TRINOMIALS & QUADRATICS - A Guide of Hints, Strategies and Simple Explanations - Algebra in Words

Algebra in Words: A Guide of Hints, Strategies and Simple Explanations (2014)

TRINOMIALS & QUADRATICS

The words “trinomials” and “quadratics” are often used interchangeably because they overlap, both in characteristics, looks and application (particularly during factoring and solving). Despite their similarities, they should not be seen as completely synonymous by definition. Because of the way many books and lessons are arranged, sometimes these are seen and used too disconnectedly or separately. This is understandable as well (when done correctly), however this may also mislead students to miss the important connection and overlap between them. This section is to help you clearly relate and differentiate the similarities and differences between them, by definition and use.

A trinomial is an expression containing three different terms, often with at least one squared variable. I say “often” because when you are introduced to factoring (using the Trial & Error or Reverse-FOIL method, for instance), you are factoring trinomials into binomials. By definition, trinomials can be comprised of any three terms to any power, but trinomials are very often directly associated with factoring into two binomials as the segue-way to solvingquadratic equations.

A trinomial sometimes overlaps as a quadratic expression and may be part of a quadratic equation. Although an expression may be both a trinomial and a quadratic expression, they are not synonymous by definition. A trinomial is a quadratic expression when the highest power (degree) of any term is 2. When they do overlap, they can be simplified (factored) the exact same way. Also, not all quadratic expressions are trinomials, as you will read next.

A quadratic equation is:

-An equation containing a squared variable (like x2), yielding a maximum of two solutions [but could contain one solution (that occurs twice), or no solution].

-It must contain a squared variable and thus is considered a 2nd degree equation.

-Although a quadratic equation can contain a term of x (to the unwritten power of 1), it can never contain a term of a power higher than 2.

-Also, a quadratic equation, when graphed, always makes a parabola (a U-shaped curve).

A quadratic equation appears in the standard form:

ax2 + bx + c = 0

There are a few things you should understand about the equation written above.

It may also be written as:

y = ax2 + bx + c = 0,

in which y = 0, as above, or:

f(x) = ax2 + bx + c = 0,

because quadratics (which make parabolas) are considered to be “functions.” Specifically, they are functions of x.

(I do not delve into “functions” in this book, but if you’re wondering, an equation is considered to be a “function” if its graph crosses the y-axis once.)

It is written in descending order and standard form in this case. For a quadratic equation, “standard form” means all terms are on one side of the equal sign, and set equal to zero (on the other side). Descending order means the terms are arranged from the highest to the lowest power, from left to right.

All quadratics are not always originally presented in descending order or standard form. If and when they are not, you should rearrange each one into standard form and descending order before simplifying and solving.

The letters a, b and c are representative of numbers, not variables (more on that down the page). Also, “a” cannot be zero. If “a” is zero, it is no longer a quadratic equation; it is then a linear equation.

A quadratic equation contains a trinomial expression when a, b and c are all non-zero numbers. However, sometimes, either the coefficient b, or constant c, or both, are zero (remember, “a” cannot be zero). This is worth highlighting because this is where students often run into trouble. I believe they run into trouble at first because, when b or c is zero, the equations just look differently, and usually the solving method is different. For that reason, there a segment dedicated to those specific cases. I will show what they look like, explain their graphical significance, how to solve them, and their expected solutions. This is continued in: Quadratics With Zero. But first you should understand the solutions to quadratic equations.

What Are “Solutions” to Quadratic Equations?

It is good you know what solutions to quadratic equations are, to give you a better, overall perspective. (The variable of a quadratic equation is usually x, but can be other letters). When the variable is x, the solutions are “x-intercepts,” which are simply the points on a graph where the parabola crosses (or the single point which touches) the x-axis; the x-intercepts are defined the same, no matter what type of equation or graph they come from. And x-intercepts are points (ordered pairs) at which y = 0, which is why you set your quadratic equation equal to zero at first (in other words, making sure it is in standard form). This is also why, if you solve by factoring, you set each factor equal to zero; or, if you solve by the quadratic formula, this is why the formula is set equal to zero. You will learn about this more in the next section.

All quadratic equations produce a parabola when graphed. The solutions are the x-intercepts of the graph. As you will see in the next few sections, there are a number of ways to solve quadratic equations. If you solve by factoring, you will get one or two solutions; those solutions will be either integers or fractions. If you solve by the quadratic formula, your answers may come out to be integers, fractions or radicals (which could be converted to decimals for graphing).

