2a-1\right)-\left(2a-1\right)$$

Factor out common term 2a-1 by using distributive property.

$$\left(2a-1\right)\left(-3a-1\right)$$

To find equation solutions, solve 2a-1=0 and -3a-1=0.

$$a=\frac{1}{2} a=-\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$1+a+a^{2}\left(-6\right)=0$$

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.

$$-6a^{2}+a+1=0$$

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

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

Square 1.

$$a=\frac{-1±\sqrt{1-4\left(-6\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$a=\frac{-1±\sqrt{1+24}}{2\left(-6\right)}$$

Add 1 to 24.

$$a=\frac{-1±\sqrt{25}}{2\left(-6\right)}$$

Take the square root of 25.

$$a=\frac{-1±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$a=\frac{-1±5}{-12}$$

Now solve the equation a=\frac{-1±5}{-12} when ± is plus. Add -1 to 5.

$$a=\frac{4}{-12}$$

Reduce the fraction \frac{4}{-12} to lowest terms by extracting and canceling out 4.

$$a=-\frac{1}{3}$$

Now solve the equation a=\frac{-1±5}{-12} when ± is minus. Subtract 5 from -1.

$$a=-\frac{6}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

$$a=\frac{1}{2}$$

The equation is now solved.

$$a=-\frac{1}{3} a=\frac{1}{2}$$

Steps for Completing the Square

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

$$1+a+a^{2}\left(-6\right)=0$$

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

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

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.

$$-6a^{2}+a=-1$$

Divide both sides by -6.

$$\frac{-6a^{2}+a}{-6}=-\frac{1}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$a^{2}+\frac{1}{-6}a=-\frac{1}{-6}$$

Divide 1 by -6.

$$a^{2}-\frac{1}{6}a=-\frac{1}{-6}$$

Divide -1 by -6.

$$a^{2}-\frac{1}{6}a=\frac{1}{6}$$

Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{1}{6}a+\frac{1}{144}=\frac{1}{6}+\frac{1}{144}$$

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

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

Factor a^{2}-\frac{1}{6}a+\frac{1}{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}.

$$\left(a-\frac{1}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$a-\frac{1}{12}=\frac{5}{12} a-\frac{1}{12}=-\frac{5}{12}$$

Add \frac{1}{12} to both sides of the equation.

$$a=\frac{1}{2} a=-\frac{1}{3}$$