Some of the questions below are not the regular type of questions you will see in your assignments and tests. Don't worry if you haven't come across questions of this nature before; there's always a first time for everything. They are designed to make you get used to questions that require proofs, which is a useful skill to acquire, especially if you are going to do math beyond high school. We hope to add more questions, as the opportunity arises.
In real life, an example is better than precepts;
In mathematics, a proof is better than examples.
abstractproofs), they often give examples instead. This is normal, being their first introduction to proof-related problems.
frustrationwith this experience, together with our teeming fascination with the high school, are part of what motivated these proof-type exercises.
For a quadratic in the usual standard form $ax^2+bx+c$, the discriminant is the familiar quantity $$b^2-4ac$$ that appears in the quadratic formula. It is used to predict the nature of the roots without actually computing the roots.
Observe that in order to have $b^2-2ac< 0$, $a$ and $c$ must have the same signs to begin with.
This exercise can be used
as an alternative test for complex roots in a quadratic equation, deploying $b^2-2ac$
instead of the traditional $b^2-4ac$.
However, being (that $b^2-2ac< 0\implies b^2-4ac< 0$ is) a one-sided implication,
this approach has some limitations.
For example, it can happen that $b^2-2ac>0$ while $b^2-4ac< 0$ (check this for the quadratic $2x^2+4x+3$), so it is not entirely reliable.
As will be shown later, where the use of $b^2-2ac$ is more appropriate is when the degree is beyond $2$; that is, from cubics onwards.
In those cases, the computation of the actual discriminant comes with increasing complexity, whereas $b^2-2ac$ is easy to calculate. So, if by
chance $b^2-2ac< 0$, then we can infer that the polynomial contains complex roots, and the need to compute the actual discriminant may be
obviated.
Finally,
since the implications $$b^2-ac< 0~\implies b^2-4ac< 0 \quad\textrm{and}\quad b^2-3ac< 0~\implies b^2-4ac< 0$$ are also true,
we will show in the next
exercise why we have settled for $$b^2-2ac< 0~\implies b^2-4ac< 0,$$ instead of the other two.
What happens if $b^2-2ac=0$? We have mixed results, with all but one indicating complex roots. So, let's isolate this one case: it occurs
when the two roots of the quadratic are zero, which only happens for the parent quadratic (equation $x^2=0$).
With that said, the contrapositive of the above statement is: if $b^2-2ac \leq 0$, then not all the roots are real
. (Bear
in mind what was said about $b^2-2ac=0$.)
Since a quadratic has only two roots and complex roots occur in conjugate pairs, $b^2-2ac< 0$ for a quadratic means that the quadratic has
only complex roots.
This was why we chose $b^2-2ac< 0$ above.
The converse is not true in both cases.
One sees that the behaviour of the discriminant -- and the
quadratic itself -- is, to some extent, influenced by $b$. Take $b$ for behaviour.
In particular, a perfect square trinomial cannot have a missing constant term; otherwise, it will degenerate to the form $y=ax^2$, which doesn't directly fit into the trinomial definition.
When put in the form $ax^2+bx+c$, the quadratics that go through the origin are precisely the ones with a root $-\frac{b}{a}$ (and of course $0$ as a second root).
For a perfect square trinomial, this reduces to $br^2+2cr=0$ or $bs^2+2cs=0$, since $r=s$.
So we have $r(br+2c)=0$, whence $r=0$ or $r=\frac{-2c}{b}$.
Recall that the $x$-coordinate of the vertex of any quadratic $ax^2+bx+c$
is given by $x=\frac{-b}{2a}$. For a perfect square trinomial, the roots sit on the vertex, so $$\frac{-b}{2a}=\frac{-2c}{b}. $$
Re-arranging, we obtain $b^2=4ac$, the usual condition for a perfect square trinomial.
Actually, the first result applies to all quadratics, not only the ones in factored form.