Also, you may find that a quadratic equation has either one, two, or “no real” solutions. Here is a quick summary of each scenario:

·   One solution means that the parabola only touches the x-axis once; it does not cross the x-axis. You may think of it as “sitting on” the x-axis. Another more technical way to say it is, “The x-axis is tangent to the vertex of the parabola.”

This will occur when the trinomial factors into a binomial squared. It is also good to know that a binomial squared comes from a “perfect square trinomial.”

·   Two solutions means the parabola crosses the x-axis twice.

·   Finally, you may find “no real solutions”. This means that the parabola does not cross or touch the x-axis, but don’t be fooled. Just because it doesn’t cross or touch the x-axis doesn’t mean it doesn’t exist… it still exists, and can still be graphed. This conclusion can only be made through use of the quadratic formula. If a quadratic equation is prime (which only means it can’t be factored), this is still not grounds for saying “no solution”… it may just mean the solutions are radicals. But it may also mean there are “no real” solutions. The reason this is specifically answered as “no real” solution, instead of no solution, is because this is often the result of the square root of a negative number. For more on that, see: The Square Root of Negative One, and: Prime vs. No Solution.

These concepts and a closer look at the solving methods behind them are discussed in more detail in the next few sections.  

Solving Quadratic Equations

Trinomials and quadratic equations can be solved in three general ways:

1. Factor & Solve

1b. Take the Square Root of Both Sides

2. Use the Quadratic Formula

3. Graph & Check

When using factoring to solve, there are actually three different factoring methods you can use, so you might say there are five or six total possible ways to solve quadratic equations, (although, as you will see, the factoring method(s) won’t always work).

You also see “Taking the Square Root of Both Sides” in the list as “1b.” This is an alternative method to certain cases in which factoring can be used as well. There is a type of equation which can be solved in two ways, either by “Factor & Solve” or by “Taking the Square Root of Both Sides”. This is explained in: When both b & c are 0: ax2 = 0.

I re-wrote the list of ways to solve quadratic equations below, with the more specific, sub-methods included, so you get a concise list of methods and choices you can use.

1. Factor & Solve:

·   Trial & Error/Reverse FOIL Method

·   The ac/Grouping Method

·   The Complete the Square Method

1b. Take the Square Root of Both Sides

2. Use the Quadratic Formula

3. Graph & Check

In the following sections, I will go over the when as opposed to the how (see your text book for the “how”). I have good reasons for this. The textbooks usually do a good job of showing you how to implement the methods, and the steps are not really that complicated; a lot of practice is the key to becoming good at factoring and solving quadratic equations. However, the books don’t usually answer a question many students have, which is, “when is the best time to use each method?” I’m going to answer that question, as well as give more of a top-down perspective on solving quadratic equations.

When dealing with quadratics, you should also get accustomed to starting them (or preparing them) the same way, no matter which method you use to solve them. You should always:

·   Look for a GCF to factor out (this is a big one students often forget to do), and

·   Arrange into descending order and standard form.

Factor & Solve

  Usually, you should try to factor and solve a quadratic equation (before using the quadratic formula) because it’s faster and involves fewer steps (if it’s able to be factored). Factoring and solving can lead you to the answer(s), however if you can’t factor, this leaves your answer inconclusive. If a quadratic is prime (can’t be factored), it doesn’t necessarily mean there is no solution, but you must then use the quadratic formula to come to that conclusion (to either find the answers or find that there is no real solution).

  But many quadratic equations can be factored. There are three general ways to factor, but more importantly, there are better times to use each method and clues to dictate when those times are. Although I do not teach you how to do each method (as I stated, your textbooks do a good job at that), I will highlight the clues and tell you the best time to use each method.

  Before factoring, you must go through a series of steps to set up and prepare your equation, no matter which method of factoring you will use. These are very important, and students often forget one or all of these because you don’t always have to do them:

·   Simplify as much as you can by combining like-terms, if necessary.

·   Arrange all terms into Descending Order according as: ax2 + bx + c = 0.

·   Put into Standard Form by moving all terms to one side (the left) and setting them equal to zero.

·   Look for a Greatest Common Factor. Sometimes you can factor and solve successfully if you forget to do this, but it will often leave all numbers larger, and the problem more tedious. If the GCF is a number, you can factor it out, then remove it (because if you divide both sides by it, the zero divided by it on the other side eliminates it, and the zero remains zero).

o However, if the GCF is a variable or a power of “x”, it won’t be eliminated, but it will equal zero as one of your solutions. Factoring out a variable may allow you to properly factor using one of the factoring methods (including to factor again) that you otherwise wouldn’t be able to do. Actually, when the GCF contains a variable or power of x, the problem might not have been a quadratic to begin with. Factoring that out may leave you with a quadratic that you can then factor by quadratic methods.

