Statistics as the Logic of Science

\documentclass{article} \usepackage[utf8]{inputenc} % !BIB TS-program = biber \usepackage[backend=biber,style=numeric, citestyle=authoryear]{biblatex} \addbibresource{blog.bib} \usepackage{amssymb} \usepackage{dirtytalk} \usepackage{csquotes} \usepackage{amsmath} \usepackage{calc} \usepackage{textcomp} \usepackage{mathtools} \usepackage[english]{babel} \usepackage{fancyhdr} \usepackage{url} \def\UrlBreaks{\do\/\do-} \usepackage{breakurl} \usepackage[breaklinks]{hyperref} \usepackage{graphicx} \graphicspath{ {images/} } \usepackage{wrapfig} \usepackage{float} \usepackage[T1]{fontenc} \usepackage{outlines} \usepackage{enumitem} \setenumerate[1]{label=\null} \setenumerate[2]{label=\null} \setenumerate[3]{label=\roman*.} \setenumerate[4]{label=\alph*.} \newcommand{\midtilde}{\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}} \usepackage{CJKutf8} \pagestyle{fancy} \fancyhf{} \title{Statistical Analysis by Communicative Functionals: \\ Lecture 2 – Statistics as The Logic of Science} \author{Justin Petrillo} \begin{document} \maketitle The question of $science \ itself$ has never been its particular object of inquiry but the existential nature, in its possibility and thereby the nature of its actuality. Science is power, and thus abstracts itself as the desired meta-good, although it is always itself about particularities as an ever-finer branching process. Although a philosophic question, the \textit{question of science} is inherently a political one, as it is the highest good desired by the society, its population, and its government. To make sense of science mathematically-numerically, as statistics claims, it is the scientific process itself that must be understood through probability theory as \say{The Logic of Science.} \footcite{QTSTATS} \section{Linguistic Analysis of the Invariants of Science: The Laws of Nature} The theory of science, as the proof of its validity in universality, must consider the practice of science, as the negating particularity. The symbolic language of science, within which its practice and results are embedded, necessarily negates its own particularity as well, as thus to represent a structure universally. Science, in the strict sense of having achieved already the goal of universality, is $de-linguistified$. While mathematics, in its extra-linguistic nature, often has the illusion of universal de-linguistification, such is only a semblance and often an illusion. The numbers of mathematics always can refer to things, and in the particular basis of their conceptual context always do. The non-numeric symbols of mathematics too represented words before short-hand gave them a distilled symbolic life. The de-linguistified nature of the extra-linguistic property of mathematics is that to count as mathematics, the symbols must themselves represent universal things. Thus, all true mathematical statements may represent scientific phenomena, but the context and work of this referencing is not trivial and sometimes the entirety of the scientific labor. The tense of science, as the time-space of the activity of its being, is the $tensor$, which is the extra-linguistic meta-grammar of null-time, and thus any and all times. \section{The Event Horizon of Discovery: The Dynamics between an Observer \& a Black Hole} The consciousness who writes or reads science, and thereby reports or performs the described tensor as an action of experimentation or validation, is the transcendental consciousness. Although science is real, it is only a horizon. The question is thus of its nature and existence at this horizon. What is knowable of science is thereby known as \say{the event horizon}, as that which has appeared already, beyond which is mere a \say{black hole} as what has not yet revealed itself – always there is a not-yet to temporality and so such a black hole can always be at least found as all that of science that has not and cannot be revealed since within the very notion of science is a negation of withdrawal (non-appearance) as the condition of its own universality (negating its particularity). Beginning here with the null-space of black-holes, the physical universe – at least the negative gravitational entities – have a natural extra-language – at least for the negative linguistic operation of signification whereby what is not known is the \say{object} of reference. In this cosmological interpretation of subjectivity within the objectivity of physical space-time, we thus come to the result of General Relativity that the existence of a black-hole is not independent of the observer, and in fact is only an element in the Null-Set, or negation, of the observer. To ‘observer’ a black-hole is to point to and outline something of which one does not know. If one ‘knew’ what it was positively then it would be not ‘black’ in the sense of not-emitting light \textit{within the reference frame (space-time curvature) of the observer}. That one $cannot$ see something, as receive photons reflecting space-time measurements, is not a property of the object but rather of the observer in his or her subjective activity of observation since to be at all must mean there is some perspective from which it can be seen. As the Negation of the objectivity of an observer, subjectivity is the $negative \ gravitational \ anti-substance$ of blackholes. Subjectivity, as what is not known by consciousness, observes the boundaries of an aspect (a negative element) of itself in the physical measurement of an ‘event horizon.’ These invariants of nature, as the conditions of its space-time, are the laws of dynamics in natural science. At the limit of observation we find the basis of the conditionality of the observation and thus its existence as an observer. From the perspective of absolute science, within the horizon of universality (i.e. the itself as not-itself of the black-hole or Pure Subjectivity), the space-time of the activity of observation (i.e. the labor of science) is a time-space as the hyperbolic negative geometry of conditioning (the itself of an unconditionality). What is a positive element of the bio-physical contextual condition of life, from which science takes place, for the observer is a negative aspect from the perspective of transcendental consciousness (i.e. science) as the limitation of the observation. Within Husserlian Phenomenology and Hilbertian Geometry of the early 20th century in Germany, from which Einstein’s theory arose, a Black-Hole is therefore a Transcendental Ego as the absolute measurement point. Our Solar System is conditioned in its space-time geometry by the MilkyWay galaxy it is within, which is conditioned by the blackhole Sagittarius A* ($SgrA*$). Therefore, the unconditionality of our solar space-time (hence the bio-kinetic features) is an unknown of space-time possibilities, enveloped in the event horizon of $SgrA*$. What is the inverse to our place (i.e. space-time) of observation will naturally only exist as a negativity, what cannot be seen. \section{Classical Origins of The Random Variable as The Unknown: Levels of Analysis} Strictly speaking, within Chinese Cosmological Algebra of 4-variables ($\mu$, X,Y,Z), this first variable of primary Unknowing, is represented by $X$, or Tiān (\begin{CJK*}{UTF8}{gbsn}天\end{CJK*}), for ‘sky’ as that which conditions the arc of the sky, i.e. “the heavens” or the space of our temporal dwelling ‘in the earth.’ We can say thus that $X=SgrA*$ is the largest and most relevant primary unknown for solarized galactic life. While of course X may represent anything, in the total cosmological nature of science, i.e. all that Humanity doesn’t know yet is conditioned by, it appears most relevantly and wholistically as $SgrA*$. It can be said thus that all unknowns ($x$) in our space-time of observation are within \say{\textit{the great unknown}} ($X$) of $SgrA*$, as thus $x \in X$ or $x \mathcal{A} X$ for the negative aspectual ($\mathcal{A}$) relationship \say{x is an aspect of X}. These are the relevant, and most general (i.e. universal) invariants to our existence of observation. They are the relative absolutes of, from, and for science. Within more practical scientific judgements from a cosmological perspective, the relevant aspects of variable unknowns are the planets within our solar system as conditioning the solar life of Earth. The Earthly unknowns are the second variable Y, or Di (\begin{CJK*}{UTF8}{gbsn}地\end{CJK*}) for “earth.” They are the unknowns that condition the Earth, or life, as determining the changes in climate through their cyclical dynamics. Finally, the last unknown of conditionals, Z, refers to people, Ren (\begin{CJK*}{UTF8}{gbsn}人\end{CJK*}) for ‘men,’ as what conditions their actions. X is the macro unknown (conditionality) of the gravity of ‘the heavens,’ Y the meso unknown of biological life in and on Earth, and Z the micro unknown of psychology as quantum phenomena. These unknowns are the subjective conditions of observation. Finally, the 4th variable is the “object”, or Wu (\begin{CJK*}{UTF8}{gbsn}物\end{CJK*}), $\mu$ of measurement. This last quality is the only $real$ value in the sense of an objective measurement of reality, while the others are imaginary in the sense that their real values aren’t known, and can’t be within the reference of observation since they are its own conditions of measurement within \say{the heavens, the earth, and the person}. \footcite[p.~82]{CHIN-MATH} In the quaternion tradition of Hamilton, ($\mu$, X,Y,Z) are the quaternions, ($\mu$, i,j,k). Since the real-values of X,Y,Z in the scientific sense can’t be known truly and thus must always be themselves unknowns, they are treated as imaginary numbers ($i=\sqrt{-1}$) with their ‘values’ merely coefficients to the quaternions $i,j,k$. These quaternions are derived as quotients of vectors, as thus the unit orientations of measurement’s subjectivity, themselves representing the space-time. We often approximate this with the Cartesian X,Y,Z of 3 independent directions as vectors, yet such is to assume Euclidean Geometry as independence. \printbibliography \end{document}
Cardinality and Countably Infinite Sets

