## The Bias of 1-D Opinion Polling: Explaining the Low Polling Numbers for Candidate Beto O’Rourke

In Electoral Opinion Sampling, whether of candidates, policies, or values, it is commonplace to ask subjects yes/no questions, where someone either choses one person out of a list of candidates or says whether or not he or she agrees or disagrees with a political statement. Such a question though only has one layer of information and disregards the unique nature of an opinion, which includes not only the final choice – voting for a candidate or policy – but also the reasoning behind the choice, the “why?” behind the claim. Thus, only the surface of the opinion manifold is measured through the yes/no questions of mass politics. This creates a bias in our statistical understanding of the population’s political views since it collapses the distribution of opinions into a single norm, leaving us with the impression of polarization, where people are either on the right or left of an issue with little sense of the commonalities or overlaps. Thus, when the political sphere appears polarized, it is more of a problem in measurement, than in the actual underlying viewpoints. To resolve this social-political problem of polarization, where the nation can’t seem to come to a common viewpoint, we must look at the depth of the opinion manifold by mapping out a system of opinions rather than a single norm.

We can use Game Theory to represent an opinion as an ordering of preferences, i.e. A < B < C < D < E < F. Where each choice-element of the preference set must be strictly ordered in relationship to each other, leaving a ranked list of choices, one has a strict ordering of preferences. This was used to represent opinion in Arrow’s Theorem of Social Choice. Yet, without any allowable ambiguity, the result proves an equivalence between the aggregate social choice methods of dictatorship (one person chooses the social good) and democracy (the majority chooses the social good). This explains the critical political observation that mass politics – based upon superficial opinions – often becomes fascist – where one personality dominates the national opinion at the exclusion of immigrant or marginal groups. This game-theoretic error of restricting preferences is equivalent to the recently noted behavioral-economic error of excluding irrationalities (i.e. risk-aversion) from micro utility-maximization. Instead, we can represent opinion as a partial ordering of preferences, rather than a strict ordering. Thus, an opinion is represented as a tree graph, algebraically by A >> B, B >> D, B >> E, A >> C, & C >> F, or a tree data structure, formatted as {A: (B: (D,E), C: (F))} (i.e. JSON). The relationship of inclusion (>>, i.e. A >> B) can be interpreted as ‘A is preferred over B’ or ‘B is the reason for A,’ depending on whether one is looking at the incomplete ranking of choices or the irrationality of certain value-claims. In micro-economics, this yields a non-linear hyperbolic functional relationship between individual opinion and the aggregate social choice, rather than a reductionist linear functional relationship. In a hyperbolic space, we can represent each opinion-tree as a hyper-dimensional point (via a Kernel Density Estimation) and perform commonplace statistical tools, such as linear-regression or the multi-dimensional Principal Component Analysis, resulting in hyper-lines of best-fit that describe the depth of the aggregate social choice.

This method of deep-opinion analysis is particularly useful for understanding electoral dynamics still in flux, as with the Democratic Primaries, where there are too many candidates for people to have a strictly ranked preference of them all. In such an indeterminate thermodynamic system (such as a particle moving randomly along a line of preferences), there is an element of complexity due to the inherent stochastic ‘noise’ as people debate over each candidate, change their minds, but ultimately come to deeper rationalities for their opinions through the national communication mediums. Instead of trying to reduce this ‘noise’ to one Primary Candidate choice so early in the democratic process when the policies of the party are still being figured out – similar to waiting to measure a hot quantum system (i.e. collapsing the wave-function of the particle’s position) while it is still cooling into equilibrium – we can instead represent the probabilistic complexity of the preference distributions. In preference orderings of democratic candidates, this means that while the underlying rationality of an opinion (deep levels of a tree) may not change much during an election cycle, with small amounts of new information the surface of the top candidate choice may change frequently. In order to make a more predictive electoral-political models, we should thereby measure what is invariant (i.e. deep-structure), always missed in asking people for their top or strictly-ranked preferences. While a single candidate may consistently be people’s second choice, he or she could end up still polling at 0%. If this ordering isn’t strict, i.e. always less than the top choice but above most others, then the likelihood of this ‘2nd-place candidate’ being close to 0% is even higher. Without the false assumption of deterministic processes, it is not true that the surface measurement of the percent of the population willing to vote for a candidate is equivalent to the normative rationality of that candidate – the 0% candidate may actually represent public views very well although such cannot be expressed in the 1-dimensional polling surveys. Thus, while the actual electoral voting does collapse the chaotic system of public opinion into a single choice aggregated over the electoral-college, such measurement reduction is insignificant so early in a democratic process with fluctuating conditions. As a thermodynamic rule, to measure a high-entropic system, we must use hyper-dimensional informational units.