·   Make sure the coefficient of the leading term (the “a” connected to x2) is positive. You can’t factor if it’s negative (it can be negative, though, when using the quadratic formula). If it is negative, treat -1 as a GCF of each term. By factoring it out, you will simply change the sign of each term, and the zero on the other side isn’t affected.

·   (Optional) You may choose to eliminate all fractions first, if there are fractions. You are usually taught that it is a good rule of thumb to begin any type of problem by removing fractions by multiplying by the LCD. You don’t have to though, and sometimes you can even factor them into binomials, but you will account for them in the final solving steps if you don’t remove them first.

Trial & Error/Reverse FOIL Method

Various books have different ways of naming factoring methods. I use both these names here because I think they accurately describe the process they’re used for. To do this method, you must simplify (combine like-terms), arrange all terms into standard form (move all terms onto one side), and put into descending order. You then pay attention to the factors of the first and last terms of the trinomial, write out the two binomials (or, once you get good enough at it, keep them in your head) then FOIL these factors to see if you arrive back at the original trinomial. If it works, you’ve found your factors; then set each binomial equal to zero and solve. Keep trying factors until you find the ones that work (that’s what’s trial and error about it). Also, I call it “Reverse-FOIL” because you are starting with the product (the original trinomial), then coming up with factors to try, then FOILing them to check. But as I stated, my intention is not to focus on the how, but the when. So when is the best time to us this method?

It is common to first approach a trinomial/quadratic expression with this method because, if it can be done (easily), it can be the quickest method with fewest steps. An equation with an expression of this type has the best likelihood to be solved by starting with the Trial & Error Method when “a” and “c” are relatively small. I use the word “small” loosely, and everyone’s interpretation of “small numbers” may be a little different. My use of “small number” could be taken two ways. It could mean a number with few factors, or it could be taken more literally, meaning between 1 and about 15. Either way, the smaller the number, the fewer factors it will tend to have.

Ideally, prime numbers or numbers with few factors will yield the fewest possible combinations of factors. The fewer factor combinations there are, the fewer the choices there are to try (multiply and test). There’s not really much else to say about this method. As the “a” and “c” numbers approach larger values, there could be so many possible combinations of factors to trial that it can become very time consuming. When the size and possibilities of factors seems overwhelming, this is a good time to do the “ac/Grouping Method.”

The ac/Grouping Method

You will often first be exposed to the “Grouping Method” when you are learning factoring (before learning to solve quadratic equations and equations with trinomials). The Grouping Method is simply a method of factoring that is introduced to teach you how to factor four terms [if there are enough similarities (common factors) among pairs of terms]. Books don’t often call the “ac Method” the “ac/Grouping Method;” they usually call it one or the other. This ac/Grouping Method is comprised of two main parts:

·   First using the a and c to find the correct factors (by multiplying a and c, then looking at all the two-factor combinations of that product, called “ac”), then

·   Writing out the four associated terms and using the grouping method to factor (into two binomials),

then solving.

When is the best time to use this method? It is worth noting that there are often two groups of students: those who prefer the Trial & Error/Reverse FOIL Method, and those who prefer this ac/Grouping Method. The reason I’m telling you this is because you can always skip the Trial & Error Method and start by using this method every time (in other words, you don’t have to try the Trial & Error Method first, then go on to this method next if Trial & Error doesn’t work out) if you like this method better. Some students prefer this method because it can be quicker on average, because it removes a lot of guess work (trial and error) and the time spent on all the error factor combinations.

Regardless, a good time to use this method is when the a and c factors are large and/or have many factors. To give you an idea of what might be “large” or “having many factors,” if a = 18 and c = -24, there could be many factor combinations from them. The “18” has three pairs of factors {(1∙18), (2∙9), (3∙6)}, and the “-24” has eight pairs of factors {(-1∙24), (1∙ -24), (-2∙12), (2∙ -12), (-3∙8), (3∙ -8), (-4∙6), (4∙ -6)}. The negative sign doubles the number of factor combinations, because either factor could be negative.

Start by multiplying a and c; this gives you the product “ac,” (it will be an actual number). You are then to look at every possible two-integer-factor combination of the product ac. You may consider setting this up in the following way: make two columns: one with the heading “factors,” and the other “b.” The point is to find the numbers that when multiplied give you the product of “a” times “c”, and when added give you the middle term of the original trinomial, “b.” When the “a” and “c” numbers of a trinomial are large or have many factor possibilities, this method will help you quickly find the combination needed to complete the “factoring by grouping” method. Again, don’t forget to solve your binomials once you factor into them.

