Algebra 1

What makes an expression quadratic

What do the area of a square, the height of a thrown ball and the number of handshakes in a room have in common?

You need

Degree, terms and standard formAlgebra 1

Unlocks

The shape of a parabolaAlgebra 1 Solving quadratics by factoringAlgebra 1 Solving quadratics with square rootsAlgebra 1

Where you are 4 in Quadratic expressions and parabolas

Algebra9

Quadratics12

Quadratic expressions and parabolas4

Graphing from standard form The shape of a parabola The vertex and the axis of symmetry

What makes an expression quadratic

Why this exists

A linear expression such as 3x + 2 changes by the same amount every time x goes up by 1, and its graph is a straight line. A great many quantities do not behave like that. The area of a square of side x is x^2. The height of a ball thrown upward, t seconds after it leaves the hand, is a number times t^2 plus a number times t plus a starting height. The number of handshakes when n people each shake hands once is \displaystyle \frac{1}{2}n^2 - \frac{1}{2}n. Each of these, written out, has its variable squared and no higher power.

They form the next family after the linear ones, and the family is called quadratic, from the Latin quadratus, a square. Recognizing a member matters because the whole family behaves alike. Every quadratic graphs as a parabola, every quadratic equation has at most two solutions, and one short list of methods solves all of them. Before any of that, you have to be able to tell whether the expression in front of you is one, and it is often in disguise: written as a product, or with a square that cancels away.

Definition

A quadratic expression in x is a polynomial in x of degree exactly two. Once it has been multiplied out and like terms combined, the highest power of x in it is x^2.

Every quadratic expression can be written in standard form, highest power first:

ax^2 + bx + c, \qquad a \neq 0

where a, b and c are numbers. ax^2 is the quadratic term and a is the leading coefficient; bx is the linear term; c is the constant term.

The condition a \neq 0 is the definition: it says the x^2 is really there. Nothing is asked of b and c, and either or both may be zero, so x^2, 5x^2 - 3 and x^2 + 4x are all quadratic.

Three things stop an expression from being quadratic:

a power higher than two, as in x^3 - x^2, which is cubic
no x^2 left after simplifying, as in x(x + 1) - x^2 = x, which is linear
x in a place a polynomial does not allow: in a denominator as in \displaystyle \frac{1}{x^2}, under a root as in \displaystyle \sqrt{x}, or in an exponent as in 2^x. These are not polynomials, so they have no degree at all.

A quadratic equation is an equation that can be rearranged into ax^2 + bx + c = 0 with a \neq 0.

Intuition

Squared means area. Figure 1 draws (x + 2)(x + 3) as a rectangle x + 3 wide and x + 2 tall, cut along the two places where the x ends.

Figure 1(x + 2)(x + 3) = x^2 + 5x + 6. Only the shaded piece is x by x.

There are four pieces. The shaded square has both of its sides equal to x, so its area is x^2. Each strip has one side that grows with x and one that does not, so their areas are 3x and 2x, which are linear. The corner has no side that grows, so its area is the constant 6. Adding the pieces,

(x + 2)(x + 3) = x^2 + 5x + 6

The x^2 term is, at heart, one length that grows with x times another length that grows with x. Multiply three such lengths and you have a volume, x^3, which belongs to a different family.

The square shows up in the way the values change, too. Figure 2 builds squares of side 1, 2, 3 and 4 out of tiles. Each square is the one before plus an L-shaped border, and each border is two tiles longer than the last: 3, then 5, then 7.

Figure 2Each square adds a border two tiles longer than the last, so the steps between the areas go up by 2 every time.

A linear expression grows by the same amount at every step. A quadratic grows by amounts that themselves go up by the same amount at every step.

Results

A product of two linear expressions is quadratic. If p \neq 0 and r \neq 0,

(px + q)(rx + s) = pr\,x^2 + (ps + qr)\,x + qs

and the leading coefficient pr is not 0.

Second differences are constant. List the values of ax^2 + bx + c at x = 0, 1, 2, 3, \dots and subtract each value from the next. Those first differences are

a + b, \quad 3a + b, \quad 5a + b, \quad 7a + b, \quad \dots

and the differences between them, the second differences, are all the same number:

2a, \quad 2a, \quad 2a, \quad \dots

A linear expression has constant first differences, so its second differences are all 0. Since a \neq 0, a quadratic’s second differences are never 0. Going the other way, a table at x = 0, 1, 2, 3, \dots whose second differences are constant and not zero is matched by exactly one quadratic: the second difference is 2a, the first of the first differences is a + b, and the value at x = 0 is c.