Cardinality and Countably Infinite Sets

Cardinality is a term used to describe the size of sets. Set A has the same cardinality as set B if a bijection exists between the two sets. We write this as |A| = |B|. One important type of cardinality is called “countably infinite.” A set A is considered to be countably infinite if a bijection exists between A and the natural numbers ℕ. Countably infinite sets are said to have a cardinality of אo (pronounced “aleph naught”).

Remember that a function f is a bijection if the following condition are met:

1. It is injective (“1 to 1”): f(x)=f(y)⟹x=y

2. It is surjective (“onto”): for all b in B there is some a in A such that f(a)=b.

A set is a bijection if it is both a surjection and in injection.

A set is a bijection if it is both a surjection and an injection.

Example 1. Show that the set of integers ℤ is countably infinite.

To show that ℤ is countably infinite, we must find a bijection between ℕ and ℤ, i.e. we need to find a way to match up each element of ℕ to a unique element of ℤ, and this function must cover each element in ℤ.

We can start by writing out a pattern. One pattern we can use is to count down starting at 0, then going back and “picking up” each positive integer. This follows the pattern {0, -1, 1, -2, 2, -3, 3…}. We match to ℕ to ℤ as follows:

0 1 2 3 4
0 -1 1 -2 2

Notice that each even natural number is matched up to it’s half. It follows the function f(n) = n/2.

