Solve 14/a-a=3

$$1 \frac { 4 } { a } - a = 3$$

Solve for a

$$a=-4$$

$$a=1$$

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.

$$1\times 4-aa=3a$$

Multiply a and a to get a^{2}.

$$1\times 4-a^{2}=3a$$

Multiply 1 and 4 to get 4.

$$4-a^{2}=3a$$

Subtract 3a from both sides.

$$4-a^{2}-3a=0$$

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

$$-a^{2}-3a+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba+4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-4=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

Rewrite -a^{2}-3a+4 as \left(-a^{2}+a\right)+\left(-4a+4\right).

$$\left(-a^{2}+a\right)+\left(-4a+4\right)$$

Factor out a in the first and 4 in the second group.

$$a\left(-a+1\right)+4\left(-a+1\right)$$

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

$$\left(-a+1\right)\left(a+4\right)$$

To find equation solutions, solve -a+1=0 and a+4=0.

$$a=1 a=-4$$

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.

$$1\times 4-aa=3a$$

Multiply a and a to get a^{2}.

$$1\times 4-a^{2}=3a$$

Multiply 1 and 4 to get 4.

$$4-a^{2}=3a$$

Subtract 3a from both sides.

$$4-a^{2}-3a=0$$

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.

$$-a^{2}-3a+4=0$$

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

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

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-\left(-3\right)±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$a=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-1\right)}$$

Add 9 to 16.

$$a=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$a=\frac{-\left(-3\right)±5}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$a=\frac{3±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{3±5}{-2}$$

Now solve the equation a=\frac{3±5}{-2} when ± is plus. Add 3 to 5.

$$a=\frac{8}{-2}$$

Divide 8 by -2.

$$a=-4$$

Now solve the equation a=\frac{3±5}{-2} when ± is minus. Subtract 5 from 3.

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

Divide -2 by -2.

$$a=1$$

The equation is now solved.

$$a=-4 a=1$$

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.

$$1\times 4-aa=3a$$

Multiply a and a to get a^{2}.

$$1\times 4-a^{2}=3a$$

Multiply 1 and 4 to get 4.

$$4-a^{2}=3a$$

Subtract 3a from both sides.

$$4-a^{2}-3a=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-a^{2}-3a=-4$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\left(-\frac{3}{-1}\right)a=-\frac{4}{-1}$$

Divide -3 by -1.

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

Divide -4 by -1.

$$a^{2}+3a=4$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+3a+\left(\frac{3}{2}\right)^{2}=4+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+3a+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$a^{2}+3a+\frac{9}{4}=\frac{25}{4}$$

Factor a^{2}+3a+\frac{9}{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}.

$$\left(a+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$a+\frac{3}{2}=\frac{5}{2} a+\frac{3}{2}=-\frac{5}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$a=1 a=-4$$