So, here is the general statement,
which we invite you to VERIFY: for any quadratic $ax^2+bx+c$ with roots $r,s$ (rational, irrational, or complex),
the discriminant $\Delta$ can be given in terms of the roots as $\Delta=a^2(r-s)^2$.
Beware that this discriminant
expression -- though appearing like a perfect square -- might in fact be negative, when the roots $r,s$ are
complex numbers. From this it can be seen that the discriminant can always be dressed up like a perfect square, even when it is not. Don't be carried away by the external appearance of things; it is their internal
constitution that matters most.
Thus, the discriminant can also be expressed in terms of just one of the roots.
Except for perfect square
trinomials, the above statement is biconditional
.
Never underrate quadratics in which $a=-\frac{1}{4}$; they are quite special.
Consequently, quadratics of the
above form can be factored -- as $(x+a)(a^{n-1}x+1)$.
The case $a=1$ yields a perfect square trinomial, the
only perfect square trinomial in this quadratic family, when $a$ is a real number.
An interesting case occurs when $a=2$, as the resulting discriminant is the square of a
Mersenne number.
The requirement that $a> 1$ is just
to exclude the zero discriminant. Also, limiting $a$ to positive integer values is merely for convenience.
A proof that uses the principle of mathematical induction will be needed. The first value of $k$ to try, based
on the conditions of the question, is $k=3$.
The associated discriminant in this case is $(a^6-1)^{2}=(a-1)^2(a+1)^2(a^2+a+1)^2(a^2-a+1)^2$.
Again, one uses the principle of mathematical induction in this exercise.
Again, one uses the principle of mathematical induction in this exercise.
Equivalently, the original quadratic being a perfect square trinomial implies that the new one is. This way, if the factorization of the first quadratic is $(px+q)^2$, then the second quadratic factors as $[pq(x+1)]^2$.
Again, one sees that if any of the two quadratics has equal roots, then so does the other. And if the former factors as $a(x-r)(x-s)$, then the latter factors as $ac(x+1+\frac{b}{c}r)(x+1+\frac{b}{c}s)$.
Thus, any solution to $acx^2+(b^2-2ac)x+ac=0$ is a ratio of the solutions to $ax^2+bx+c=0$.
Since every $x$ coordinate moves by the same unit under a horizontal translation, the above exercise can be
applied to the roots of a quadratic equation in particular. Thus, if the roots of two quadratic equations differ by the same amount, then
their discriminants are equal.
To see this, assume that the quadratic equation $ax^2+bx+c=0$ has roots
$\alpha$ and $\beta$. VERIFY that
the quadratic equation whose roots are $\alpha+k$ and $\beta +k$, for some
constant $k$, is $$ax^2+(b-2ak)x+(c-bk+ak^2)=0,$$ and that the discriminant of this new equation is the same as the old one, both $b^2-4ac$.
The case when $k=\frac{b}{a}$ is worth a separate mention, as it leads to a different transformation.
The above exercise may be taken as a characterization of horizontal reflection of quadratic functions.
The result assumes, implicitly, that the given quadratics have real roots. If the roots are complex numbers, it
still holds.
The condition that $A$ has the same sign as $a$ is to remove the possibility of the second quadratic being a
reflection of the first in both
the $y$ and the $x$ axes.
Beautiful, beautiful thing.
Notice that this is a partial converse to Question 1 above.
In general, if two quadratics have the same
leading coefficient, then we need both a horizontal translation and a vertical translation to obtain one from the other. With
equal discriminants, however, the need for a vertical translation is removed.
Thus, in terms of transformations, we can
distinguish a special family of quadratics: the ones having the same discriminants and the same leading coefficients.
If we take the discriminant as an operation $D$ on the set of quadratic functions, then $D$ enjoys $D(f(rx))=D(rf(x))=r^2\times D(f(x))$, where $f(x)=ax^2+bx+c$ and $r\neq 0$
Translating just the roots of a quadratic by a certain amount
is equivalent to a horizontal translation of the entire quadratic, by the same amount.