The odd numbers follow the function f(n) = -(n+1)/2

We can write this as a piecewise function as:

$f(n) =

\begin{cases}

n/2, & \text{if n is even} \\

-(n+1)/2, & \text{if n is odd}

\end{cases} $

Now we need to check if our function is a bijection.

Injectivity: Suppose the function is not injective. Then there exists some natural numbers x and y such that f(x)=f(y) but x≠y.

For even integers, x/2 = y/2 ⟹ x=y

For odd integers, Then (-x+1)/2 = -(y+1)/2 ⟹ x=y

These are contradictions, so the function is injective.

Surjectivity: Suppose the function is not surjective. Then there is some integer k such that there is no n in ℕ for which f(n) = k.

For k≥0, k=n/2 ⟹2k=n ⟹ n is an even number in ℕ

For k<0, k=-(n+1)/2 ⇒ -2k-1 = n ⇒ n is an odd number in ℕ

These are contradictions, so the function is surjective.

Since f is both injective and surjective, it is a bijection. Therefore, |ℤ| = |N| =אo. ∎

This result is often surprising to students because the set ℕ is contained in the set ℤ.

Example 2. For A = {2n | n is a number in ℕ}, show that |A| = |ℤ| (the set of integers has the same cardinality as the set of even natural numbers).

We can either find a bijection between the two sets or find a bijection from each set to the natural numbers. Since we already found a bijection from ℤ to ℕ in the previous example, we will now find a bijection from A to ℕ.

One function that will work is f(n) = n/2. Checking that it is a bijection is very similar to Example 1.

Since |A| = |ℕ| and |ℤ| = |ℕ|, then |A| = |ℤ| = אo.∎

There are many sets that are countably infinite, ℕ, ℤ, 2ℤ, 3ℤ, nℤ, and ℚ. All of the sets have the same cardinality as the natural numbers ℕ. Some sets that are not countable include ℝ, the set of real numbers between 0 and 1, and ℂ.

Georg Cantor was a pioneer in the field of set theory and was the first to explore countably infinite sets

Georg Cantor was a pioneer in the field of set theory and was the first to explore countably infinite sets

 

Completing the Square

Completing the Square

by: HT Goodwill

What Is Completing the Square?

Completing the Square is a technique in algebra that allows us to rewrite a quadratic expression that was in standard form:

$$ax^2 + bx + c$$

into vertex form:

$$a(x-h)^2 + k$$

Why Do We Complete the Square?

It should be noted that many times, students think (are taught?) that completing the square is just a way to solve a quadratic equation that has the form:

$$ax^2 + bx + c = 0$$.

However, in other areas of mathematics, we sometimes need to express a quadratic in the vertex form.

Leading up to How To

Suppose we have a perfect square binomial that we expand using FOIL:

$$\begin{align*}
(x + B)^2 & = (x + B)(x + B)\\
& = x^2 + Bx + Bx + B^2\\
& = x^2 + 2Bx + B^2
\end{align*}$$

If we reverse this string of equations, we see that any quadratic that has this pattern: \(x^2 + 2Bx + B^2\) which is a perfect square and we can factor it as \((x + B)^2\).

To Complete the Square, we adjust a quadratic expression so that it exhibits the perfect-square pattern. Here are some examples:

Introductory Examples

Example 1: Complete the square on \(x^2 + 14x + 40\)

Match up the quadratic expression with the perfect square pattern, starting with the \(x^2\) term.

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + 14x + 40
\end{align*}$$

Step 1: Matching The Quadratic Terms

The quadratic terms match since they have the same coefficient.

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + 14x + 40\\
\mbox{Match?} & \quad
\end{align*}$$

Step 2: Matching the Linear Terms

The value of \(2B\) is assigned be us and is whatever we need it to be so we can match the perfect-square pattern. In this example, they match, if we assign \(2B = 14\).

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + { 2B}x + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + {14}x + 40\\
\mbox{Match?} & \quad \quad\quad
\end{align*}$$

Step 3: Matching the Constant Term

In order to match the perfect square pattern, the constant term has to be equal to \(B^2\). Since we defined \(2B = 14\) in Step 2, we know that \(B = 7\) which implies \(B^2 = 7^2 = 49\).

To adjust our constant term, we add zero (which won’t change the value of anything) and choose to rewrite the zero as \(B^2 – B^2\).

$$\begin{align*}
x^2 + 14x + 40 & = x^2 + 14x + 0 + 40 &&\mbox{Adding } 0.\\
& = x^2 + 14x + { (49 – 49)} + 40 &&\mbox{Since } B^2 – B^2 = 49-49 = 0\\
& = x^2 + 14x + 49 – 49 + 40\\
& = (x^2 + 14x + 49) – 49 + 40&&\mbox{Regrouping}
\end{align*}$$

Matching up to the pattern:

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad (x^2 + 14x + 49) – 49 + 40\\
\mbox{Match?} & \quad \quad\quad \quad\quad
\end{align*}$$

Since the first three terms (the ones in the parentheses) match the perfect-square pattern, we know those three terms will factor into \((x+B)^2\).

$$\begin{align*}
(x^2 + 14x + 49)- 49 + 40
& = (x+7)^2 – 49 + 40\\
& = (x+7)^2 – 9
\end{align*}$$

Answer: \(x^2 + 14x + 40 = (x+7)^2 – 9\)

Example 2: Complete the square on \(x^2 – 9x + 3\).

