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Quadratic Roots
A general quadratic equation has the form
where . The quadratic formula gives the roots of this equation as
Hence the two roots, and , can be designated as
and
By averaging them and recognizing that the parabola is symmetric we can easily see that its vertex is at
Furthermore we can show that
and
We could also get this by recognizing that the quadratic can be factored over the complex numbers as
Multiplying this out we get
then equating this with the original equation above.
Example: Find a quadratic with integer coefficients with the specified roots.
i.) and .
ii.) and
Solutions:
i.) We could create factors and then multiply these out but its somewhat easier to use the sum and product rules above.
If and the
and
So the quadratic
has those roots. Notice we are choosing
since its easiest.
ii.) Letting
and we get
and
So the quadratic
has the desired roots. However, its coefficients are not integers, so we can choose
and write
to get a quadratic with integer coefficients.
The Discriminant
We can get more out of the quadratic formula. We define the discriminant to be
This is the part under the square root. If we assume that the coefficients of the quadratic are real, then
If we get two distinct real roots.
If we get a single repeated real root.
If we get two non-real complex conjugate roots.
Blue Question: How do the above results change if we don't assume that the coefficients are all real?
The discriminant can be a useful tool.
Example: Find the equations of a line which passes through the point
and touches the graph of only once.
The line we seek looks something like the picture below. We're looking for a line tangent to the graph.
We don't know know slope of the line, but we do know it passes through the point so we can write its equation as
and substitute this into the equation of the parabola.
Now get all the terms on one side and write in standard quadratic form:
(*)
We want to find
so that this equation has a single solution. That means its discriminant is 0.
So
or
There are two lines that fit the description:
and
shown at left.
Notice how the discriminant focused on only what was needed.
We didn't need the points of intersection. We just need to ensure
that the equation (*) had a single solution for some
.
Interesting websites about osculating circles:
http://poncelet.math.nthu.edu.tw/d2/curves/bow-osc.html
https://php.radford.edu/~ejmt/Resources/CurveSimulator/CurveSimulator.html
There is a Mathematica notebook posted below which will let you look at the osculating circles of a function of your choice.
STEP 2001, Math III, #3: Consider the equation
,
where
and are real numbers.
(i) Show that the roots of the equation are real and positive if and only if
and , and sketch the region of the plane in which these conditions hold.
(ii) Sketch the region of the
plane in which the roots of the equation are real and less than 1 in magnitude.
A solution to this problem is posted below.
Purple Question: (no calculator) Find the sum of all real solutions of
STEP 2005, Math I, #3: In this question
and are distinct, non-zero real numbers, and is a real number.
(i) Show that , if
and are either both positive or both negative, then the equation
has two distinct real solutions.
(ii) Show that the equation
has exactly one real solution if
Show that this condition can be written
and deduce that it can only hold if .
Completing the square
Completing the square is a technique taught in Algebra 2 classes to enable students to either graph quadratics or to prove
the Quadratic Formula. There are many other uses for this technique. Here is an example.
Example: Prove that
.
Solution: As always, there are several ways to approach this problem. The approach we will use here is to rewrite the
expression on the left by completing the square---twice.
We introduce some terms and rewrite:
The last statement is easy to see since we are adding two squared quantities to 2, the minimum value of the expression
on the left is 2 and hence is greater than 0.
STEP 2004, Math II, #2: Prove that, if , then there is no value of for which
(*)
Find the solution set of (*) for
.
For , the sum of the lengths of the intervals in which satisfies (*) is denoted S. Find S in terms of
and deduce that S <
.
Sketch the graph of S against
.
Solution of this problem is posted below.
STEP 2000, Math III, #4: The function is defined by
Prove algebraically that the line
always intersects the curve at least once if , but that there are values
of
for which there are no points of intersection if .
Find the equation of the oblique asymptote of the curve
. Sketch the graph in the two cases (i) ; and
(ii)
. (You need not calculate the turning points.)
The solution to STEP 2000, Math III, #4 is posted at the bottom of this page.
Also at the bottom of this page you will find a Mathematica notebook to accompany the problem.
STEP 1999, Math II, #2: Consider the quadratic equation
(*)
where
, and
(i) For the case where
, , , , find the set of values of for which equation (*) has no real roots.
(ii) Prove that if
and , then (*) has no real roots for any value of .
(iii) If
, and , show that (*) has real roots if and only if or .
A solution to this problem is posted below.
STEP 2002, Math III, #5: Give a condition that must be satisfied by
, , and for it to be possible to write the
quadratic polynomial
in the form , for some .
Obtain an equation, which you need not simplify, that must be satisfied by
if it is possible to write
in the form , for some and .
Hence or otherwise write as a product of two quadratic factors.
A solution to this problem is posted below.
Quadratics in Two Variables
Quadratics in two variables are equations of the form
These equations have interesting geometric representations on the xy-plane.
Our emphasis here will be to factor them.
Green Question: Solve the inequality:
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