Well, this should be expected, because a quadratic is largely dependent on its roots.
Scaling just the roots of a quadratic by a certain factor is equivalent to a horizontal stretch or compression of the entire quadratic, by the same factor.
Exponentiation
of the roots does, however, have effect on the leading term.
Observe that $k$ doesn't have to be an integer in the above exercise; in that case, the conclusion has to be modified
slightly, as follows: linear operations on the roots of a quadratic have a corresponding linear effect
on the coefficients.
Here we investigate how the discriminant of a quadratic ($ax^2+bx+c$) behaves if the coefficients $a,b,c$ form an arithmetic or
a geometric progression. Then we investigate other number patterns.
This is purely for the interest of theory; you'll rarely see such numbers
in actual numerical problems involving quadratics. What to note here is that as the coefficients grow
, the discriminant tends to be negative.
Thus, if the coefficients in a quadratic equation form a geometric progression, the quadratic will not have real
solutions.
For example, the quadratic equation $x^2+3x+9=0$ does not have real solutions because the coefficients $1,3,9$ are in a
geometric progression; in fact, our one-sided discriminant $$b^2-2ac=3^2-2(1)(9)=-9< 0,$$ as well as the usual discriminant
$b^2-4ac=3^2-4(1)(9)=-27< 0$.
Observe that the usual discriminant is a multiple of $3$ in this case. This is ALWAYS true if the
integer coefficients in a quadratic form a geometric sequence.
We will encounter a similar result pertaining the discriminant of a cubic later.
By the way, quadratics
in which the coefficients form a geometric progression have a special cubic connection.
Simply find what the values of $A$ and $B$ should be, using the given information about $a,b,c$. Then carry out the
expansion $(ax^2+bx+c)(Ax+B)$ to obtain a DIFFERENCE OF CUBES or a SUM OF CUBES.
Thus, without recourse to the cubic discriminant,
we understand why cubic equations of the form
$x^3-n^3=0$ and $x^3+n^3=0$ ALWAYS have ONLY ONE real solution: their factorizations contain a quadratic part in which the coefficients form a geometric
progression.
The case in which the coefficients form an arithmetic sequence is more acccommodating than when they form
a geometric sequence, in which case the discriminant is non-negotiably negative.
Here's a little detail on how accommodating the arithmetic case can be.
Suppose the coefficients $a,b,c$ are all positive as well as the common difference $d$, then the discriminant is positive when $d>\frac{\sqrt{3}}{2}b$,
and negative when $d<\frac{\sqrt{3}}{2}b$.
For example, the coefficients in the quadratic $x^2+2x+3$ form an arithmetic sequence with a common difference
of $1$; its discriminant -- $2^2-4(1)(3)=-8$ -- is negative because $d=1< \sqrt{3}=\frac{\sqrt{3}}{2}(2)$, that is $d< \frac{\sqrt{3}}{2}b$.
Alright. Next question please.
Zero discriminant, integer coefficients, and arithmetic sequence cannot co-exist in the same quadratic environment
.
Can we make the above statement biconditional? Not yet, in view of the next exercise.
This result is similar to the preceding one, but we have decided to separate them because the quadratics in question are two different
families; in particular, when $d=-a$, the middle term is missing (because the coefficients become $a,0,-a$),
giving rise to quadratic equations of the form $ax^2-a=0$.
Nevertheless, there is a transformation that connects these two quadratic families.
In the case when $d=7a$, the prototype is $f(x)=x^2+8x+15$, while $g(x)=x^2-1$ is the parent for the case when $d=-a$.
The two quadratics have the same leading coefficient and the same discriminant, so in view of a previous result (see the section
on transformation of quadratics), there is a horizontal transformation that connects these two quadratics.
In fact, let $$h=\frac{8-0}{2}=4,~\textrm{then}~g(x+h)=(x+4)^2-1=x^2+8x+15=f(x).$$
Similarly, $f(x-4)=(x-4)^2+8(x-4)+15=x^2-1=g(x)$.