As before, we match up our quadratic with the perfect-square pattern. We’ll do this one a bit more quickly.

Step 1: Quadratic and Linear Terms

Both quadratic terms have a coefficient of \(1\) so they already match. The linear terms will match if we assign \(2B = -9\). So without any real work or effort, we match the first two terms already.

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2-9x + 3\\
\mbox{Match?} & \quad \quad\,\,
\end{align*}$$

Step 2: Constant Term

Since we assigned \(2B = -9\), we divide by 2 to get \(B = -\frac{9}{2}\). And we know our constant term needs to be \(B^2 = \left(-\frac{9}{2}\right)^2 = \frac{81}{4}\). and we introduce it the same way we did before, by adding zero.

$$\begin{align*}
x^2 – 9x + 3
& = x^2 – 9x + 0 + 3 && \mbox{Adding }0\\
& = x^2 – 9x + \left(\frac{81}{4} – \frac{81}{4}\right) + 3 && \mbox{Since } B^2 – B^2 = \frac{81}{4}-\frac{81}{4} = 0\\
& = x^2 – 9x + \frac{81}{4} – \frac{81}{4} + 3\\
& = \left(x^2 – 9x + \frac{81}{4}\right) – \frac{81}{4} + 3 && \mbox{Regrouping}
\end{align*}$$

Making sure we match the pattern, we have

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad \left(x^2 + 14x + \frac{81}{4}\right) – \frac{81}{4} + 40\\
\mbox{Match?} & \quad\,\,\,\quad\quad\quad\quad
\end{align*}$$

Since the three terms in the parentheses match the perfect-square pattern, we know those three terms factor as a perfect square.

$$\begin{align*}
\overbrace{\left(x^2 – 9x + \frac{81}{4}\right)}^{\mbox{Perfect Square}\frac{81}{4} + 3
& = \overbrace{\left(x – \frac 9 2\right)^2}^{\mbox{Factored}} – \frac{81}{4} + \frac{12}{4}\\
& = \left(x – \frac{9}{2}\right)^2  – \frac{69}{4}
\end{align*}$$

Answer: \(x^2 – 9x + 3 = \left(x – \frac{9}{2}\right)^2 – \frac{69}{4}\)

Example 3: Dealing With Leading Coefficients. Complete the square on \(2x^2 + 12x – 5\).

Solution In order to complete the square, we first need to reduce the leading coefficient to 1.

Step 1: Factor out the leading coefficient out of the first two terms.

$$2x^2 + 12x – 5 = 2[x^2 + 6x] – 5$$

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Step 2: Complete the Square \textit{Inside the Braces}\vskip 2mm

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\qquad Quadratic Term: \textit{Inside} the braces, our quadratic term has a coefficient of 1.\vskip 2mm

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\qquad Linear Term: \textit{Inside} the braces, we set \({\color{blue} 2B = 6}\).\vskip 2mm

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A quick check and we see that we are matching the perfect-square pattern \textit{inside the braces}.

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\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\,\,\,\, x^2 + {\color{blue}2B}x + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[x^2 + {\color{blue}6}x] – 5\\
\mbox{Match?}&\quad\quad \checkmark\quad\,\,\,\checkmark
\end{align*}

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Let’s move on to the constant term.\vskip 2mm

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\qquad Constant Term: \textit{Inside} the braces we have \({\color{blue} 2B = 6}\) which means \(B = 3\) and so \({\color{red} B^2 = 3^2 = 9}\). As before, adjust our constant term by adding {\color{red} 0}, but now it is \textit{inside the braces}.

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\begin{align*}
2[x^2 + {\color{blue} 6}x] – 5
& = 2[x^2 + {\color{blue} 6}x + {\color{red} 0}] – 5 && \mbox{Adding } {\color{red} 0} \mbox{ \textit{inside} the braces}\\
& = 2[x^2 + {\color{blue} 6}x + {\color{red} (9 – 9)}] – 5 && \mbox{Since } B^2 – B^2 = {\color{red} 9 – 9 = 0}\\
& = 2[x^2 + 6x + 9 – 9] – 5\\
& = 2[(x^2 + 6x + 9) – 9] – 5 && \mbox{Regrouping \textit{inside} the braces}
\end{align*}
\vskip 2mm

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Still keeping our attention focused \textit{inside} the braces, we check to see if we match the perfect-square pattern.

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\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[(x^2 + 6x + 9) – 9] – 5\\
\mbox{Match?}&\quad\quad \checkmark\quad\,\,\,\checkmark\quad\,\,\checkmark
\end{align*}

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Since the three terms inside the parentheses match the pattern, we know those three terms form a perfect square, so they will easily factor into \((x + B)^2\).

