*Quadratics* are expressions whose largest exponent, or highest power, equals $2$. A quadratic equation connects two expressions, one on either side of the equals sign, at least one of which has a term with a variable squared. A quadratic function is structured in the usual notation, with “$f(x)$” (or similar function notation) set equal to a quadratic. In ordinary circumstances, quadratic equations have two solutions; another way to say the same thing is that their graph, a parabola, has two $x$-intercepts. There are exceptions to this rule, exceptions we’ll cover in this module, but your normal expectation should follow the pattern that the number of solutions to an equation equals the highest exponent found in that equation. In the case of a quadratic, that’s $2$!

A pirate ship fires a cannonball at a merchant ship but the cannonball misses high, sailing over the merchant ship and eventually splashing into the ocean. The flight of the cannonball can be modeled by the function $h(x)=(−½)x_{2}+2x+5$, where $x$ is the time, in seconds, and $h$ is the height above the water at which the ball is found at time $x$. After how many seconds does the cannonball attain its maximum height above the water?

A. 0.5 seconds

B. 2 seconds

C. 4 seconds

D. 5 seconds

The CLT may test your knowledge of quadratics through a word problem; this question featuring a projectile is a typical example. Let’s start by reviewing standard form of quadratic (or parabola, speaking in graphing terms): $ax_{2}+bx+c$, where $a$, $b$, and $c$ are coefficients. There will always be a negative on the “$a$” term in a projectile question, indicating an “upside down” parabola used to model where the flight begins, where it reaches its maximum, and where it hits the ground (or, in this case, the water).

Though seemingly complicated at first, this question is actually straightforward if you know two things: 1) that “maximum” corresponds to the *vertex* of the parabola (as does *minimum* in an upward-facing parabola) and 2) the formula for the x-coordinate of the vertex: $−b/2a$. Since this question is asking for the time, or $x$ value, we have only to apply the formula: $(2)(−½)−2 =2$. The answer is **B**: 2 seconds.

But there are a few other important things to note as we study this question. Any quadratic equation in a real-life situation will reveal, in addition to the maximum (or minimum), two other points in the object’s flight: the starting point and the point at which the object hits the ground (or water). The starting point is the $y$-intercept, just as it is in a linear equation. And the $y$-intercept is transparent in a quadratic: it is simply the c term. So answer D here is a trap answer of sorts, since $5$ is a significant number in this situation. But it’s not the time at which the maximum is reached; rather, 5 (feet) is the height above the water at which the cannonball began its journey out of the cannon.

Finding the “landing point” for the cannonball involves finding the other intercept, the $x$-intercept, and that process is more complicated. The first step is to determine whether the quadratic can be *factored* (see mini-lesson below). Once we have set the quadratic equal to zero (required to find the solution or $x$-intercept), the best way to proceed is to multiply the equation by $−2$ to get rid of the leading coefficient ($−½$). That yields $x_{2}−4x−10$. As you can see if you remember your factoring principles, this equation will not factor.

That leaves with the Swiss Army Knife of quadratics, useful in every situation: the *quadratic formula*. Does a song run through your head as you recite it? “The quadratic formula’s -b, plus or minus the square root of ($b_{2}−4ac$), all over 2a.” If you apply the formula in this case, you get something quite messy. (Want to try it? Check your answer below.) Thankfully, we are not asked for the time at which the cannonball hit the water!

(Spoiler: the quadratic formula, once simplified in this case, yields $4±14 $. Taking only the “plus” answer tells us that the cannonball hits the water between $5$ and $6$ seconds after being fired. Not an easy quantity to figure out without a calculator!)

To understand what it takes to factor a quadratic with leading coefficient of $1$ (so in the form $x_{2}+bx+c$), it helps to review the opposite of factoring, that is, multiplying (or expanding) binomials such as $x+2$ and $x−4$. Commonly known as “FOIL”–an abbreviation for “first, outer, inner, last,” expanding involves multiplying the two binomials in four steps so that each term in the first binomial gets multiplied by each term in the second. Expanding $x+2$ and $x−4$ yields $x_{2}−4x+2x−8$, which, after combining like terms in the middle, simplifies to $x_{2}−2x−8$.