Observe that we have used $h=\frac{b-B}{2a}$, where $b,B$ are the middle coefficients
and $a$ is their common leading coefficient. This is the amount by which two quadratics with the same discriminant and the same leading coefficient
get horizontally transformed to each other.
By the way, is $d=7a$ and $d=-a$ the only case when the discriminants of the
resulting quadratics are equal? No way.
Let's consider two numerical examples, while noting that of necessity we have $d_{1}\neq d_{2}$ above:
the two sequences already have the same first terms, so they cannot afford to have the same common difference.
Take the quadratic $x^2+3x+5$, in which the coefficients $1,3,5$ form an arithmetic sequence with common difference $d_{1}=2$. Its discriminant is
$3^2-4(1)(5)=9-20=-11$.
In order to construct another quadratic with the same discriminant, we need a $d_{2}$ that satisfies $d_{1}+d_{2}=6$;
that is, $2+d_{2}=6$. We get $d_{2}=4$. The resulting quadratic is $x^2+5x+9$, with accompanying discriminant $5^2-4(1)(9)=25-36=-11$.
Second example. Let's take the quadratic $7x^2-x-9$, in which the coefficients $7,-1,-9$ form an arithmetic sequence with common difference
$d_{1}=-8$. Its discriminant is $(-1)^2-4(7)(-9)=1+252=253$.
Another quadratic with the same discriminant (as $7x^2-x-9$)
is obtained by first constructing
a three-term arithmetic sequence with first term $7$ whose common difference $d_{2}$ satisfies $d_{1}+d_{2}=6(7)$; that is, $-8+d_{2}=42$. Since
$d_{2}=50$, the resulting quadratic is $7x^2+57x+107$, with accompanying discriminant $57^2-4(7)(107)=3249-2996=253$.
Beautiful, useless thing.
Perfect square discriminant, integer coefficients, and arithmetic sequence, are in short supply.
In fact, up to a horizontal translation, there is only one quadratic family in which the coefficients are integers that form an arithmetic sequence
with the quadratic having a perfect square discriminant. It is the family we encountered earlier, namely $a(x^2+8x+15)$
(or its horizontal translation $a(x^2-1)$).
For the next set of questions in this section, assume throughout that the coefficients $a,b,c$ of the quadratic $ax^2+bx+c$ are all positive AND that $a\geq 1$.
The first few perfect squares are $1,4,9,16,\cdots$.
The first few perfect cubes are $1,8,27,64,\cdots$.
The fist few triangular numbers are $1,3,6,10,\cdots$. The $n$th triangular number is $\frac{n(n+1)}{2}$.
Fibonacci numbers are generated via a recurrence relation: $$F_{n}=F_{n-1}+F_{n-2},\quad\quad F_{0}=0,~F_{1}=1.$$ So the first few Fibonacci numbers are $0,1,1,2,3,5,8,\cdots$. For our purposes, we don't want $0$ as the first Fibonacci number, in accordance with our restriction above where we stipulated that $a\geq 1$.
The first few Lucas numbers are $2,1,3,4,7,\cdots$. In general, Lucas numbers are generated via a recurrence relation: $$L_{n}=L_{n-1}+L_{n-2},\quad\quad L_{0}=2,~L_{1}=1.$$
When it comes to quadratics, one of the popular high school questions is to construct a quadratic equation
with given roots.
In this section, we'll modify this tradition slightly, asking you to construct quadratic equations
with given discriminants.
The questions here can be played as a GAME, and so they are suitable for a classroom activity.
Two players (or two groups of students) will be needed. Player A will choose a random integer, and player B will produce a quadratic
whose discriminant is the given integer (or will provide a proof that such a quadratic does not exist). Then player B will choose an integer,
and player A will produce a quadratic whose discriminant is the chosen integer (or will give a proof that such a quadratic is not possible).
