Quadratic Formula Problems

Recall that for a, b, c ∈R, a ≠ 0, the solutions to ax ² + bx + c = 0 can by found using

1.1 Problem 1
Find the dimensions of a window with area 10 ft² and perimeter 13 ft.

1.1.1 Solution

Let h denote the height of the window and w the width. Note that h,w > 0, since we
cannot have negative dimensions , nor can we have 0 as a dimension, since the area is
non-zero.

The area is 10 ft² and perimeter 13 ft
<->2h + 2w = 13 and hw = 10 by the definitions of area and perimeter
<->2h + 2w = 13 and w = 10/h since h > 0
<->2h + 2(10/h) = 13 by substitution
<->2h² + 20 = 13h multiplying both sides by h
<->2h² − 13h + 20 = 0 subtracting 13h from both sides
<-> ( quadratic formula )
<-> h = 4 or h = 5/2 .
Hence, the dimensions of the window are

1.1.2 Some things to note

• It is important that you include implication arrows in your arguments.
• Implication arrows denote the equivalance of statements, not numbers. You cannot
inter change <-> and =.
• It is important to justify your steps in your write up. Some of the justifications here
(for example, subtracting 13h from both sides) are clear enough that they need not be
stated, however it is generally better to include more justification than not enough.

1.2 Problem 2
Which real numbers exceed their reciprocals by exactly 1?

1.2.1 Solution
A real number x exceeds its reciprocal by exactly 1

1.2.2 Some things to note
• The fact that x ≠ 0 is implied by the problem - since the reciprocal of x is not defined,
0 is clearly not a possible candidate for a solution. Furthermore, we have shown the two
possible values for x for which this is true, both of which are not equal to 0. Hence, it is
not necessary to state that x ≠ 0.
• When arriving at a solution that is not particularly intuitive ( like this one ), it will
generally be in your best interest to take the time and check to make sure it satisfies the
original condition, just to avoid errors in calculation . It is easy to show that both of these
do.

1.3 Problem 3

Find the solutions to

1.3.1 Solution

Cpheck in the original euqation:

<->1 = −1, which is false.
So 2 is not a solution to the original statement.
Hence, our only solution is x = 5.

1.3.2 Some things to note
•The reason we wound up with an extraneous root is that the implication in the first step
( squaring both sides) only goes one way . As mentioned, we showed if
then x = 5 or x = 2, however we did not show that if x = 5 or x = 2 that
Hence, we have not guarenteed that both of those are solutions, just that if x is a solution,
it will be 2 or 5.
• This is an example of why one must be careful when using biconditionals (double sided
arrows).

2 Polynomials: Definitions and Propositions

Definition 1. A polynomial is a function f : R -> R defined by
where k is a non-negative integer and are real numbers called coefficients .
Examples of Polynomials:

Definition 2. The degree of a polynomial is the largest d such that We denote
the degree of a polynomial p(x) by deg(p(x)). If p(x) = 0, the constant zero function, we
say that it has no degree.

Examples of Degree:

deg(f(x)) = 2
deg(g(x)) = 17
deg(h(x)) = 4
deg(p(x)) = 0.

Proposition 1. If a, b ∈Z with b ≠ 0, then there exists a unique integer pair k, r such
that a = kb + r and

Examples:
If a = 16, and b = 3, then 16 = 5 · 3 + 1, so k = 5, r = 1.
If a = 36 and b = 12, then 36 = 3 · 12, so k = 3, r = 0.
If a = 154 and b = 35, then 154 = 4 · 35 + 14, so k = 4 and r = 14.

We see in these examples that r just refers to the remainder when we divide a by b .
A similar statement is true for polynomials.

Proposition 2. If p₁(x) and p₂(x) are polynomials and p₂(x) ≠ 0, then there exists a
unique pair of polynomials q(x) and r(x) such that p₁(x) = q(x)p₂(x) + r(x), and either
r(x) = 0, or deg(r(x)) < deg(p₂(x)).

Note that it is important that p₂(x) ≠ 0, because as is the case with real numbers, we
still cannot divide by 0.

3 Long Division with Polynomials

3.1 Divide 2x² + 3 by 2x − 1

Thus, we see that the remainder is 7/2 , so We can easily
check that this is true. Here, q(r) = x + 1/2 and r(x) = 7/2 . Note that deg(r(x)) = 0 < 1 =
deg(2x − 1).

3.2 Divide x³ − 1 by x + 1

So the remaineder is −2, which means that x³ −1 = (x² −x+1)(x+1)+(−2). Here
q(x) = x² − x + 1 and r(x) = −2. Note that deg(r(x)) = 0 < 1 = deg(x + 1).