To factor a quadratic, we literally undo this process, like movie playing in reverse. If given $x_{2}−2x−8$, we consider what two integers would multiply to make $−8$ (the “last” portion of FOIL) and what two integers would add to make $−2$ (the “outer” plus the “inner” portion of FOIL). (Note that the “first” part of FOIL doesn’t really come into play because the leading coefficient is $1$, and that can only be $1×1$ if we’re using integer factors). Looking at $8$ as a multiple, we could use some combination of $1$ and $8$ or $2$ and $4$. However, the multiple is really $−8$; for a negative product, we need one negative integer and one positive integer.

We can’t decide among these options until we consider the middle term’s coefficient: $−2$. Knowing that this coefficient comes from adding two coefficients together, we establish that the negative coefficient must have a larger than absolute value than the positive coefficient; that is, the negative coefficient must be larger if you don’t consider the negative. (Think of a tug of war between a negative and positive number where the negative is winning.) So we could use $−8$ and $1$ as a pair, or else $−4$ and $2$. But only the second of these pairs, when added together, produced $−2$. That’s our winner.

With this in mind, we can construct our two binomials to be multiplied together. The leading terms have coefficients of $1$, so in that case it’s always $(x+?)(x+?)$. Plugging in $−4$ and $2$, we have $(x−4)(x+2)$, which can also be written in reverse since multiplication is commutative: $(x+2)(x−4)$.

Finally, we should consider different positive/negative sign combinations from the one we have above. Here are the other three possibilities:

Two positive signs: both factors must have positive signs

A negative sign, then a positive sign: both factors must have negative signs

A positive sign, then a negative sign: one factor must have a negative and the other a positive, with the positive “winning,” that is, having a larger absolute value.

Try factoring these three quadratic expressions:

- $x_{2}+7x+12$

(spoiler)

$(x+3)(x+4)$ or $(x+4)(x+3)$

- $x_{2}−6x+5$

(spoiler)

$(x−1)(x−5)$ or $(x−5)(x−1)$

- $x_{3}+3x−10$

(spoiler)

$(x+5)(x−2)$ or $(x−2)(x+5)$

On somewhat rare occasions, the CLT will present a quadratic expression or equation with a leading coefficient greater than 1. There are two different ways this can occur, each corresponding to its own solution strategy:

- Sometimes the expression can be turned into a “traditional” quadratic by factoring out the greatest common factor. For example: consider $4n_{2}+16n+12$. It looks hard to factor, but if we divide out the GCF, we have $4(x_{2}+4x+3)$, and the quadratic inside the parentheses now factors cleanly into $(x+1)(x+3)$. Always look out for the possibility of a GCF!
- If there is a leading coefficient and no GCF, there are ways to factor such a quadratic, but this situation is rare enough on the CLT that it’s not worth going into detail about those here (look up “slide and divide” online if you’re curious). In these situations, work out your quadratic formula muscles! This has the benefit of solidifying the formula in your mind. The other benefit is that the quadratic (again, in rare cases) may not be factorable into integer factors at all, so if you default directly to quadratic formula, you avoid the frustration of trying to factor something that isn’t factorable! (You may also figure out that the quadratic isn’t factorable if you use the UnCLES method and look at the answers, discovering that some or all of them have square roots present. A sure sign that the quadratic formula is your ace in the hole!)

The other variation to a quadratic is a quadratic disguised as a *cubic* expression (third degree, with the highest exponent being a $3$). But sometimes there is a *variable GCF* present; that is, it’s possible to factor out a variable. Consider $r_{3}−11r_{2}+30r$. It looks scary at first, but if you factor out an r, you’re left with a factorable quadratic, which becomes $(r−5)(r−6)$. (In this case, you would note that $0$ is also a root of the cubic expression, for the factored-out $r$, as well as $5$ and $6$.)