Complete the Square

Completing the Square is definitely in a category of its own. You may not even consider it factoring by the same definition as the other factoring methods, since it is more-so like a manipulation technique, involving factoring as a step.

This method is mainly used when conventional factoring doesn’t work, because the

c-number may not factor into integers, but it can still be manipulated.

It’s important to remember that to begin, the leading coefficient must always be positive 1, so before proceeding, if the coefficient is anything other than 1, divide each term by the “a” coefficient, and this will make the coefficient of the leading term “1” (and the other terms will change proportionally). Also, as you make your new “c-number,” don’t forget to add it to the other side, to maintain the equality. Students often forget to do the two things mentioned in this paragraph.

This creates a “perfect square trinomial” (on one side). A perfect square trinomial is a special case, in which the coefficient of the leading term will be an unwritten “1,” (which is a perfect square), the new c-number you made will be a perfect square number [this new c-number may also be referred to as , which is the formula for how to make it], and the coefficient b will be exactly two times the square root of the new c-number.

According to this special case, “A perfect square trinomial factors to a binomial squared.”

But this slightly deviates from a regular problem where you’re given a perfect square trinomial to solve. A perfect square trinomial will factor into a binomial squared, and when set equal to zero and solved will give you one answer. This is because a perfect square trinomial makes a parabola which touches (but doesn’t cross) the x-axis, thus it has one x-intercept (solution). However, when you complete the square, your

(what becomes a) binomial squared equals a number on the other side, and once it is solved, will result in two answers (not one).

The Quadratic Formula

The Quadratic Formula is:

in which “a,” “b” and “c” refer to the numbers from the

standard form equation: ax2 + bx + c = 0.

The thing about this formula/method is that it always works. It will work when a quadratic equation can or can’t be factored. Even if there is no solution, the endpoint of this method will reveal that.

About half the time, your solution(s) from using the quadratic equation won’t be integers or even rational numbers. In those cases, the best way to express your answers will be in radical form (as opposed to decimal form). If you ever wonder why simplifying radicals is drilled into your minds, it’s so you can use and navigate through the Quadratic Formula from beginning to end. But getting to the very end… the very last step is where students commonly make a mistake. Simply put, they often forget to finish it.

Note: “Standard form” is slightly different for quadratic equations than linear equations.

The Part Everyone Forgets (The Last Step of the Quadratic Equation)

Sometimes, at this point, your answer is completely simplified… but sometimes it’s not. You should never assume it is completely simplified until you attempt this last step. Consider you’ve gone through the Quadratic Formula and get to this last step:

Look for a GCF in the numerator (and factor it out), and factor the denominator. In this example, the GCF in the numerator is 3, which should be factored out. In the denominator, 6 factors into 3 and 2. Now it can be seen that the numerator and denominator have a common factor of 3:

Cancel the common factor of 3 out of the numerator and denominator, then re-write the simplified answer. If there is a radical in your answer and you must graph it (remember, an answer is an x-intercept point), convert your radical to a decimal and reduce everything to one number. Also, take extra care not to do the last step improperly, as many often do, as explained in: The Wrong Way To Simplify a Rational Expression

Graph & Check

Graph & Check is much different for quadratic equations than it is for linear equations. For example, Graph & Check for linear equations is used to find a solution of a system of two linear equations (the point where two lines cross). The circumstances (solutions) are different for quadratic equations because quadratics make parabolas. Therefore the way to find points to be graphed is different, because you can’t just find and graph any three random points for a quadratic with a guarantee that they will be representative of the complete (parabolic) shape, as you would for a linear equation.

Also, the context of the word “solutions” is different for linear (systems) and quadratic equations. Solutions to quadratic equations are x-intercepts. [Just to be clear, you can (and do) find the x-intercept of a linear equation (you just don’t call it the “solution”), and you could also plot two parabolas on the same graph and find their points of possible intersection… but again, those points aren’t the primary contextual use of “solutions,” and it’s something you aren’t commonly asked to do].

Here are the minimal points you need to graph a quadratic equation:

·   The vertex

·   The x-intercept(s), AKA the “solutions”

·   The y-intercept

·   Any additional points

Let’s look at these points in greater depth.

Regarding the y-intercept, every quadratic (parabola) has one. If you have the equation written in descending order and standard form, it’s the number which is “c” from

ax2 + bx + c = 0. Every parabola will cross the y-axis (once), regardless of if it crosses the x-axis.

The vertex must always be found, as this is the (either maximum or minimum) point where the graph shifts from the positive to negative direction (or vice versa). This is its inflection point.