Why it is true

The product. Multiply every term by every term, as in expanding binomials:

\begin{aligned}(px + q)(rx + s) &= px \cdot rx + px \cdot s + q \cdot rx + q \cdot s \\ &= pr\,x^2 + (ps + qr)\,x + qs\end{aligned}

Neither factor has a power of x above the first, and powers add when they multiply, so nothing above x^2 can appear. The x^2 term does appear, because p and r are both non-zero and so their product pr is too. This is one case of a general rule: the degree of a product is the sum of the degrees, here 1 + 1 = 2.

The differences. Take the step from any x to x + 1:

\begin{aligned} &\bigl(a(x + 1)^2 + b(x + 1) + c\bigr) - \bigl(ax^2 + bx + c\bigr) \\ &\quad = a(x^2 + 2x + 1) + bx + b + c - ax^2 - bx - c \\ &\quad = 2ax + a + b \end{aligned}

The first difference is itself a linear expression in x, with 2ax as its only part that changes. Each time x goes up by 1, it goes up by 2a, so every second difference is 2a. The constant c cancels at the first subtraction and b at the second, which is why only a is left. Putting x = 0, 1, 2, \dots into 2ax + a + b gives the list a + b, 3a + b, 5a + b above. In Figure 2, where the expression is x^2, the border added to a square of side x has 2x + 1 tiles.

For the other direction, the values at x = 0, 1 and 2 fix a, b and c through the three facts above, and the quadratic they give has the same second difference as the table at every later step, so it reproduces every later row.

Worked examples

Example 1

Reading off a, b and c

Write each quadratic in standard form and name a, b and c.

(a) 7 + 2x^2 - 5x \qquad (b) 4 - x^2 \qquad (c) \displaystyle \frac{x^2}{2} + 3x

Solution

Standard form lists the terms from the highest power down, and each coefficient keeps the sign in front of its term.

(a) 2x^2 - 5x + 7, so a = 2, b = -5 and c = 7.

(b) -x^2 + 4. There is no x term, which is the same as a term 0x, and -x^2 means -1 \cdot x^2. So a = -1, b = 0 and c = 4.

(c) \displaystyle \frac{x^2}{2} is \displaystyle \frac{1}{2}x^2, and there is no constant, so the standard form is \displaystyle \frac{1}{2}x^2 + 3x with \displaystyle a = \frac{1}{2}, b = 3 and c = 0.

Example 2

Quadratic or not

Decide which of these are quadratic expressions.

(a) 3x - x^2 + 1 \qquad (b) x^3 + 4x^2 \qquad (c) \displaystyle \frac{5}{x^2} - 2 \qquad (d) \displaystyle x^2 + \sqrt{x} \qquad (e) \displaystyle \sqrt{2}\,x^2

Solution

(a) Quadratic. In standard form it is -x^2 + 3x + 1, with a = -1. The order the terms were written in does not matter, and a negative leading coefficient is allowed.

(b) Not quadratic. The highest power is 3, so it is cubic. Having an x^2 term is not enough; there must be nothing above it.

(c) Not quadratic. \displaystyle \frac{5}{x^2} is 5x^{-2}, a negative power, so this is not a polynomial and has no degree.

(d) Not quadratic. \displaystyle \sqrt{x} is x^{1/2}, which is not a whole-number power, so again this is not a polynomial.

(e) Quadratic, with \displaystyle a = \sqrt{2} and b = c = 0. The rules are about the powers of x, not about the coefficients, which may be any real numbers.

Example 3

Simplify before deciding

Decide whether each expression is quadratic.

(a) (2x - 1)(x + 3) \qquad (b) x(x + 4) - x^2 \qquad (c) (x + 5)^2 - (x - 5)^2

Solution

The degree is read only after multiplying out and combining like terms, so do that first.

(a) (2x - 1)(x + 3) = 2x^2 + 6x - x - 3, which is 2x^2 + 5x - 3. Quadratic, with a = 2, as the product rule in Results predicts for two linear factors.

(b) x(x + 4) - x^2 = x^2 + 4x - x^2 = 4x. Linear. It looked quadratic, but the x^2 cancels.

(x^2 + 10x + 25) - (x^2 - 10x + 25) = 20x

Linear. Both squares have leading coefficient 1, so their x^2 terms cancel.

Example 4

Quadratic equations

Decide which equations are quadratic. Write each quadratic one in the form ax^2 + bx + c = 0, and solve any that is not.

(a) n(n + 1) = 42 \qquad (b) (x - 3)^2 = x^2 + 5 \qquad (c) 5x^2 = 20

Solution

To sort an equation, move everything to one side and simplify; what is left decides it.