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\begin{align*}
2\big[{\color{blue}\overbrace{\left(x^2 + 6x + 9\right)}^{\mbox{Perfect Square}}} – 9\big] – 5
& = 2\big[{\color{blue}\overbrace{\left(x + 3\right)^2}^{\mbox{Factored}}} – 9\big] – 5
\end{align*}
\vskip 5mm

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Step 3: Adjusting the Final Form\vskip 2mm

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Start by multiplying the leading coefficient \textit{through the square braces} (NOT the parentheses!), and simplify.

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\begin{align*}%
{\color{blue} 2}\big[\left(x + 3\right)^2 – 9\big] – 5
& = \big[{\color{blue} 2}(x+3)^2 – {\color{blue} (2)} 9\big] – 5 && \mbox{Multiply through braces}\\
& = \big[2(x + 3)^2 – 18\big] – 5\\
& = 2(x + 3)^2 – 18 – 5 && \mbox{Braces no longer needed}\\
& = 2(x+3)^2 – 23 && \mbox{Combine the constants}
\end{align*}
\vskip 5mm

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\textbf{Answer:} \(2x^2 + 12x – 5 = 2(x+3)^2 – 23\)
\vskip 1.5cm

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\textbf{\large Quick Example}\vskip 2mm

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\textbf{Example 4: A Quick Example}\vskip 2mm

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Complete the Square on \(3x^2 – 4x + 8\).\vskip 5mm

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\textbf{Solution:}\vskip 2mm

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\begin{align*}%
3x^2 – 4x + 8 & = 3\left[x^2 – \frac 4 3 x\right] + 8 && \mbox{Factor out leading coefficient}\\[6pt]
& = 3\left[x^2 {\color{blue}- \frac 4 3} x\right] + 8 && \mbox{Identify } {\color{blue} 2B = -\frac 4 3} \longrightarrow {\color{red} B^2 = \left(-\frac 2 3\right)^2 = \frac 4 9}\\[6pt]
& = 3\left[x^2 {\color{blue}- \frac 4 3} x + {\color{red} \left(\frac 4 9 – \frac 4 9\right)}\right] + 8 && \mbox{Add } {\color{red} 0}\\[6pt]
& = 3\left[\left(x^2 {\color{blue}- \frac 4 3} x + {\color{red}\frac 4 9}\right)\,\,{\color{red}- \frac 4 9}\right] + 8 && \mbox{Regrouping}\\[6pt]
& = 3\left[\left(x – \frac 2 3\right)^2 – \frac 4 9\right] + 8 && \mbox{Factor}\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + 8 && \mbox{Distribute}\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + \frac{24} 3\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3 && \mbox{Combine like terms}
\end{align*}
\vskip 5mm

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\textbf{Answer:} \(3x^2 – 4x + 8 = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3\)
\vskip 1.5cm

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\textbf{\large The Quick Method}\vskip 2mm

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The method shown in the examples can be made more efficient if we recognize that the pattern is always the same. For a simple quadratic with a leading coefficient of \(1\), the completed square form looks like this:

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\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-1) grid (2,1);
% \draw (-2,-1) grid (2,1);
% Equation
\node {%
\(%
x^2 + {\color{blue} 2B}x + c = (x + {\color{blue} B})^2 – {\color{blue} B^2} + c
\)};
% Arrows
\node (linear) [inner sep=0pt] at (-1.625,-0.15){};
\node (BLow) [inner sep=0pt] at (0.75, -0.15){};
\node (BUp) [inner sep=0pt] at (0.75, 0.2){};
\node (B2) [inner sep=0pt] at (1.75,0.2){};
\draw [-latex] (linear.south) to [out=270,in=270] (BLow.south);
\draw [-latex] (BUp.north) to [out=90,in=90] (B2.north);
% Labels
\node at (-0.375, -0.625) {\(\div 2\)};
\node at (1.5, 0.75) {\scriptsize Minus the Square};
\node at (3,-1) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}

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Inside the final parentheses we always end up with \(x + B\), where \(B\) is half of the coefficient of the original \(x\) term.\vskip 2mm

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Next, we subtract \(B^2\) \textit{outside} the parentheses. Let’s try it with one of our previous examples to see it in action.\vskip 5mm

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\textbf{Example 5: Simple Quick Method}\vskip 2mm

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Use the quick version of Completing the Square on \(x^2 + 14x + 40\).