As for the x-intercepts, you may find:

·  two x-intercepts, if the parabola crosses the x-axis,

·  one x-intercept, if the (vertex of the) parabola touches but doesn’t cross the x-axis (as is the case for perfect square trinomials), or

·  no x-intercepts, if the parabola neither touches nor crosses the x-axis. For this, you will get “no solution”, but the parabola may still exist.

This is where “any additional points” comes in. If you found the vertex, the y-intercept, and the x-intercepts, you can successfully sketch a decent representation of the parabola, by graphing the points and drawing the line through the points in a smooth, curved way (not in a rigid way as if you were playing “connect the dots”). Finding more points will just help you make a more accurate and complete curve.  

Quadratics with Zero

This section is dedicated to showing you that the following cases are still considered quadratics (and therefore also second degree equations. Actually, quadratics in which b or c are zero are called “incomplete quadratics”). The coefficient “a” must be a number other than zero, otherwise, the equation would no longer be a quadratic. Also, keep in mind that when b = 0, the “bx” term equals zero and will not be written. Likewise, when c = 0, it won’t be written. However, a zero will appear (only) if it is alone on one side of the = sign. Even though one or two terms within a quadratic can be zero, you may have to solve them differently than if all terms are non-zeros. Here, we look at those cases.

When c is 0: ax2 + bx = 0

When c = 0,

ax2 + bx + 0 = 0

which will be shown as:  

ax2 + bx = 0

A few examples are:

A) 4x2 + 2x = 0,

B) 3x2 + x = 0,

C) -x2 + 5x = 0, or

D) 7x2 – 3x = 0.

In either case, you will always expect two solutions: x = 0, and x = another #.

Note: In cases where “c is 0,” then “x = 0” is always one of the solutions.

From a graphical perspective, anytime c is 0 in a quadratic equation, the resulting parabola will always cross through the origin (0, 0), as well as another point along the x-axis. As in any quadratic equation, “c” represents the y-intercept, which in this case is 0. In this case, the origin (0, 0) is both the y-intercept and one of the x-intercepts.

These are the steps to solving:

·   Factor out the GCF… which will include “x” and possibly a number as well.

·   Set the outside factor(s) equal to zero, and in this case, this “x” automatically equals zero.

·   Set what is inside the parentheses equal to 0 and solve for x.

Using the first example: 4x2 + 2x = 0,

factor out x and 2 (the GCF is 2x) from both terms:

2x(2x + 1) = 0,

set the outside factors equal to zero:

2x = 0, divide both sides by 2, and thus x = 0.

Set what’s inside the parentheses equal to zero and solve for x:

(2x + 1)à 2x + 1 = 0.

Subtract 1 from both sides:    

2x + 1 – 1 = 0 – 1

Giving: 2x = -1.

Divide both sides by the coefficient 2, and x =

The solutions are x = 0 and x = .

When Both b & c are 0: ax2 = 0

When both b and c = 0,

ax2 + 0x + 0 = 0,

which will be shown as: ax2 = 0.

Some examples are:

2x2 = 0,

x2 = 0, or

-3x2 = 0.

In all cases, the only solution is x = 0 (because if you divide both sides by the coefficient in front of x2, then take the square root of both sides, you will get “0”).

Graphically, this will produce a parabola whose vertex is the origin (0, 0), with the line “x = 0” as the vertical line of symmetry. There is only one solution because (the vertex of) the parabola of this type touches but does not cross the x-axis. Here, the origin (0, 0) is both the y-intercept and (the only) x-intercept.

When b is 0: ax2 + c = 0

When only b = 0,

ax2 + 0x + c = 0

which will be shown and seen as: ax2 + c = 0

Some examples are:  

E) 2x2 – 2 = 0,

F) 9x2 – 4 = 0,

G) x2 – 36 = 0,

H) 4x2 + 25 = 0, or

I) 3x2 – 5 = 0.

Of the given examples, all except “4x2 + 25 = 0” are considered to be:

“The difference of two squares.” Example H is discussed a few pages later.

“The Difference of Two Squares”

We will look at each example, specifically, but before that, it’s important you see the two ways in which problems like these can be solved, so you can notice the pattern in the examples to follow. Either approach is started in the same way.

First, look for a GCF. If there is a GCF, factor it out, then proceed to divide both sides by it. Since zero is on the right, any (non-zero) number you divide by it will equal zero. At this point, there are two ways you can proceed to solve.

One way is by moving the constant to the other side of the equal sign, then taking the square root of both sides. The other is by factoring into binomials. We will look at each method in more detail. Either choice is completely valid. It’s really up to you to decide which method you prefer. To reiterate, the two ways are:

A1. Factor into conjugate pair binomials, or

B1. Take the square root of both sides.

There is a time that “taking the square root of both sides” will be preferable, as I will show in the following examples.