(a) Multiply out: n^2 + n = 42, so n^2 + n - 42 = 0. Quadratic, with a = 1, b = 1 and c = -42. No square was written, but n is multiplied by an expression containing n.

(b) Multiply out the left side:

\begin{aligned} x^2 - 6x + 9 &= x^2 + 5 \\ -6x + 9 &= 5 \\ -6x &= -4 \\ x &= \frac{2}{3} \end{aligned}

Subtracting x^2 from both sides removed every square, so this equation is linear, with one solution. Check. \displaystyle \left(\frac{2}{3} - 3\right)^2 = \left(-\frac{7}{3}\right)^2 = \frac{49}{9} and \displaystyle \frac{4}{9} + 5 = \frac{49}{9}.

Example 5

When the leading coefficient depends on a number

For which values of k is (k^2 - 9)x^2 + (k - 3)x + 5 a quadratic expression in x? What is it for the other values?

Solution

The expression is quadratic exactly when its leading coefficient, k^2 - 9, is not zero, so find where it is zero:

k^2 - 9 = 0 \qquad\Longrightarrow\qquad k = 3 \quad\text{or}\quad k = -3

So it is quadratic for every value of k except 3 and -3.

At k = 3 it becomes 0x^2 + 0x + 5 = 5, a constant. At k = -3 it becomes 0x^2 - 6x + 5 = -6x + 5, which is linear. Whether an expression is quadratic is a question about a alone.

Example 6

Counting handshakes

In a room of n people, everyone shakes hands with everyone else exactly once. The number of handshakes H is:

\begin{array}{ccccccc} n & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline H & 0 & 0 & 1 & 3 & 6 & 10 \end{array}

Show that H is quadratic in n and find it.

Solution

The inputs go up by 1, so take differences. The first differences are 0, 1, 2, 3, 4 and the second differences are 1, 1, 1, 1: constant and not zero. So one quadratic H = an^2 + bn + c matches the table.

Read off its coefficients from Results. The second difference is 2a = 1, so \displaystyle a = \frac{1}{2}. The value at n = 0 is c = 0. The first of the first differences is a + b = 0, so \displaystyle b = -\frac{1}{2}. Therefore

H = \frac{1}{2}n^2 - \frac{1}{2}n = \frac{n(n - 1)}{2}

Check. At n = 5: \displaystyle \frac{25}{2} - \frac{5}{2} = 10.

The factored form explains why the count had to be quadratic. Each of the n people shakes n - 1 hands, and that counts every handshake twice, once from each end. So H is half the product of two linear expressions in n, and a room of 10 people makes \displaystyle \frac{10 \cdot 9}{2} = 45 handshakes.

Boundaries

Degree is read after simplifying. The written form can mislead either way: x(x + 4) - x^2 looks quadratic and is linear, while n(n + 1) shows no square and is quadratic.

Quadratic in which variable? x^2 + xy is quadratic in x but only linear in y. The area \pi r^2 is quadratic in r. Always say which letter is the variable.

Only whole-number powers count. x^{-2} and x^{1/2} are not polynomial terms, so an expression containing them has no degree and is not quadratic, however much it looks like one.

The table test needs equally spaced inputs. At x = 0, 1, 3, 4 the differences do not compare. If the inputs go up by h each time instead of 1, the second differences are still constant, equal to 2ah^2. And a table only speaks for its own rows; it cannot prove a rule continues past them.

Two variables at once is a different object. x^2 + y^2 also has degree two, but taken in both letters together it is a second-degree expression in two variables. Set equal to a number, x^2 + y^2 draws a circle. Second-degree equations in two variables draw the conics, and the parabola is only one of them.

Where this goes next

Once an expression is known to be quadratic, its graph is a parabola opening up when a > 0 and down when a < 0, folded about its axis of symmetry and turning at its vertex.

Solving a quadratic equation has its own tools, each suited to a shape of equation. When it is a product equal to zero, the zero product property splits it in two. When b = 0, the square root property finishes it in one step. For any quadratic at all, completing the square and the quadratic formula work, and the discriminant b^2 - 4ac tells you from a, b and c alone how many real solutions to expect.

The second differences grow into fitting a quadratic to data, and the same questions asked about the whole family become quadratic functions.

Exercises

Standard form

Write each quadratic in standard form and name a, b and c.

4x^2 - 9x + 1
6 - 2x + x^2
8 - 3x^2
5x - 2x^2
\displaystyle \frac{x^2 - 6x}{3}

Quadratic or not

Decide whether each expression is quadratic. If it is, write it in standard form. If it is not, say why not.

