Tag Archives: equation

Mathematical discrimination in the brave new world of Common Core

Merriam Webster
discriminating
adjective dis·crim·i·nat·ing

: liking only things that are of good quality

: able to recognize the difference between things that are of good quality and those that are not

Full Definition of DISCRIMINATING

1: making a distinction : distinguishing
2: marked by discrimination:
a : discerning, judicious
b : discriminatory

Take your pick !

In the UK we used to call it prejudice, as in “racial prejudice”.

Today the perfectly good word “discrimination” has been hijacked and the old meaning has all but disappeared.

Now you ask “What’s this got to do with math?”.

We have to look at quadratic equations for an answer.

The standard quadratic equation is ax2 + bx + c = 0
It may or may not have real roots
and the sign of D = b2 – 4ac resolves this uncertainty.

D < 0 … no real roots, D = 0 … equal real roots D > 0 … different real roots

What is this D ? It is the “discriminant” of the equation.

It is used to discriminate between equations with real roots and equations without real roots.

So WHY isn’t it in the Common Core math standards. It’s about as standard a thing as is possible?

Clearly the CCSSM authors are guilty of serious discrimination in discriminating against the word “discriminant”, in order not to be accused of discrimination language. Neither “discrimination” nor “discriminant” appear in the CCSSM doc, and the result is that the poor kids ( in the little darlings sense) have to complete the square from scratch every time they have a quadratic equation ( instead of once only in order to get the discriminant formula).

So sad !

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What is Algebra really for ?

An example tells a good tale.

Translation of a line in an x-y coordinate system:

Take a line  y = 2x -3, and translate it by 4 up and 5 to the right.

Simple approach : The point P = (2, 1) is on the line (so are some others!). Let us translate the point to get Q = (2+5, 1+4), which is Q = (7, 5), and find the line through Q parallel to the original line.  The only thing that changes is the c value, so the new equation is  y = 2x + c, and it must pass through Q.  So we require  5 = 14 + c, giving the value of c as -9.

Not much algebra there, but a horrible question remains – “What happened to all the other points on the line ?”

We try a more algebraic approach – with any old line  ax + by + c = 0, and any old translation, q up and p to the right.

First thing is to find a point on the line – “What ? We don’t know ANYTHING about the line.”

This is where algebra comes to the rescue. Let us suppose (state) that a point P = (d, e) IS on the line.

Then ad + be + c = 0

Now we can move the point P to Q = (d + p, e + q)  (as with the numbers earlier), and make the new line pass through this point:  This requires a new constant c (call it newc) and we then have  a(d + p) + b(e + q) + ‘newc’ = 0

Expand the parentheses (UK brackets, and it’s shorter) to get  ad + ap + be + bq + ‘newc’ = 0

Some inspired rearranging gives  ‘newc’ = -ap – bq – ad – be, which is equal to -(ap + bq) – (ad + be)

“Why did you do that last step ?” – “Because I looked back a few lines and figured that  (ad + be) = -c, which not only simplifies the expression, it also disposes of the unspecified point  P.

End result is:  Translated line equation is  ax + by + ‘newc’ =0,  that is,  ax + by + c – (ap + bq) = 0

and the job is done for ALL lines, even the vertical ones, and ALL translations. Also we can be sure that we know what has happened to ALL the points on the line.

I am not going to check this with the numerical example, you are !

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The angle bisector theorem and a locus

The theorem states that if we bisect an angle of a triangle then the two parts of the side cut by the bisector are in the same ratio as the two other sides of the triangle. So I thought, what if we move the point, G in the gif below, so that GB always bisects the angle AGC, then where does it go ? Or, more mathematically, what is its locus. Geometry failed me at this point so I did a coordinate geometry version of the ratios and found that the locus was a circle ( a little surprised !). The algebra is fairly simple.

gif bisector locus

What was a real surprise was that if we take the origin of measurements to be the other point where the circle cross the line ABC, then point C is at the harmonic mean of points A and B. There is YOUR problem from this post.

Getting the construction to show all this was tricky. It is shown below.

locus angle bisector point full

I did the construction on my geometry program, Geostruct (see my Software page).

Line bb passes through A. Point D is on line bb. Point E is on this line and also on the line through C parallel to line cc joining B and D, So AD and DE are in the same ratio as AC and CB. When point D is moved the ratios are unchanged (feature of the program). So the point of interest is the intersection of a circle centre A, radius AD, and a circle centre C and radius DE (= FC)

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“Observe and make use of structure”. Observe would be a start. A tale from the chalkface.

Here’s my little story:
It was a class of day release students on a Higher National Certificate course in engineering. I reached a point in one class with a relationship between p and q, p = kq, with k a constant. “What’s its graph look like”, I asked. Deathly silence. “Ok, let’s try x and y”. Result y = kx. Same question, same response. “Well, what about y = 3x ?”. Same question, same response. So I wrote y = 3x + 2. Their eyes lit up, and they unanimously shouted “It’s a straight line!”.

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Subtraction in algebra – let’s use algebra !

I have seen some heavy handed ways of explaining the identity

a – (b + c) = a – b – c

Let us use algebra. Give the left hand side a name, say d .Then

a – (b + c) = d

This is an equation, so add (b+c) to each side and get

a = d + (b + c), then a = d + b + c as the parentheses are now superfluous.

Now subtract  b  from each side

a – b = d + c

Now subtract  c  from each side

a – b – c = d

so  a – (b + c) =  a – b – c

or is this too simple ? Look, no messing with p – q = p + -(q) stuff,

and no appeal to the famous distributive law.

You can do this, and other stuff, with numbers as well.

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Read this : Negative numbers, by A. N. Whitehead

Alfred North Whitehead, professor of mathematics and philosophy, and famous for his collaboration with Bertrand Russell on their joint effort, the Principia Mathematica, also wrote a book, “Introduction to Mathematics”, in 1911, for High School students and others who really wanted to know what math was all about.
The section on negative numbers is so relevant to the teaching of that topic today that it is a MUST READ. Click the link to download this section.

Alfred North Whitehead: Introduction to Mathematics

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