The Democratic Primary candidate Beto O’Rourke is precisely such a hidden 2nd-place candidate thus far, who is indeed was polling close to 0% (now he is at 4%) in the primary although the votes he received in his Texas Senate run alone would place him near 3.5% of total votes across both parties, assuming no one in any other state voted for him and Trump was substituted for Sen. Ted Cruz. Due to risk-aversion, there is a tendency to vote for candidates who may win and avoid those who may lose. This causes initial polling measurements of elections to be skewed towards the more well-known candidates, since deciding upon the newer candidates early-on appears as a losing-strategy until they have gained traction. Yet, this presents a problem of risk-taking ‘first-movers’ in the transition of a new candidate to the front-line. Such explains only part of the low-polling for Beto, since Pete Buttigieg is also effected by the same time-lag for newcomers. When a candidate introduces a change to the status quo, we would expect a similar behavioral risk-aversion and resultant time-lag while the social norm is in the process of changing. While Pete’s gay-rights policy is already the norm for the Democratic Party, Beto’s Immigration-Asylum policy is not, given Obama’s record of a high-number of deportations, and thus we would expect Beto’s polling numbers to grow more slowly at first than Pete’s. Complex information to support this hypothesis is available by comparing the differential polling between the General Election and the Primary Election – Beto was found to be the Democratic Candidate most likely to win against President Trump, yet out of the top 7 primary candidates, he is the least likely to be picked for the primary, even though most Democrats rank ‘winning against Donald Trump’ as their top priority. This inconsistency is explained through the irrationality of vote preferences as only partially order-able (i.e. not-strict) thus far. Within the Primary race, people who may support Beto’s policies will not yet choose him as their candidate because of the newcomer and status-quo time-lag biases, although they believe he may be most likely to win over the long-run of the general election. In the General Election, Beto is the 2nd-place candidate across both parties under a rule of

## The Bias of Underreporting in Experimental Design: Ranked-Choice Voting & The Case of the Latino Vote

The ‘hidden 2nd-place bias’ from my previous blog post <a href=“blog 1”> is due to a bias in the question-asking of political candidate polling, neglecting a necessary ambiguity in the answer when it can’t be precisely known.  This ambiguity is essential to the social sciences since an ethical social life is generally defined by a plurality – to attempt to erase this plurality by reductionist measurement will lead to unethical states of affairs by cutting out the voices and perspectives of those marginalized within the plural sociality.  Yet, ambiguity goes against our precepts of science as a deterministic process that yields an exact answer, although quantum physics has revealed an indeterminacy at base of physical processes.  Thus, social processes must be conceived through bio-physics as having emergent complexity that creates a chaotic indeterminacy although with stable supra-structures, and such is a condition of freedom in society and for societies.  While previously we addressed the bias in 1-dimensional polling questions – those that require a strict ordering of preferences – here we address the bias of underreporting by marginalized groups.  Both effect the result in the same way and indeed stem from the same norm-reducing force – marginalized preferences are suppressed while the norm is exaggerated.

The purpose of ranked-choice voting is to allow the marginalized vote to be seen in the aggregate, rather than be reduced by the normalizing force of single-choice voting (‘dominant group conformity’).  The more complex the available choice of the vote, the more complexity will be preserved about the actual beliefs of the population in all its variation.  Thus, latino voices are heard more clearly in ranked-choice than single-choice.  The ranked-choice voting is where there is a strict ordering of candidate preferences.  As we demonstrated previously, the (minority) latino preference for Beto is more clearly visible when he is allowed as a second choice within a ranked-choice voting system.  Still, more complexity can be allowed is we use partially ordered choice sets for voting, as represented in a tree diagram, where instead of A<<B<<C<<D, we have A<{B,C},B<D,C>D, indicating that B and C are indistinguishable on the same level, although both are less than A and greater than D, i.e. A<<{B,C}<<D.