Example E: 2x2 – 2 = 0

This is a classic example of factoring out the GCF first, which here is 2.

2(x2 – 1) = 0

Divide both sides by 2, which gives you

(x2 – 1) = 0

At this point, you can go forward with either method. I’m going to demonstrate both, to prove that each is valid and yields the same outcome. But first, I’m going to show “factoring into conjugate pair binomials.”

As “x2 – 1” is the difference of two squares, it can easily be factored into

(x – 1)(x + 1) = 0

Set each factor in parentheses equal to zero, and solve for x.

x – 1 = 0    x + 1 = 0

x - 1 + 1 = 0 + 1  x + 1 – 1 = 0 - 1

x = 1    x = -1

so x = +/- 1

There is a very important lesson in this example, which is that

“the difference of two squares can be factored into conjugate pair binomials.”

It should also be expected that “the difference of two squares” will always yield two opposite solutions (however there is one technical exception to this, explained in: Clarification: When the Solution is 0).

Conjugate Pair Binomials

It’s good to be familiar with conjugate pair binomials, visually, by definition, by name, and by common use.

Conjugate pair binomials are the result of factoring “the difference of two squares.”

They are known as conjugate pair binomials because…

·   they are conjugate in that they are joined and connected in some way,

·   they come in pairs (as conjugates do), and

·   they are binomials (each set of parentheses contains two terms).

They appear as two parentheses with the same first and second terms, but opposite signs in between them.

When conjugate pairs are multiplied, the result is “the difference of two squares.”

The advantage of identifying and factoring the difference of two squares into conjugate pair binomials is that it is quick and involves few steps.

Also the use of conjugate pair binomials are an essential part of “rationalizing the denominator” when the denominator contains a binomial with at least one radical. This topic is one that is usually covered near the end of the semester, often displaced from the lessons which introduce factoring quadratic equations and “special cases.” During this time displacement, students sometimes forget to see the connection of this concept. Additionally, upon learning it, students often do not realize the relevance for which it will be needed later.

More on the procedure for multiplying conjugate pair binomials and graph-related information is covered in the Special Case subsection: The Difference of Two Squares.

The other way to solve for x, starting from “x2 – 1” is by taking the square root of both sides

Taking the Square Root of Both Sides

We start back with “x2 – 1” from example E to see that it can also be solved by taking the square root of both sides. It is set equal to zero.

x2 – 1 = 0

In this case, we proceed by moving the constant (here, -1) to the other side:

x2 – 1 + 1 = 0 + 1, making it x2 = 1.

Take the square root of both sides:

Remember that taking the square root of a number gives the positive and negative root number (because if you square a positive or negative number, you always get a positive result), so it can be said that x equals plus or minus 1, written as

x = +/- 1.

The next two examples demonstrate “factoring the difference of two squares into conjugate pair binomials.”

Example F: 9x2 – 4 = 0

Factored: (3x – 2)(3x + 2) = 0

Notice that 3x is the square root of 9x2 and 2 is the square root of 4. Set each set of parentheses equal to zero and solve for x.

3x – 2 = 0    3x + 2 = 0

3x – 2 + 2 = 0 + 2   3x + 2 – 2 = 0 – 2

3x = 2    3x = -2

divide both sides by 3   divide both sides by 3

x =

Example G: x2 – 36 = 0

Factored: (x – 6)(x + 6) = 0

x – 6 = 0    x + 6 = 0

x – 6 + 6 = 0 + 6  x + 6 – 6 = 0 – 6

x = 6 and -6

Example I: 3x2 – 5 = 0

This is an interesting example, one of which you are sure to encounter. It’s important to remember that you can take the square root of any (positive) number or term, but only when you take the square root of a perfect square will your result be integers (non-radical or non-decimal numbers). In the three examples before this one, the terms were perfect squares and therefore could be factored (into binomials), however in this example, the 3 and the 5 are not perfect squares (however, the x2 still is)… so they can’t be factored using integers. Therefore, when one or both of the terms involved are not perfect squares, it is often preferred to approach solving by moving the constant to the other side of the equation and taking the square root of both sides, as seen in the following steps.

3x2 – 5 + 5 = 0 + 5

3x2 = 5

Divide both sides by 3, giving:

x2 = ,

take the square root of both sides:

and

x = +/-

The Sum of Two Squares

In any quadratic equation in which b = 0, any time the c-number is added, the polynomial in the equation is considered prime, and has no real solution. (Please see the two notes below).

You should recognize equations such as these as:

“the sum of two squares.”

and you should remember:

“The sum of two squares is prime.”

