Introduction to Quadratics

What makes an equation quadratic

A quadratic equation is one that can be written in the form

where $a$, $b$, and $c$ are fixed numbers and $a \ne 0$. This is called standard form. Every term is moved to one side so the other side is $0$, and the terms are written in order of decreasing power, the squared term first. The three numbers have names. We call $a$ the leading coefficient, $b$ the coefficient of the $x$ term, and $c$ the constant term.

The one requirement you cannot drop is $a \ne 0$. The word quadratic comes from quadratus, the Latin for “squared,” and it is the $ax^2$ term that makes an equation quadratic. If $a$ were $0$, that term would vanish and leave $bx + c = 0$, an ordinary linear equation of the kind you already solve. So $a \ne 0$ is exactly what separates a genuine quadratic from a linear equation. There is no such rule for $b$ or $c$: either or both may be zero, and the equation is still quadratic as long as the $x^2$ term survives.

Many quadratics do not arrive in standard form. Part of the skill is recognizing one however it is dressed up, then rearranging it. To do that, move every term to the left with the same balancing steps you used on linear equations. Then collect like terms until the right side reads $0$.

The middle two rows look incomplete, yet both are quadratic. Rewriting $2x^2 = 8$ as $2x^2 - 8 = 0$ shows $b = 0$, and rewriting $x^2 = 7x$ as $x^2 - 7x = 0$ shows $c = 0$. A missing $x$ term or a missing constant is fine; a missing $x^2$ term is not.

What it means to be a solution

A solution of an equation, also called a root, is a value of the variable that makes the equation true. For a quadratic in standard form, a root is a number you can put in place of $x$ so that $ax^2 + bx + c$ works out to exactly $0$. Nothing about that idea is new; it is the same meaning of “solution” you have used since your first linear equation. What is new is how many solutions to expect.

A linear equation like $3x + 5 = 20$ has exactly one solution. A quadratic can have two solutions, one solution, or no real solution at all. You will see why once you have a method in hand, but hold on to the possibility from the start. Finding one root does not mean you are done, because a second one may be waiting. The surest way to test a candidate is to substitute it and check.

The zero-product property

Substitution tells you whether a guess is a root, but it will not hand you the roots to begin with. For that you need a method, and the first one rests on a single fact about $0$ so familiar you may never have said it out loud. That fact is this: if a product of numbers equals $0$, then at least one of the numbers must be $0$. This is the zero-product property. Multiplying by $0$ always gives $0$, and nothing else does, so two nonzero numbers can never multiply to $0$. Here is the reason in full.

The property turns a factored quadratic into two easy linear equations. Suppose an equation arrives already broken into a product of factors set equal to zero, such as

The left side is a product of the two factors $x - 2$ and $x + 3$. By the zero-product property that product is $0$ exactly when one of the factors is $0$, so the equation splits into two:

Each piece is a linear equation you can solve at a glance, giving $x = 2$ or $x = -3$. Those are the two roots. Notice how to read a root straight off a factor: the factor $x - 2$ is zero when $x = 2$, and the factor $x + 3$, which is $x - (-3)$, is zero when $x = -3$. The number shown in the factor and the root have opposite signs, and that sign flip is the most common place to slip.

These two forms, the factored $(x - 2)(x + 3) = 0$ and the standard form, are the same equation wearing different clothes. Expanding the product the way you did in the chapter on expanding and factoring,

so $(x - 2)(x + 3) = 0$ is just $x^2 + x - 6 = 0$ rewritten. The factored form is the useful one, because it puts the roots on display, while the expanded standard form hides them. Turning a standard-form quadratic back into a product of factors is the subject of the next few lessons. Here you are handed the factored form and asked only to finish the job.

Solving by taking square roots

The zero-product property needs a factored form. A second method handles quadratics with no middle $bx$ term directly. Consider

for some number $k$. You are looking for every number whose square is $k$. Square roots almost answer this, but there is a subtlety that trips up nearly everyone, and it comes from a fact about squaring you already know. A negative times a negative is positive, so a number and its opposite have the same square. For instance $3^2 = 9$, and $(-3)^2 = 9$ as well.

That means $x^2 = 9$ has two solutions, $x = 3$ and $x = -3$, not just the positive one. We record both at once by writing

read ”$x$ equals plus or minus the square root of $k$.” The $\pm$ is not decoration; it captures the second root that a plain square root would hide. How many solutions you actually get depends on the sign of $k$. When $k$ is positive, there are two real square roots, $\sqrt{k}$ and $-\sqrt{k}$, so the equation has two solutions. When $k$ is zero, the only number whose square is $0$ is $0$ itself, so there is exactly one solution, $x = 0$. In that case the two roots have merged into a single repeated one. When $k$ is negative, there is no real solution at all. That is because squaring any real number, positive or negative, gives a result that is zero or positive and never negative. (Numbers whose square is negative do exist, but they are not real numbers, and they wait for a later chapter.) This is the first place you can see all three possibilities the definition promised: two roots, one root, or none.

When the square sits on a group

The same move works when the thing being squared is not $x$ alone but a whole expression. An equation like

is solved by taking square roots of both sides, again keeping the $\pm$:

Now $x - h$ is pinned down to two values, and a single balancing step finishes the job: $x = h \pm\sqrt{k}$. The reasoning is exactly the same as before. Whatever sits inside the square, treat it as one quantity, and that quantity has to be one of the two numbers whose square is $k$.

A first look at the graph

Every quadratic has a picture, and it explains at a glance why the number of solutions can be two, one, or none. Take the expression $ax^2 + bx + c$, and for each value of $x$ plot the point whose height $y$ equals that expression. The points trace a smooth U-shaped curve called a parabola. Solving $ax^2 + bx + c = 0$ means finding the $x$ values where $y = 0$, and $y = 0$ is the horizontal axis. So the real roots of a quadratic are exactly the points where its parabola crosses the $x$-axis.

A parabola can meet the $x$-axis in two places, giving two roots, or just touch the axis at its lowest (or highest) point, giving one repeated root. The parabola can also float entirely above or below the axis, giving no real roots. That is the same three-way split you met with $x^2 = k$, now drawn as a picture. The curve above is $y = x^2 - 4$, and it crosses the axis at $x = -2$ and $x = 2$, the two roots of $x^2 - 4 = 0$. Drawing parabolas carefully, locating the lowest point, and reading off their features is the work of a later chapter. For now the picture is only here to make the count of solutions something you can see rather than only trust.

Common mistakes