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\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (0,-1) grid (5,1);
% \draw (0,-1) grid (5,1);
% Equation
\node at (1.2,-0.25){%
\(%
x^2 + {\color{blue} 14}x + 40 = (x + {\color{blue} 7})^2 \underbrace{- {\color{blue} 7^2} + 40}_{\mbox{\scriptsize Add Together}} = (x + 7)^2 – 9
\)};
% Arrows
\node (linear) [inner sep=0pt] at (-1.625,-0.15){};
\node (BLow) [inner sep=0pt] at (0.75, -0.15){};
\node (BUp) [inner sep=0pt] at (0.75, 0.2){};
\node (B2) [inner sep=0pt] at (1.75,0.2){};
\draw [-latex] (linear.south) to [out=270,in=270] (BLow.south);
\draw [-latex] (BUp.north) to [out=90,in=90] (B2.north);
\draw [-latex] (2.85,-0.575) .. controls (3,-0.625) and (5,-0.75)..(5,-0.25);
% Labels
\node at (-0.375, -0.625) {\(\div 2\)};
\node at (1.5, 0.75) {\scriptsize Minus the Square};
\node at (5,-1) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\vskip 5mm

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\textbf{Answer:} \(x^2 + 14x + 40 = (x + 7)^2 – 9\)

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Note: This is the same answer we got in Example 1.\vskip 1.5cm

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\textbf{Example 6: Quick Method with Leading Coefficient}\vskip 2mm

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The quick version can also be used when the leading coefficient isn’t \(1\). Like before, we just factor out the leading coefficient first, and then work inside the braces.\vskip 2mm

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Let’s complete the square on \(5x^2 + 80x + 13\).

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\noindent\textbf{Solution}\vskip 2mm

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Step 1) Factor out the leading coefficient.

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\[%
5x^2 + 80x + 13 = 5[x^2 + 16x] + 13
\]

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Step 2) Now use the quick method to complete the square on the \textit{inside} of the braces.

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\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-2) grid (4,2);
% \draw (-2,-1) grid (4,1);
% \draw [red, step = 2] (0,-1) grid (3,1);
% Equation
\node at (1.2,0){%
\begin{minipage}{\textwidth}
\begin{align*}%
5[x^2 + {\color{blue} 16}x] + 13
& = 5[x^2 + {\color{blue} 16}x] + 13\\[5mm]
& = 5[(x + {\color{blue} 8})^2 – {\color{blue} 64}] + 13\\[6mm]
& = 5(x + {\color{blue} 8})^2 – 320 + 13\\
& = 5(x + {\color{blue} 8})^2 – 307%
\end{align*}%
\end{minipage}
};
% Arrows
\node (linear) [inner sep=0pt] at (2.375,0.95){};
\node (BLow) [inner sep=0pt] at (2.23, -0.05){};
\node (BUp) [inner sep=0pt] at (2.23, 0.25){};
\node (B2) [inner sep=0pt] at (3.23,-0.05){};
\draw [-latex] (linear.south) to [out=270,in=90] (BUp.south);
\draw [-latex] (BLow.south) to [out=270,in=270] (B2.south);
% Labels
\node at (2.625, 0.625) {\scriptsize \(\div 2\)};
\node at (3, -0.55) {\scriptsize Minus the Square};
\node at (4.5,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\vskip 5mm

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\textbf{Answer:} \(5x^2 + 80x + 13 = 5(x + 8)^2 – 307\)

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\vskip 1.5cm

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\textbf{\large Solving Equations}\vskip 2mm

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When solving a quadratic equation, I find the technique of completing the square is not very efficient, except in the simple cases where the leading coefficient is \(1\) and the linear coefficient is even.\vskip 2mm

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But that doesn’t mean we shouldn’t know \textit{how} to solve more complicated quadratic equations with this technique.\vskip 5mm

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\textbf{Two Approaches}\vskip 2mm

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When solving an equation through completing the square, there are two basic approaches: (1) Complete the square first, \textit{then} solve the equation, or (2) Solve the equation \textit{as} you complete the square.\vskip 5mm

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\textbf{Example 7: Solving a Quadratic Equation by Completing the Square}\vskip 2mm

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Use Completing the Square to solve \(4x^2 + 3x – 10 = 0\).\vskip 5mm

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\textbf{Solution (Method 1)}\vskip 2mm

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For this approach, we’ll first complete the square, then solve the equation.\vskip 2mm

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Step 1) Complete the Square on \(4x^2 + 3x – 10\).\vskip 2mm

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Using the techniques we’ve discussed above, we get the following.

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\[%
4x^2 + 3x – 10 = 4\left(x + \frac 3 8\right)^2 – \frac{169}{16}
\]
\vskip 5mm

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Step 2) Solve the equation.

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\begin{align*}
4x^2 + 3x – 10 & = 0\\[6pt]
4\left(x + \frac 3 8\right)^2 – \frac{169}{16} & = 0 && \mbox{Complete the Square}\\[6pt]
4\left(x + \frac 3 8\right)^2 & = \frac{169}{16} && \mbox{Add the constant}\\[6pt]
\left(x + \frac 3 8\right)^2 & = \frac{169}{64} && \mbox{Divide by \(4\)}\\[6pt]
x + \frac 3 8 & = \pm\sqrt{\frac{169}{64}} && \mbox{Take the square root}\\[6pt]
x + \frac 3 8 & = \pm\frac{13} 8 && \mbox{Simplify}\\[6pt]
x & = – \frac 3 8 \pm\frac{13} 8 && \mbox{Subtract}
\end{align*}

