# Equation Solving IV

A+S 101-003: High school mathematics from a more advanced
point of view

1. We’ll proceed with quadratic equations today. We’ll come back to
stuff
after

quadratics.

2. We need a volunteer to take notes and send them to the email list as an
attachment.

3. You received Amy Curl’s wonderful notes on last week’s discussion. (If you
need a

hard copy, ask Dr S.) Here are a couple of things Amy brought to light:

• Multiplying both sides of an equation by an expression involving a variable
can

result in a change of the domain of the new equation. (Think of multiplying

by , as in the example Brad put forth in
Seminar 3; 0 is lost to the domain

of the resulting equation. If 0 is a solution to the original equation, we might

a solution.) What to do in real class time about this? Maybe. put in

parentheses next to the equation that the domain has been messed around

with and when giving the solution set there will have to be accounting for

that.

• Students aren’t always clear about the meaning of “solution” to an equation.

Dr. Jones’ example was interesting (and students do this stuff all the time) :

To solve x ^{2} −3x+2 = 0, students isolate the variable (a great tactic in linear

equations) and solve for x with
. Is that wrong? Hmm.

In the classroom when solving an equation, it will help to provide students

with a clear idea of your expectations.

4. A problem presented last time (but slightly gussied up). Here it is again:

**Problem **Let E be an equation. Which of the following operations to E
results

in an equivalent equation ? If one uses the operation under consideration, what

notations/comments could be made next to the resulting equation?

(a) Adding the same constant to both sides of the equation.

(b) Adding the same algebraic expression to both sides, assuming that involves

only the variables in that algebraic expression are in E also .

(c) Adding the same algebraic expression to both sides of E, assuming the
algebraic

expression contains variables not in E.

(d) Squaring both sides of E.

(e) Cubing both sides of E.

(f) Taking the square root of both sides of the equation.

(g) Multiplying both sides of E by the same constant.

(h) Multiplying both sides of E by the same algebraic expression, assuming the

algebraic expression involves only variables contained in E.

Quadratic equations in the real numbers

The algorithm for solving a linear equation with one variable calls for
isolating the variable.

Last time we briefly discussed the appropriateness of isolating variables in the
quadratic

equation case. We saw that it might not lead to a (numerical) solution to a
quadratic but

that students could be expected to do stuff like

x^{2} = −2x + 5.

x^{2} + 2x = 5 (isolate the variable)

x(x + 2) = 5 ( factor out the variable )

Voila!

Of course there’s not all that much wrong with the above procedure. The only
problem:

we expected to have a real- number solution (s).

**Note. We should** expect to see such procedures/solutions from high school
students.

As Dr. J pointed out during the first seminar, we ask high school students to
solve

x + 2y = 7 for y in terms of x, so a solution that involves x isn’t unheard of.
Of course

here with the equation x^{2} = −2x + 5 we expect real number solutions-it’s
important to

clarify what is meant by “solution to the equation(s)”.

Okay, so we assume that the student knows what we mean by “solution set” to the

given quadratic equation. Obviously, isolating the variable is not the cleanest
way to solve.

So how do we find the numeric solutions to a quadratic?

So, let’s take up a simpler quadratic equation:

x^{2} = 4

This one “talks to us”. It says to me “ a number squared is 4”. Arithmetic facts
lead

me to the two solutions {−4, 4}. No prob. Should the “square root word” be
invoked here?

There’s some confusion about the meaning of “square root” .these guys can be
negative,

versus the square root function.a function with range the non- negative reals .
Let’s leave

that for another time.

How about

(x − 1)^{2} = 4?

Or,

(2x − 3)^{2} = 4.

So how can be solve

x^{2} = −2x + 5?

**Real-world problem** You’re designing a small rectangular city garden. The length

is one less than twice the width. What dimensions should the garden be built so
that its

area is 6 square yards? I leave you with the following
questions: Could you do a clean,

student-friendly derivation of the quadratic formula? In your most honest
moment, do

you think presenting such a derivation, or having students discover it with you,
might be

of interest to you and to your students?

It might do some good to bring home to our students that ingenuity was required
to solve

quadratic equations. Some human(s), somewhere, quite cleverly finessed the
“isolating the

variable doesn’t work” obstruction. Do our students just want to see the
“formula” and

get it over with? Might they be able to appreciate that formula if it was
presented as a

pretty decent accomplishment? Or let them discover themselves how to do the
“finessing”

?

Question. Do you think it would be helpful/useful to derive the quadratic
formula,

perhaps jointly with your students? Have you ever tried yourself to derive it
and/or

thought about its derivation and its validity?

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