2x^2 + x^3 - 1
(x - 5)(x + 5)
x(3 - 2x)
\displaystyle \frac{3}{x^2} - x
(x + 4)^2 - x^2
(2x + 1)^2 - 3x^2
x^2 + 2^x

Quadratic equations

Decide whether each equation is quadratic. If it is, write it as ax^2 + bx + c = 0. If it is not, solve it.

x(x - 3) = 10
(x + 2)^2 = x^2 + 12
3x^2 = 2x^2 + 5x
(x - 1)(x + 1) = 2x^2
(2x + 1)(x - 4) = 2x^2 + 3

Leading coefficients and tables

For which values of k is (k + 4)x^2 - 6x + 1 quadratic in x? What is it otherwise?
For which values of m is (m^2 - 1)x^2 + (m - 1)x + 2 quadratic in x? What is it for each other value?
At x = 0, 1, 2, 3, 4 a rule gives y = 1, 3, 9, 19, 33. Show that a quadratic matches the table, and find it.
At x = 0, 1, 2, 3, 4 a rule gives y = 2, 6, 18, 54, 162. Can a quadratic match every row?

Reasoning

Explain why the definition insists that a \neq 0 but allows b and c to be zero.
Explain why the product of a linear expression and a quadratic expression is never quadratic.
Show by an example that the sum of two quadratic expressions need not be quadratic.
A table with inputs one apart has every second difference equal to 0. Explain why no quadratic can match it.

Answers

a = 4, b = -9, c = 1. It is already in standard form.
x^2 - 2x + 6, so a = 1, b = -2, c = 6.
-3x^2 + 8, so a = -3, b = 0, c = 8.
-2x^2 + 5x, so a = -2, b = 5, c = 0.
\displaystyle \frac{1}{3}x^2 - 2x, so \displaystyle a = \frac{1}{3}, b = -2, c = 0. Dividing each term by 3 gives \displaystyle \frac{x^2}{3} - \frac{6x}{3}.
Not quadratic. The x^3 makes it degree three.
Quadratic: x^2 - 25. Multiplying out, x^2 + 5x - 5x - 25, and the linear terms cancel.
Quadratic: -2x^2 + 3x. Multiplying out gives 3x - 2x^2.
Not quadratic. \displaystyle \frac{3}{x^2} = 3x^{-2}, so it is not a polynomial.
Not quadratic: it is 8x + 16, which is linear. (x^2 + 8x + 16) - x^2 = 8x + 16.
Quadratic: x^2 + 4x + 1. (4x^2 + 4x + 1) - 3x^2 = x^2 + 4x + 1, and only part of the x^2 cancels.
Not quadratic. 2^x has x in the exponent, so it is not a polynomial.
Quadratic: x^2 - 3x - 10 = 0. Multiply out, then subtract 10.
Not quadratic; x = 2. x^2 + 4x + 4 = x^2 + 12, so 4x = 8. Check: (2 + 2)^2 = 16 = 2^2 + 12.
Quadratic: x^2 - 5x = 0. Subtract 2x^2 + 5x from both sides; here c = 0.
Quadratic: x^2 + 1 = 0. x^2 - 1 = 2x^2, so -x^2 - 1 = 0; multiplying both sides by -1 gives the same equation with a = 1.
Not quadratic; x = -1. 2x^2 - 7x - 4 = 2x^2 + 3, so -7x = 7. Check: (-1)(-5) = 5 and 2 + 3 = 5.
Quadratic for every k \neq -4. At k = -4 the leading coefficient is 0 and the expression is -6x + 1, which is linear.
Quadratic for every m except 1 and -1, where m^2 - 1 = 0. At m = 1 it is the constant 2. At m = -1 it is -2x + 2, which is linear.
y = 2x^2 + 1. First differences 2, 6, 10, 14; second differences 4, 4, 4, constant. So 2a = 4 and a = 2; c = 1; a + b = 2, so b = 0. Check: 2(4^2) + 1 = 33.
No. First differences 4, 12, 36, 108; second differences 8, 24, 72, which are not constant, and any quadratic would make them constant. (Each value is 3 times the one before: y = 2 \cdot 3^x.)
If a = 0 the x^2 term is gone and what is left, bx + c, is linear or constant, so a \neq 0 is exactly what makes the degree two. b and c do not affect the highest power, so the expression stays degree two whatever they are.
Multiplying adds degrees: the highest power in the product is x \cdot x^2 = x^3, with coefficient the product of two non-zero leading coefficients, so it is not zero. The product is cubic.
For example, (x^2 + x) + (-x^2 + 1) = x + 1, which is linear. The x^2 terms cancel whenever the two leading coefficients are opposites.
A quadratic’s second differences all equal 2a. If they are 0 then a = 0, and an expression with a = 0 is not quadratic. (A table like that is matched by a linear rule.)