A partially ordered vote is more inclusive than a strict ordering (ranked-choice) vote, which is more inclusive than a single vote.  For now, it is best to conceive of strict-ordering as a more democratic voting system, while partial-ordering as an improved sampling method since it would perhaps be unrealistic as a voting rule, although still useful for administrative decision-making.  Yet, when we consider quantum computer in the next blog, we will consider a democratic polity where everyone can submit quantum-votes in real-time and thus recover the concept of partial-ordering as the foundation for a democratic voting system.  The concept of measuring “quanta” as the infinitesimally small quantities is the same as measuring the marginalized peoples whose rights are minimal, i.e. quantum entities.  The thermodynamic principle at work is that biophysical diversity entails a dynamic nature to a system – the more entropy allowed in a system, and therefore representational state ambiguity, will allow for greater diversity to be measured.

It should be no surprise that Republicans are against ranked-choice voting systems since much of their policy relies upon conformity to a shrinking notion of nationalism, that only includes latinos if they forsake their race.  Thus, republican latinos vote to exclude marginalized populations, such as those of the same race/ethnicity, because they believe such will benefit their particular interest even if it is against the interest of their more general group of belonging.  In fact, as marginalized groups, latinos are systematically threatened for voting, and especially if they vote in the interest of their group/race/class rather than the interest of the dominant/oppressor class.  This explains why latino votes are either i) not-cast due to voter intimidation by deportation or questioning of citizen-status, as Trump has exemplified; ii) divided across districts such they cannot sum to any significant race/class voting-block due to gerrymandering; or iii) vote against their interest through a form of internalized racism against oneself.  These methods of electoral system manipulation nullify or invert the vote of the most marginalized population in the USA, latinx.  While it is true that racism in the US would be undermined finally by a ranked-choice or partially-ranked-choice voting system, it will always have pushback from Republicanism that seeks the reductionist unity of population information over the plural complex diversity.

## Navier-Stokes is hard?

Navier-Stokes is hard?
Author: James Lowman

I was recently asked the question; “How can the Navier-Stokes equations both describe our observable world and not be known to always have solutions in 3D?

For those that don’t know, the Navier-Stokes (NS) equations represent a mathematical model that describes fluid. It is derived from quantities in physics that are conserved. Those are:

• Conservation of mass
• Conservation of momentum
• Conservation of energy

The Navier-Stokes equations are yet to be considered solved. There is a prize association called “The Millennium Prize Problems” which offers one million dollars to anyone that can solve one of the seven problems they have listed, and the existence and smoothness of a solution to the Navier-Stokes equations is on that list.

So the question I was asked seems completely reasonable. How can these equations describe physical phenomena with knowledge of a solution? The answer touches on an interesting, and modern, intersection in science.

We use a number of assumption to simplify a NS system in order to approximate a solution. These approximations are usually handled by a computational fluid dynamics software package. They are high powered suites that allow for customized physics simulations. The assumptions most often used to simplify the NS equations are things like the fluids being in-compressible, the turbulence affects are averaged, or that the fluid is Newtonian. Suffice to say, even with a high powered computational engine, more often then not a scaled down and simplified version of the NS equations are all that we can easily approximate.

So if the solutions to the fluid problems are all approximate, how can we know if they are describing real physics? To compare a NS system solution to an actual physical fluid flow, we would need to perform an experiment and collect data on such a flow. This results in flawed data, because the act of introducing measurement tools into fluid domains involves disrupting the fluid. The best measurements of fluid dynamics themselves are approximations.

So what we have ended up with is approximate mathematical approximations of fluids approximating approximate experiments. Its not hard to see why the student’s question needed clarification.

The answer is: Because it works.

That’s a pretty hand-waved explanation, that no student should accept. But for anyone interested, the derivation of Navier-Stokes is a fun derivation that follows from those conservation laws that were listed, and experiments can be found in open source publications that show how much effort was placed on collecting data with minimal interference. The interested party might be able discern that experimental data matches model solutions (reasonably well), and we can start to believe that Navier-Stokes delivers on the promise of predicting real world physics.

Author: James Lowman

Time and time again,  while tutoring students,  I encounter a resistance to using  an  unfailing  tool  called  the  quadratic  equation.   This  simple  algebraic mathematical statement allows a student to find the roots of any clumsy second order polynomial with ease. I can only assume that, while the quadratic equation is drilled into memory, its useful nature is under-reported in high school mathematics.