When faced with a problem such as this, the acceptable answer responses are:

“prime,” and “no real number solutions.”

*Please Note: If you look at the standard form of a quadratic equation: ax2 + bx + c = 0, or in this case: ax2 + c = 0, you probably notice that c is added, which may lead you to think that every equation such as this is prime, but this is not enough information to make this judgement. To be clear, the form ax2 + c = 0 is correct, but whether the equation is prime all depends on the sign of the actual number that is represented by c. In other words, if the number plugged-in for c is negative, you have the difference of two squares, which is not prime. Or, if the number for c is positive you have the sum of two squares, which is prime. There is technically one exception to this, explained in:Clarification: When the Solution is 0.

**Also Please Note: The first sentence of this section states, “… the equation is considered prime, and has no solution.” This statement must be properly understood. Although the statement may seem to insinuate that “prime” is synonymous with “no solution,” by definition, this is not true. Since the “sum of two squares” is so common, you can predict its outcome of “prime” and “no solution” automatically, but the fact that they have both these outcomes is merely coincidental. (This explanation is continued in: Prime vs. No Solution). With that in mind, remember that “the sum of two squares” can still be graphed and will produce a parabola.

Now let’s look at two more examples, J and K. Notice each example is similar to an example from before, but they both have a negative sign in the leading coefficient. What we’re going to do is simplify each example first, and then decide if what remains is prime or factorable. To simplify, we must do a number of things:

·   Identify and factor out a GCF (if there is one)

o Divide both sides the by the GCF (if there is one)

·   Move the constant to the other side of the equal sign

·   Ensure the coefficient in front of x2 is positive 1

o This may have been taken care of in a previous step, otherwise…

o You can do it now by dividing both sides by the coefficient in front of x2

Look at Example J:

-4x2 + 25 = 0

Let’s simplify by going through the procedure just previously mentioned:

·   Is there a GCF? No.

·   …So there is no GCF to divide both sides by.

·   Move the constant, 25, to the other side by subtracting it from both sides:

-4x2 + 25 - 25 = 0 – 25

-4x2 = –25

·   Ensure the coefficient in front of x2 is positive 1. In this case, we will divide both sides by -4, which cancels out both negative signs, giving:

x2 =

Take the square root of both sides… which we can do, because the sign of is positive.

The solutions are: x = +/- .

Next, let’s look at Example K:

-3x2 – 5 = 0

And follow the steps to simplify, as we did in the last example.

·   Is there a GCF? In this case, yes. It is “-1”; factor that out:

-(3x2 + 5) = 0

·   Divide both sides by the GCF “-1”, which makes it:

3x2 + 5 = 0

If we pause here, we see that we have the sum of two squares, which is enough information for us to stop solving, and answer with: “prime; no real-number solutions”

Note: If you’re wondering if 3 and 5 are square numbers, you could say “sort-of” because they are the squares of the square root of 3 and the square root of 5, respectively; however 3 and 5 are not perfect squares. Any positive number is a square of another number, but a perfect square is the result of squaring an integer. Let’s proceed with the remaining two steps anyway to see what happens if we continue to simplify and solve.

·   Move the constant 5 to the other side by subtracting 5 from both sides, giving

3x2 = -5

·   Divide both sides by 3 to ensure the coefficient in front of x2 is 3, giving

x2 = -5

At this point, when you go to take the square root of both sides, you should realize that you can’t take the square root of a negative number and get a real-number solution. This further proves that “-3x2 – 5” has no (real) solution.

Special Words for Special Cases

The textbooks are generally good at highlighting the special case (quadratics) in their own section, and teaching how to solve them. And students are generally good at factoring and solving them when they are in their own isolated areas and when they know what type they’re dealing with. But once the special cases are mixed into general types of problems, students sometimes forget the signals in identifying them.

Identifying them is the first crucial step. This section contains a few sentences and key words that will help you identify the type of special case, and tell you what outcome to expect in terms of factoring, graphing, and the shortcut for multiplying. If you memorize these words, it will help you figure out the factors and answers more quickly. Some parts may seem redundant from some earlier parts of the book, but that’s okay, the repetition is good for you.

  I cover two main special cases in the coming pages. Please note that there is another common special case involving “the sum and difference of two cubes,” which I don’t cover in this book (because my primary focus is linear and quadratic equations, and squares and square roots).

Perfect Square Trinomial

Remember that: “A perfect square trinomial factors into a binomial squared.”

Likewise, “A binomial squared, when multiplied out, gives a perfect square trinomial.”

You will notice in a perfect square trinomial that:

·   the first term (ax2) and the last term (c) are perfect squares,

·   the coefficient b will be the product of

o (the square root of the first term, ax2),

o (the square root of the last term, c), and

o 2;

·   the sign in front of c will always be positive.

o It is always positive because the last number in the binomial is squared.