- We’ve used it already in this lesson, but make sure to memorize the formula for the x-coordinate of the vertex from standard form: $−2ab $.
*vertex form*: although $−b/2a$ is your primary tool for finding the vertex, it’s important to recognize vertex form when it arises: $y=a(x−h)_{2}+k$, where $(h,k)$ are the coordinates of the vertex. Note there is a minus before the $h$ so you’ll need to switch the sign; if you have $x−3$ inside the parentheses, for instance, the x-coordinate is $3$, not $−3$.- Two handy shortcuts if you are asked for the sum of the solutions or the product of the solutions to a quadratic. The sum is $−b/a$ and the product is $c/a$.
- It’s important to memorize the quadratic formula more generally (get a song in your head to help!), but the part of the quadratic formula under the radical is helpful in determining the number of solutions to a quadratic equation. It’s called the
*discriminant*and is $b_{2}−4ac$. See the explanation under “discriminant” above for how this work. - The difference of squares is very helpful: $(a−b)(a+b)=a_{2}−b_{2}$
- It’s also worth memorizing the two “perfect square quadratics”:

$(a+b)_{2}(a−b)_{2} =a_{2}+2ab+b_{2}=a_{2}−2ab+b_{2} $

Which of the following is equivalent to $(x+4)(x−4$)?

A. $x_{2}−16$

B. $x_{2}+16$

C. $x_{2}+4x−16$

D. $x_{2}−4x−16$

(spoiler)

The answer is **A**. This problem presents the *difference of squares* (see “Flashcard Fodder”). We may expand these two binomials by multiplying the first terms ($x×x=x_{2}$), the outer terms ($x×−4=−4x$), the inner terms ($4×x=4x$), and the last terms ($4×−4=−16$). Adding these four terms together cancels the two middle terms ($4x$ and $−4x$), leaving just $x_{2}−16$.

However, it is usually more efficient to memorize the form of the difference of squares so that you know to square the first term, square the last term, and put a minus in between. Using a math term you may have encountered, we multiply a binomial expression by its *conjugate* to get the difference of squares. In this case, the conjugate of $x+4$ is $x−4$, and vice versa.

Which of the following is a solution to the equation $x_{2}−5x+6=0$?

A. 5

B. 2

C. -3

D. -6

(spoiler)

The answer is **B**. We need to factor the expression on the left using the method shown in the mini-lesson. The two integers pairs of factors of six are 1) $1$ and $6$; 2) $2$ and $3$. Next, we note the signs: with a negative for adding and a positive for multiplying, we must use two negative factors. That means were adding two negative numbers to make $−5$, so $−1$ and $−6$ won’t work. The factoring must be $(x−2)(x−3)$. We set each of these factors individually equal to zero and discover that since $x−2=0$, $x$ could equal $2$, and since $x−3=0$, $x$ could also equal $3$. Only the first of these options is provided in the answers.

As an alternative solving solution, you could simply plug in the answers. If you can do this quickly, plugging in the answers (known as *backsolving*) can sometimes be the fastest approach. Plugging in the right answer, B, gives us $2_{2}−5(2)+6=0$. Since it true that $4−10+6=0$, this is the answer. Backsolving is also a great checking method; if you have any doubt that B is the answer, and provided you have time to do so, we encourage you, with any equation question, to plug in your answer to make sure it works.

Which of the following is equivalent to $2x(x−3)+6(3+x)$?

A. $2(x_{2}+6x−9)$

B. $2(x_{2}+9)$

C. $2x(x_{2}+3x−9)$

D. $2x(x+9)$

(spoiler)

The answer is **B**. Here we need to first distribute both terms outside the parentheses: multiplying through by $2x$ yields $2x_{2}−6x$ and multiplying through by $6$ yields $18+6x$. Combining those binomials together, we get $2x_{2}−6x+18+6x$. The $−6x$ and $6x$ cancel when added together, leaving $2x_{2}+18$. But the answers are put in factored form; if we notice that we can factor out a GCF from our answer, we get answer B. Alternatively, we can leave our answer as $2x_{2}+18$ and multiply all the answer choice by distribution, in which case we will find that only B matches (through D comes close!).

Which of the following is NOT a root of the expression $x_{3}+6x_{2}−7x=0$?

