If an equation is of degree 2, it means that the highest exponent is 2. Here are some examples of equations of degree 2.

f(x) = x² +3x –5

 

5x² –9 = 0

 

2x² +7x +3 = 0

 

In the section on factoring, it is shown how to solve many of these types of equations by factoring. Factoring is a good method but requires trial and error and can only easily be done on certain types of equations. The best way to solve these types of problems is by using the quadratic formula. The quadratic formula is a formula in which one plugs in certain values and gets results that are the answers. Before using the quadratic equation, make sure that the equation is in standard form. This basically means that it is set equal to 0. If the equation is not in standard form, it is usually easy to put it in standard form.

 

Standard form

ax² + bx + c = 0

 

The a will always be the number in front of the x² term. The b will be the number in front of the x and the c will be the number all by itself. Sometimes there might be an equation that does not contain the bx term. If that is the case then b = 0. If there is not a number out there by itself, c, then c = 0. This will be shown later.

 

After the equation is in standard form, find a, b, and c. Plug these values into the following formula. It is the quadratic formula.

 

 

 

 

This is read as “negative b, plus or minus, the square root of b squared minus 4 a c over 2a.” It is very important to memorize this formula by repeating this statement over and over again and be able to write the equation down from memory.

 

 

Here is a summary of the steps to solve an equation of degree 2 by using the quadratic formula.

 

1)      Get the equation in standard form.

2)      Find the values for a, b, and c.

3)      Plug in those values to the quadratic equation and get the answers.

 

 

 

 

 

Example

 

Solve x² +3x -10 = 0

 

First, see if this equation is in standard form. It is since it is set equal to 0. Find the values for a, b, and c. Since there is not a number in front of the x² term, it is understood that there is a 1 there, that would make a = 1. The number in front of the x term is 3, therefore b = 3, and c = -10. Plug these values into the quadratic equation.

 

 

 

 

Simplify underneath the radical by first doing the exponent (3²).

 

 

 

 

 

Multiply the 4(1)(-10).

 

 

 

 

 

Get rid of the parentheses and the two negative signs together.

 

 

 

Add the 9 and 40.

 

 

= 7.

 

 

Multiply the bottom.

 

 

The hard part is over. The ± sign means to look at -3+7 and -3-7. The first one, -3+7, will be solved first.

 

 

Add the top.

 

 

Divide.

 

 

This is one of the answers. Look at the -3-7.

 

Subtract the top.

 

Divide.

 

x = -5

 

This is the other answer. That means that -5 and 2 are the answers. To write these always start with the smallest number and then move on to the larger number.

x = -5, 2

 

Answer!

 

Note- The number of answers will always equal the degree of the equation. In the previous case, the equation was of degree 2, therefore there should be 2 answers.

 

 

 

 

 

 

Example

 

Solve x² –2 x = 8

 

First, see if this equation is in standard form. It is not since it is not set equal to 0. Subtract 8 from both sides.

 

x² –2 x – 8 = 0

 

It is now in standard form. Find the values for a, b, and c. a = 1, b = -2, and c = -8. Plug these values into the quadratic equation.

 

 

 

Simplify underneath the radical by first doing the exponent.

 

 

 

Multiply the 4(1)(-8).

 

 

Get rid of the parentheses and the two negative signs together, in the beginning as well as underneath the radical.

 

Add the 4 and 32.

 

 

= 6.

 

 

Multiply the bottom.

 

 

Now there are 2 problems, 2+6 and 2-6. Look at 2+6 first.

 

 

Add the top.

 

 

Divide.

 

 

This is one of the answers. Look at the 2-6.

 

 

Subtract the top.

 

 

Divide.

 

x = -2

 

This is the other answer.

 

x = -2, 4

 

Answer!

 

In the previous problem replace the 0 with a y, and it would be x² –2 x –8 = y. Notice that this equation has an x and a y in it. That means that it could be graphed. On a graph, y = 0 is on the x-axis. What this means is that when solving the equation x² –2 x –8 = 0, it is actually finding where on the graph of x² –2 x –8 = y crosses the x-axis. These are called the x-intercepts. Take the previous equation in graphing form, x² –2 x –8 = y, and graph it. Here is what the graph would look like.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Looking at the graph, the x-intercepts are at –2 and 4. This is a very important concept to understand when trying to solve these types of equations and those of a higher degree.

 

Note- The solutions to this type of problem may be called, x-intercepts, roots, or zeros. All of these names imply the exact same thing.

 

 

 

 

 

Example

 

Find the roots of 6x² = -x +2

 

See if this equation is in standard form. It is not since it is not set equal to 0. Add x to both sides, and subtract 2 from both sides.

 

6x² + x – 2 = 0

 

It is now in standard form. Find the values for a, b, and c. a = 6, b = 1, and c = -2. Plug these values into the quadratic equation.