It will resemble:

(Perfect square number)x2

+ x

+ perfect square number

Such as: 4x2 - 12x + 9 = (2x – 3)2

with the perfect square trinomial on the left, equal to its factored binomial squared on the right; and another example:

x2 + 8x + 16 = (x + 4)2

There are two ways to multiply a binomial squared. One way is by expanding it into two binomials and multiplying by the FOIL method. This way is fine, but you are highly encouraged to use the shortcut. This is the shortcut for multiplying a binomial squared, in words. I will refer to the terms of the binomial as such: (first term + last term).

1.  Square the first term.

2. Leave space (for what will be instruction #4).

3. Square the last term.

4. Multiply the first term times the last term, then double it, and write it in the space you left in instruction #2.

* A common mistake made is people treat this method like the shortcut for multiplying conjugate pair binomials, and they forget to do step 4, so they are then missing the middle (bx) term.

Do you see the connection of how the procedure for this special case is applied when completing the square?

When graphed, a perfect square trinomial is a parabola which touches (but does not cross) the x-axis. The vertex is at y = 0 and the solution to x, and (said to be) tangent to the x-axis.

The Difference of Two Squares

Remember that: “The difference of two squares factors into conjugate pair binomials.”

Likewise, “Conjugate pair binomials, when multiplied, give the difference of two squares.”

The difference of two squares is a specific case of

when b is 0; ax2 + c = 0, and “c” is a negative number. When “c” is a positive number, you have “the sum of two squares.”

Remember: “The sum of two squares is prime.”

There are two ways to multiply conjugate pair binomials. You can multiply them using the FOIL method, but you are encouraged to use the special case shortcut method. Getting used to the shortcut will be useful when you learn to rationalize denominators with radicals and binomials (not covered in this book). Plus, the shortcut is faster. This is the shortcut method in words:

1.  Square the first term,

2. Write a minus sign,

3. Square the last term.

There will be no middle (bx) term because if you FOIL the conjugate pair binomials, the product of the Outer terms plus the product of the Inner terms cancel each other out to zero. This is what makes the difference of two squares a special case.

When graphed, this makes a parabola that:

·   Has two x-intercepts because it

·   crosses the x-axis in two places,

·   has a vertex at x = 0, [at point (x, c)], and

·   the line of symmetry is the y-axis (the line: x = 0).

There is one technical exception to the statement that “the difference of two squares yields two solutions,” which is explained in: Clarification: When the Solution is 0.

Prime vs. No Solution  

As in the section: The Sum of Two Squares, when the quadratic expression [in an equation in which b (only) is zero] is prime, the result is “no (real) solution.” By definition, “prime” should not be thought as synonymous with “no solution.” The following instances can occur:

·   the polynomial can be factored and does yield (real) solutions;

·   the polynomial can’t be factored and yields no (real) solutions; and

·   the polynomial can’t be factored but does yield (real) solutions; however,

·   a polynomial which can be factored will always yield one or more (real) solutions.

Keep their definitions in mind. “Prime” means “can’t be further factored (into factors other than itself and 1).” And “No (Real) Solution” with regards to a polynomial means that the resulting graph has no x-intercepts (does not cross or touch the x-axis).

A good example is when you have a quadratic equation containing a trinomial expression, in other words, there are non-zero numbers in for a, b and c. However, as discussed earlier in: Solving Quadratic Equations, the trinomial may not be able to be factored using any combination of integer factors in the Trial & Error/Reverse FOIL Method, or the ac/Grouping Method; in this case, it would be considered “prime.” You are then to use the Quadratic Formula. You may not yet know if the Quadratic Formula will yield “real” solutions or not (but that’s why you plug in the a, b and c values and solve). If the Quadratic Formula does yield real solutions, the solutions will contain radicals of not-perfect squares (which could be converted to decimals). There is also a chance that no real solutions will result from that original prime polynomial due to a negative number in the simplified radical.

Clarification: When the Solution is 0

Before, I stated that the difference of two squares will yield two solutions, but there is one exception to this. Any quadratic equation in which b and c are zero, as seen in “When Both b & c are 0,” can also technically qualify as thedifference of two squares, such as

“4x2 – 0 = 0”, or the sum of two squares, as in “9x2 + 0 = 0”, because zero is a (perfect) square number. In cases like this,

·   there is not two solutions,

·   nor is it prime with no solutions

·   It has only one solution, that being “x = 0”.

That also makes this an exception to the statement: “The sum of two squares is prime and yields no real number solutions.”