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The two solutions are

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\[%
\begin{array}{rl}
x & = \displaystyle – \frac 3 8 + \frac {13} 8 = \frac{10} 8 = \frac 5 4\\[12pt]
x & = \displaystyle – \frac 3 8 – \frac{13} 8 = – \frac{16} 8 = – 2
\end{array}
\]
\vskip 5mm

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\textbf{Solution (Method 2)}\vskip 2mm

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This time, we’ll complete the square as we solve the equation. It should be noted that this approach is what many students think \textit{is} Completing the Square.\vskip 2mm

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\begin{align*}
4x^2 + 3x – 10 & = 0\\[6pt]
4x^2 + 3x & = 10 && \mbox{Add the constant}\\[6pt]
x^2 + \frac 3 4 x & = \frac{10} 4 && \mbox{Divide by 4}\\[6pt]
x^2 + \frac 3 4 x + \frac 9 {64} & = \frac{10} 4 + \frac 9 {64} && \mbox{Add \(B^2\)}\\[6pt]
\left(x + \frac 3 8\right)^2 & = \frac{160}{64} + \frac 9 {64} && \mbox{Factor the left side}\\[6pt]
x + \frac 3 8 & = \pm \sqrt{\frac{169}{64}} && \mbox{Take square root}\\[6pt]
x & = -\frac 3 8 \pm \frac{13} 8 && \mbox{Subtract}
\end{align*}

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Again, we see the two answers are \(x =-\frac 3 8 + \frac{13} 8 = \frac 5 4\) and \(x = -\frac 3 8 – \frac{13} 8 = -2\).
\vskip 5mm

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\textbf{Answer:} \(x = \frac 5 4\) or \(x = -2\)
\vskip 1.5cm

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\textbf{Conclusion}\vskip 2mm

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Admittedly, most high school student will Complete the Square in the context of solving quadratic equations. However, it is important that they understand that the technique is more than just an equation-solving technique. Outside of high school math, much of its use is in changing the form of a quadratic function so that it can be used in higher mathematical techniques.

Find the Vertex by Using the Quadratic Formula

The Quadratic Formula is primarily used to identify the roots (\(x\)-intercepts) of a quadratic function. What many people don’t know is that you can also easily find the vertex of the function by simply looking at the Quadratic Formula!

 

Graphing Quadratic Functions

When graphing quadratic functions by hand, we will often use the quadratic formula to obtain the roots (or \(x\)-intercepts) of the function. Then separately, we will work on finding the vertex of the function since this point defines the lowest (or highest) point on the graph, as well as the line of symmetry.

Graph of a generic quadratic function showing both roots and the vertex.

For the general quadratic function, \(f(x) = ax^2 + bx + c\), the quadratic formula identifies the roots at:

\[%
x = \frac{-b \pm\sqrt{b^2 – 4ac}}{2a}
\]
while the the vertex is found at \(\left(-\frac b {2a}, f\left(-\frac b {2a}\right)\right)\).

The \(x\)-value of the vertex, \(x = -\frac b {2a}\), is often referred to in algebra textbooks as The Vertex Formula.

 

Finding the Vertex in the Quadratic Formula

Many people notice (or are taught) that the Vertex Formula can be found nestled in the Quadratic Formula. Specifically,

\[%
\mbox{Quadratic Formula: } x = \frac{\color{blue}{-b} \pm\sqrt{b^2 – 4ac}}{\color{blue}{2a}}\longrightarrow \color{blue}{\frac{-b}{2a}} = x\mbox{-value of the vertex.}
\]

Less well known is the fact that the \(y\)-coordinate of the vertex can also be obtained easily from the quadratic formula. The vertex of a quadratic function can be written as

\[%
\mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right)
\]
where \(D\) is the discriminant of the quadratic function. Recall that the discriminant of a quadratic function is \(\color{red}{D = b^2 – 4ac}\) and it appears under the radical in the quadratic formula:

\[%
\mbox{Quadratic Formula: } x = \frac{-b \pm\sqrt{\color{red}{b^2 – 4ac}}}{2a} = \frac{-b \pm\sqrt{\color{red} D}}{2a}.
\]

Also, notice that the denominator of the \(x\)-coordinate and \(y\)-coordinate of the vertex are very similar.  The \(y\)-coordinate’s denominator is just twice as big as the \(x\)-coordinate’s!

Now, let’s try out this idea and find the vertex of a quadratic from just the quadratic formula.

Example 1:

Suppose \(f(x) = 5x^2 + 4x – 3\). Use the quadratic formula to identify the entire vertex of the function, and then verify the answer using the standard method.