 

 

 

 

Simplify underneath the radical by first doing the exponent.

 

 

 

 

Multiply the 4(6)(-2).

 

 

 

Get rid of the parentheses and the two negative signs together.

 

Add the 1 and 48.

 

 

= 7.

 

 

 

Multiply the bottom.

 

 

Now there are 2 problems, -1+7 and -1-7. Look at -1+7 first.

 

 

 

Add the top.

 

 

Divide.

 

 

This is one of the answers. Look at the -1-7.

 

 

Subtract the top.

 

 

Divide.

 

 

This is the other answer.

 

 

Answer!

 

 

 

 

Here is the graph of 6x² + x – 2 = y.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The graph crosses the x-axis at

 

 

 

 

Example

 

Find the roots of x² = -4x -13

 

First, see if this equation is in standard form. It is not since it is not set equal to 0. Add 4x to both sides, and add 13 to both sides.

 

x² + 4x +13 = 0

 

It is now in standard form. Find the values for a, b, and c. a = 1, b= 4, and c = 13. Plug these values into the quadratic equation.

 

 

Simplify underneath the radical by first doing the exponent.

 

 

Multiply the 4(1)(13).

 

 

Get rid of the parentheses.

 

 

Subtract the 16 and 52.

 

 

= 6i. If there is trouble on this step refer to the section on square roots and imaginary numbers.

 

 

Multiply the bottom.

 

 

 

Look at each term, the 4, 6i, and the 2 on bottom. Simplify since all of the terms have a common factor of 2. Divide each term by 2. This can only be done when all three terms have something in common.

 

Dividing by 1 does not change any value so simplify it out.

 

 

Since there are imaginary numbers there is no need to separate them out and look at each possible answer since they are not like terms. Leave the answer like this.

 

 

Answer!

 

Since there are imaginary numbers as answers what does this mean with the graph? In other words where will it cross the x-axis if it crosses it at imaginary places? The answer is that it will not cross the x-axis at all. Following is the graph.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Notice that it does not cross the x-axis and that is why there are imaginary numbers as roots.

 

 

 

 

 

Example

 

Find the roots of 5 x² + 6 x + 3 = 0

 

First, see if this equation is in standard form.

 

Since it is, find the values for a, b, and c. a = 5, b= 6, and c = 3. Plug these values into the quadratic equation.

 

 

 

Simplify underneath the radical by first doing the exponent.

 

 

Multiply the 4(5)(3).

 

 

 

 

 

Get rid of the parentheses.

 

 

 

 

 

Subtract the 36 and 60.

 

 

 

= 2i.

 

 

 

 

 

Multiply the bottom.

 

 

Look at each term, the -6, 2i, and the 10 on bottom. Simplify, since all of them have a common factor of 2. Divide each term by 2. This can only be done when all three terms have something in common. Do not simplify the 6 that is underneath the radical. It has to be outside of the radical to simplify it. Since it is underneath the radical, leave it alone.

Answer! This is as far as it will simplify.

 

 

Like the previous example, this has imaginary numbers as answers and therefore will not cross the x-axis.

 

 

It depends what is underneath the radical that determines the type of roots there will be. In other words, if the value underneath the radical is positive, there will be two real solutions. If the value underneath the radical is negative, there will be two imaginary solutions. Extend this a little further and if the value underneath the radical is 0, there will only be one solution. The value underneath the radical is represented in the quadratic formula by b² – 4ac. This is called the discriminant, therefore the value of the discriminant determines the nature of the roots. Here is a chart summarizing that.

 

 

Value of discriminant (b2 – 4ac)	Nature of roots	Graph
Positive	2 real	Crosses x-axis twice
Negative	2 imaginary	Does not cross x-axis
0	1 real	Crosses x-axis once

 

 

Example

 

Solve -x² + 4x - 4 = 0

 

First, see if this equation is in standard form.

 

It is in standard form. Find the values for a, b, and c. a = -1, b= 4, and c = -4. Plug these values into the quadratic equation.

 

 

Simplify underneath the radical by first doing the exponent.

 

 

Multiply the 4(-1)(-4).

 

 

 

Get rid of the parentheses.

 

 

 

 

 

Subtract the 16 and -16.

 

 

 

 

= 0.

 

 

 

 

Multiply the bottom.

 

 

Now there are 2 problems, -4+0 and –4-0. Look at –4+0 first.

 

Add the top.

 

 

Divide.

 

 

This is one of the answers. Look at the –4-0.

 

Subtract the top.

 

 

Divide.

 

x = 2

 

This is the other answer. Notice that whether adding or subtracting zero, will get the same values. Since 2 is the answer here and it is repeated, call it a double root.

x = 2

 

Answer!

 

Here is the graph of this equation showing that it only crosses the x-axis once at 2.