A. -7

B. -1

C. 0

D. 1

(spoiler)

The answer is **B**. This is one of those situations where the GCF unlocks the whole problem. You need to have eyes to see that every term contains at least one $x$! Factoring out that $x$, we get $x(x_{2}+6x−7)=0$. Working inside the parentheses, we can factor to $(x+7)(x−1)$, so the equation is now $x(x+7)(x−1)$. Don’t forget that the simple $x$ term, like the two binomials in parentheses, should be set equal to zero on its own terms. That leaves a solution of $0$, and the other two solutions must be $−7$ and $1$ since the solution always has the opposite sign from what’s inside the parentheses. Choice B, $−1$, is the odd one out and therefore the right answer. (Side note: it’s a good idea to write the word “NOT” from the question on your scratch paper; despite its being in capital letters, students overlook this kind of word all the time!)

Which of the following coordinate points is a solution to both of the equations below?

$yy =−x+2=x_{2}−4 $

A. (2,2)

B. (-2,4)

C. (2,0)

D. (-2,0)

(spoiler)

The answer is **C**. There are two ways to approach this problem. The textbook approach involves noticing that the right sides of both equations are set equal to y, so they must also be equal to each other. We can therefore say that $−x+2=x_{2}−4$. Moving all the terms to the side with $x_{2}$ (remember: always set quadratics equal to zero!), we get $x_{2}+x−6=0$. This factors to $(x+3)(x−2)=0$, meaning that $x$ must equal either $−3$ or $2$. We can plug both of these solutions back in and solve for y, but let’s be smart with the help of the UnCLES method. Looking at the answers, we can see that $−3$ is not an option for $x$, so we can only use $2$. The answer must be $A$ or $C$, and when we plug $2$ in for $x$ in either equation, we discover that $y=0$. Choice C is the winner.

Alternatively, we can backsolve here, seeing which answer fits both equations. To be efficient, we should notice that $2$ and $−2$ are both used twice as the value of $x$, so we should pick one of those to start with. It’s always easier to plug in a positive than a negative, so let’s plug in $2$ for the first equation. That yields $y=0$, so it looks like C is the answer. To be safe, we can plug $2$ in for $x$ in the second equation as well, confirming our choice.

The function $f(x)=x_{2}+4x+10$ is graphed in the real number coordinate plane. Which of the following is FALSE?

A. The graph is a parabola.

B. The axis of symmetry of the parabola is $x=2$.

C. The graph has no $x$-intercepts.

D. The graph’s $y$-intercept is at (0,10).

(spoiler)

The answer is **B**. This problem presents us with a number of challenges at first glance. The UnCLES method is especially useful here, because reading the answer choices shows us that we aren’t directly asked for the roots of this quadratic; a good thing, because it doesn’t factor! (The UnCLES method also reminds us to write down “FALSE” on our scratch paper so we stay focused on the task.)

The way this question is phrased, it’s best to go straight to the answer choices and evaluate them one by one. We hope that, by this point in the lesson, you remember that a quadratic is graphed as a parabola, so you can rule out choice A right away since it’s true! For choice B, we need to recall that the axis of symmetry has the same $x$-coordinate as the vertex. So we can use $−b/2a$ to find the vertex’s $x$-coordinate, and - whoops, that appears to be $−2$, not $2$! It looks like we’ve found our false statement, but let’s confirm by checking choices C and D.

Checking choice C takes some work and can be approached in two different ways. First, since we already know the $x$-coordinate of the vertex is $−2$, we can plug that back end and discover that the $y$-coordinate of the vertex is $6$. Since the $y$-coordinate of the vertex is above the $x$-axis and this is an upward-facing parabola (positive leading coefficient), this graph will never intersect with the $x$-axis. Choice C is true. We could also use the discriminant ($b_{2}−4ac$) to discover that the quadratic has no real solutions; the discriminant comes to $−24$, and any negative result means no real solutions.

Choice D is quicker: if we remember that the $c$ term in standard form is the same as the $y$-intercept, we can immediately identify the $y$-intercept as 10. Choice D puts the zero in the correct spot and uses $10$ for $y$, so it’s also true.

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