Solution: The quadratic formula for this function is
\[%
x = \frac{-4 \pm \sqrt{4^2 – 4(5)(-3)}}{2(5)} = \frac{-4\pm \sqrt{76}}{10}.
\]

The discriminant for this function is \(\color{red}{D = 76}\). So, the vertex should be at

\[%
\mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right) = \left(-\frac 4 {10}, -\frac{\color{red}{76}}{20}\right) = \left(-\frac 2 5, -\frac{19} 5\right).
\]

Let’s double check our \(y\)-coordinate by using the standard method (plugging \(x = -\frac b {2a} = -\frac 2 5\) into the original function).

\begin{align*}
f\left(\color{blue}{-\frac 2 5}\right)
& = 5\left(\color{blue}{-\frac 2 5}\right)^2 + 4\left(\color{blue}{-\frac 2 5}\right) – 3 && \mbox{Plugging in the \(x\)-value.}\\[6pt]
& = 5\left(\frac 4 {25}\right) – \frac 8 5 – \frac 3 1 && \mbox{Simplifying.}\\[6pt]
& = \frac 4 5 – \frac 8 5 – \frac{15} 5\\[6pt]
& = \frac{-19} 5 && \mbox{\(y\)-value of the vertex.}
\end{align*}

 

Example 2

Let’s try another example. Suppose \(f(x) = -2x^2 +12x -25\). Identify the vertex using the quadratic formula, and verify the answer with the standard method.

Solution: The quadratic formula for this function is

\[%
x = \frac{-12 \pm\sqrt{(-12)^2-4(-2)(-25)}}{2(-2)} = \frac{\color{blue}{-12} \pm\sqrt{\color{red}{-56}}}{-4}
\]

The vertex should be \(\displaystyle \left(\frac{\color{blue}{-12}}{-4}, -\frac{\color{red}{-56}}{-8}\right) = \left(3, -7\right)\).  Let’s verify the result. Using the standard method, we evaluate \(f(3)\).

\begin{align*}
f(\color{blue} 3) & = -2(\color{blue} 3)^2 + 12(\color{blue} 3) – 25\\[6pt]
& = -2(9) + 36 – 25\\
& = -18 + 11\\
& = -7
\end{align*}

Again, we see the method works.

However, the savvy math student will immediately ask, “Does it work every time? Or was there something `special’ or `unusual’ about these particular quadratic functions?”

 

Does it Always Work?

This is an important question. In order to claim something is valid in mathematics, we have to do more than look at a few examples. We have to prove its validity. So, without further ado, here is the proof of this technique.

Given: The \(x\)-coordinate of the vertex of a quadratic function is \(x = -\frac b {2a}\).

Show: The \(y\)-coordinate of the vertex of a quadratic function is \(y = -\frac D {2a}\), where \(D\) is the discriminant of the function.

Proof: Simply evaluate \(f(x) = ax^2 + bx + c\) at \(x = -\frac b {2a}\).

\begin{align*}
f\left(\color{blue}{-\frac b {2a}}\right)
& = a\left(\color{blue}{-\frac b {2a}}\right)^2 + b\left(\color{blue}{-\frac b {2a}}\right) + c\\[6pt]
& = a\left(\frac{b^2}{4a^2}\right) -\frac{b^2}{2a} + c && \mbox{Simplify a little.}\\[6pt]
& = \frac{b^2}{4a} -\frac{b^2}{2a} + \frac c 1 && \mbox{Cancel the common factor.}\\[6pt]
& = \frac{b^2}{4a} -\frac{2b^2}{4a} + \frac{4ac}{4a} && \mbox{Common denominator.}\\[6pt]
& = \frac{-b^2 + 4ac}{4a} && \mbox{Combine numerators.}\\[6pt]
& = -\frac{\color{red}{b^2 – 4ac}}{4a} && \mbox{Factor out a negative 1.}\\[6pt]
& = -\frac{\color{red}{D}}{4a}
\end{align*}

Well, how about that. It seems like any time we evaluate a quadratic at \(x = -\frac b {2a}\), we will end up with \(-\frac D {4a}\).

 

So What?

So, what does this do for us in practical terms? Suppose we were asked to graph (by hand) the function \(f(x) = 3x^2 -6x + 8\). A good first step is to find the roots, and a good method for that is to use the quadratic formula.

\begin{align*}
x & = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\\[6pt]
& = \frac{6 \pm \sqrt{(-6)^2 – 4(3)(8)}}{2(3)}\\[6pt]
& = \frac{6 \pm \sqrt{-60}} 6
\end{align*}

Oops. This quadratic function doesn’t have real-valued \(x\)-intercepts (Yes, it has complex roots, but that won’t help us graph, will it?).

But the function has a graph, we just have to find it. If only we had the vertex and one other point! Then we could graph the quadratic! But wait! We already have written out the quadratic formula, so that should tell us the vertex:

\[%
\mbox{Vertex: } \left(-\frac b {2a}, -\frac D {4a}\right) = \left(-\frac 6 6, -\frac{-60}{12}\right) = (1, 5).
\]

Now, together with the \(y\)-intercept at \((0,8)\), we can easily plot the function.

Graph of the quadratic function. The vertex was easily found using only the quadratic formula.

 

Conclusion

In conclusion, the more we understand about the tools we have at our disposal, the easier it becomes to work through the mathematics that crop up in school and careers. And the more we learn, the more we become intrigued by the idea of other possibilities.  I mean, doesn’t this insight into the quadratic formula just beg the question, “What else can we learn from the Quadratic Formula? Or the discriminant?”

 

© HT Goodwill and www.mathacademytutoring.com, 2016. Unauthorized use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to HT Goodwill and mathacademytutoring.com with appropriate and specific direction to the original content.

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