# The Major Topics of School Algebra

** Quadratic Equations **

Before approaching quadratic equations, students need some firm grounding in the

concept of a square root, which is more subtle than usually realized. Given a
positive

number s , then there is one and only one positive number r so that r^{2}
= s. The fact

that there is such an r is not trivial to prove, and, in fact, cannot be proved
in school

mathematics. But the fact that there is at most one such r, i.e., the
uniqueness of this

r, can be demonstrated with care to eighth graders and beyond. By definition,
this r

is called the square root of s, and is denoted by
. Thus by the definition of the

notation, is always ≥ 0. From the
uniqueness of the square root, one concludes the

critical fact that

for all positive a, b

This fact is usually either left unexplained, or relegated
to verification by calculator for a

few special cases. We recommend that it be carefully explained (proved), not
only because

students can learn from such reasoning, but also that they need to learn such
reasoning

to prepare them for advanced mathematics.

Now let x be a number. Then a **quadratic equation in one variable ** is an
equality of

the form (or can be brought to this form by the associative, commutative, and
distributive

laws) ax^{2} +bx+c = 0 which asks for all the real numbers x that make this
equality valid

(a, b, c are constants and a ≠ 0). A **solution**, or a **root**, of the equation is a
real number

which **satisfies** the equation, i.e., . To
** solve the equation ** is to find

all the solutions of the equation. It may be pointed out to students at the
outset that,

unlike the case of a linear equation in one variable, some quadratic equations
do not have

solutions, e.g., x^{2} + 3 = 0.

The method of solution dates back to the Babylonians some four thousand years
ago.

It consists of two steps:

(A) One can solve all quadratic equations of the form a(x + p)^{2} + q = 0, if it

has a solution.

(B) All quadratic equations can be brought to the form in (A) by the use of

the associative, commutative, and distributive laws .

Solving (A) is relatively straightforward: if it has a solution, it has to be
one of two

possibilities:

Again, the precise method of arriving at these solutions is worth repeating.
First,** if
we assume that there is a solution **, then we have an equality among numbers:

, from which we conclude that must have one of the two possible values

as above. This does not say that these values are indeed solutions of a(x + p)

^{2}+ q = 0.

For that, we have to directly verify that

This is a routine computation. The point, however, is that solving a quadratic equation

involves nothing but computation with fixed numbers. The consideration of “variables”

does not enter.

A little reflection is in order at this point. For to be solutions (i.e., points

on the number line), necessarily q/a ≤ 0, because there is no square root of a negative

number in the sense defined above. Conversely, as we have seen, if we know q/a < 0, then

are solutions. Thus we have proved, incidentally, that

a(x + p)^{2} + q = 0 has solutions
q/a ≤ 0.

The proof of (B) of course depends on the technique of** completing the square**. On

the one hand, this step is decisive for the solution, and is not one that yields
easily to

discovery learning. On the other hand, once the idea of completing the square is
accepted,

its implementation is quite routine. Indeed, if we know in advance that ax^{2} + bx
+ c can

be brought to the form a(x+p)^{2} +q for some suitable p and q, then expanding the
latter

by the distributive law gives

ax^{2} + bx + c = ax^{2} + 2apx + (ap^{2} + q)

Comparing the coefficients of both sides, we conclude that letting

and

should work. Indeed it does, as a simple computation shows. Another way, equally
valid,

of looking at this process is to ask what can be put in the box to make the
expression

within the parentheses a square:

Bearing in mind that (x^{2} + 2kx + k^{2}) = (x + k)^{2}, we easily come to the same
answer.

Recalling the earlier formula for the roots of a(x + p)^{2} + q = 0 as
, we have

now obtained the famous **quadratic formula **for the roots of ax^{2} + bx + c = 0:

(Note that this derivation requires the use of the identity
.) The earlier

reasoning about the solvability of a(x + p)^{2} + q = 0 leads to a similar
conclusion:

ax^{2} + bx + c = 0 has solutions
b^{2} − 4ac ≥ 0.

Completing the square is a basic technique in school mathematics, and its
significance

goes beyond getting the roots of a quadratic equation. When we get to the
discussion of

quadratic functions, for example, we will see how it leads to a complete
understanding of

their graphs.

It remains to bring closure to this discussion by mentioning that, if the
quadratic

polynomial ax ^{2} + bx + c (a ≠ 0) already comes in the form
for
some

numbers then solving the quadratic equation
becomes

extremely pleasant: the roots are and
. This is because the following basic
fact about

numbers implies that or
.

If pq = 0 for two numbers p and q, then one of p and q is
0.

The proof of this fact should be emphasized in the classroom: By FASM, we may
act as

if p and q are both rational numbers , which may be taken to be quotients of
integers. So

suppose p ≠ 0, we will prove that q = 0. Indeed, multiply both sides of pq = 0
by the

reciprocal of p, we get immediately q = 0, as desired.

The long and short of it is that, if a quadratic polynomial is already factored
as a

product of linear polynomials , then the corresponding quadratic equation can be
solved

by inspection. This is certainly one reason why the factoring of quadratic
polynomials

is of interest, and this fact can be used to motivate the usual exercises on
factoring such

polynomials.

It is a remarkable fact that, conversely, if we can solve a quadratic equation,
then we

also obtain a factorization of the corresponding quadratic polynomial. Let
ax^{2}+bx+c = 0

be given. Denoting the roots given by the quadratic formula by and as
above, then

we claim that the following identity in x is valid:

for all x

Why this is remarkable is that it allows us to recover the whole expression ax ^{2}
+ bx + c

completely as the product of as soon as we get to know the two
values of

x at which ax^{2} +bx+c becomes 0 (i.e.,
and ). This can

be explained as follows. From the explicit expression of the roots
of
ax^{2}+bx+c = 0

in terms of a, b, and c (i.e., the quadratic formula), we obtain the following
interesting

relations between the roots and the coefficients of a quadratic polynomial:

and

Therefore,

which is the asserted identity.

In particular, we see that factoring quadratic polynomials can be made entirely
mechanical:

just use the quadratic formula to get the roots and apply the preceding
identity.

Students should be made aware of this perspective to the factoring of quadratic
polynomials.

Many students do not see that the identity
requires

any proof. This is an example of a not-uncommon confusion between a simple
statement

(if for all x, then and are the roots)
and its not-so-

simple converse (if and
are the roots, then for

all x). From this instance, we can see the need to stress reasoning in school
mathematics.

If the quadratic polynomial is x^{2} + bx + c, i.e., if a = 1, then the preceding
relations

between the roots and the coefficients simplify to

and

These attractive relations between roots and coefficients have generalizations
to polynomials

of any degree.

The ability to solve quadratic equations greatly enlarges the range of word
problems,

which up to this point involve only linear phenomena. One particular example is
the

problem of finding the rectangle having the greatest area among all rectangles
with a

fixed perimeter.

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