For those who aren’t familiar:

$ax^2 + bx + c = 0$                (1)

$x=\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$                    (2)

If you can get an equation into the form of equation??, then you can use the Coefficients a, b, and c in order to solve for x in equation??. The ± in equation?? implies that there are two solutions for x (this being the result of solving a second order polynomial equation).  These two solutions represent the roots of equation??. The roots are the values of x that allow equation?? to equal 0 on the left hand side. Roots  are  of  interest  to  us,  because  they  are  a  quick  and easy  way  to determine factors of a quadratic (not to mention all the mathematical reasons to identify when a function is equal to zero).  Students are drilled on factoring. I rarely see a student that is unable to factor a simple quadratic, but I often encounter  those  that  hav problems  when  the  factors  fail  to  be  immediately obvious.  The quadratic equation is always the answer. I often despair that education relies entirely on problem singularity, a belief that there is only one correct way to approach a question.  When I see a student struggle  for minutes  to  try  and  factor  a  difficult  quadratic,  I  can’t  help  but wonder why they shy away from the quadratic equation.  Sometimes it is hyper- focus that keeps them from finding an alternate path forward,  but more and more I encounter fear.  Factoring is supposed to be an easy shortcut, while the quadratic equation has the confusing ± and the scary √ .  But a shortcut fails to be short when it takes time to muddle through possible variations.  Take, for example, the following quadratic:

$10x^2 + 13x – 30 = 0$        (3)

Some  people  may  be  able  to  see  immediately  that  this  equation  can  be factored.  Others  might have enough experience with factoring to only fumble through one or two permutations before generating the factored result quickly. The  rest  of  us,  myself  included,  might  struggle  with  5  or  more  permutations before coming close to a factored result.  Yet the quadratic yields the answer in 4 quick lines:

$x=\frac{-13 \pm \sqrt{169 + 1200}}{20}$              (4)

$x=\frac{-13 \pm \sqrt{1369}}{20}$                     (5)

$x=\frac{-13 \pm 37}{20}$                           (6)

so $x= \frac{6}{5}$ and $x=\frac{-5}{2}$         (7)

[mathjax]

# 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.

$$2x^2 + 12x – 5 = 2[x^2 + 6x] – 5$$

Â

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

Â

Â

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

Â

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\\
\end{align*}

Â

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

Â

\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}.

Â

\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

Â

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

Â

\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\\
\end{align*}

Â

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 $$(x + B)^2$$.

Â

\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

Â

Step 3: Adjusting the Final Form\vskip 2mm

Â

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

Â

\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

Â

\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 Â \textbf{Answer:} $$3x^2 – 4x + 8 = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3$$ \vskip 1.5cm Â \textbf{\large The Quick Method}\vskip 2mm Â 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: Â \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} Â 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 Â 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 Â \textbf{Example 5: Simple Quick Method}\vskip 2mm Â Use the quick version of Completing the Square on $$x^2 + 14x + 40$$. Â \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 Â \textbf{Answer:} $$x^2 + 14x + 40 = (x + 7)^2 – 9$$ Â Note: This is the same answer we got in Example 1.\vskip 1.5cm Â \textbf{Example 6: Quick Method with Leading Coefficient}\vskip 2mm Â 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 Â Let’s complete the square on $$5x^2 + 80x + 13$$. Â \noindent\textbf{Solution}\vskip 2mm Â Step 1) Factor out the leading coefficient. Â \[% 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 Â \textbf{Answer:} $$5x^2 + 80x + 13 = 5(x + 8)^2 – 307$$ Â \vskip 1.5cm Â \textbf{\large Solving Equations}\vskip 2mm Â 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 Â But that doesn’t mean we shouldn’t know \textit{how} to solve more complicated quadratic equations with this technique.\vskip 5mm Â \textbf{Two Approaches}\vskip 2mm Â 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 Â \textbf{Example 7: Solving a Quadratic Equation by Completing the Square}\vskip 2mm Â Use Completing the Square to solve $$4x^2 + 3x – 10 = 0$$.\vskip 5mm Â \textbf{Solution (Method 1)}\vskip 2mm Â For this approach, we’ll first complete the square, then solve the equation.\vskip 2mm Â Step 1) Complete the Square on $$4x^2 + 3x – 10$$.\vskip 2mm Â Using the techniques we’ve discussed above, we get the following. Â \[% 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*} Â The two solutions are Â \[% \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

Â

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

Â

\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*} Â 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 Â \textbf{Answer:} $$x = \frac 5 4$$ or $$x = -2$$ \vskip 1.5cm Â \textbf{Conclusion}\vskip 2mm Â 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. 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.

## 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?”

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