In order to give sufficient opportunities to the players, FIVE rounds will be needed. The winner is the player with more points.
If there's a tie, the game can be repeated as the facilitator deems fit.
The players need to be versed
in the underlying theory, which we present in the exercises below.
You may begin with the quadratic equation $x^2+3x+2=0$, whose discriminant is $1$.
Note that the coefficients of
the resulting quadratic need not be integers, but the discriminant has to be.
Note that this is related to the preceding question.
Consider $x^2+(k+2)x+(k+1)=0$.
Having $-1$ as discriminant strips the quadratic of integer coefficients.
Why does this happen?
Recall that the discriminant $\Delta$ is given by $\Delta=b^2-4ac$. If this is to equal $-1$, then $b$ MUST be an ODD number.
Put $b=2n+1$ for some integer $n$. Then
$$b^2-4ac=-1\implies (2n+1)^2-4ac=-1,\quad 4n^2+4n+2=4ac.$$ We can write $4n^2+4n+2$ as $2(2n^2+2n+1)$, where the expression in the bracket is ODD,
so $4n^2+4n+2$ is at most a multiple of $2$. Now return to the equation $4n^2+4n+2=4ac$, and complete the argument that shows that BOTH $a$ and $c$
cannot be integers. Or, you can initiate an entirely different line of argument.
Is $-1$ the only discriminant that strips the quadratic of its INTEGER coefficients? NO.
What's going on here? We'll summarize it in the next two exercises.
Clearly, if the given ODD number is positive, we want $k$ to be a member of the set $\{0,1,2,3,\cdots\}$,
and if it's a negative ODD number, then $k$ belongs to $\{\cdots,-3,-2,-1\}$.
We see why we couldn't get $-1$ as discriminant in a previous exercise: It's NOT possible to write $-1=4k+1$, with $k\in\{\cdots,-3,-2,-1\}$.
This explains why we couldn't get $2$ as discriminant in a previous exercise.
So, it is IMPOSSIBLE to get $\pm 2,\pm 6,\pm 10,\pm 14,\cdots$ as discriminants if ALL the coefficients in a quadratic equation are integers.
Higher powers of $2$ beyond $8$ are also possible. So we can obtain quadratic equations (with INTEGER coefficients) whose discriminants are $16k, 32k, 64k,\cdots$, for any positive integer $k$.
A quadratic in the form $ax^2+bx+c$ can be factored over the rational numbers (which includes the integers) if its discriminant
$b^2-4ac$ is a perfect square.
If all the coefficients $a,b,c$ are provided, then the factorability property can be decided -- by computing the discriminant.
In this section, any two of the coefficients $a,b,c$ will be FIXED (i.e., given) while the remaining one will be FREE. We investigate admissible
values of the free coefficient that will guarantee factorability.
This is a problem on existence as well as enumeration. So we need to construct
and count
.
In the present case, one possible value of $B$ is $a+c$, giving the factorization
$$ax^2+(a+c)x+c=(x+1)(ax+c).$$
What's the second permissible value of $B$?
If $a>1$ and $c> 1$, we can improve upon the above result.
Two such values are $B=a+c$ and $B=ac+1$, with accompanying factorizations $(x+1)(ax+c)$ and $(x+c)(ax+1)$, respectively.
Recall that a prime number has only TWO DIVISORS: $1$ and itself.
A number that is not prime is said to be composite.
Take the quadratic $x^2+Bx+5$, for example. In order to be able to factor this over the integers, there are only two admissible values of $B$, namely $\pm 6$.
Use the previous exercise and the fact that a perfect square has an odd number of divisors.
If $a$ and $c$ have opposite signs, then the maximum among all the allowable values of $B$ is $-ac-1$, and the minimum is $ac+1$.
Let's start with a choice of $C$ that is likely to complete the square
for the quadratic -- this is a popular
high school quadratic question. It is $C=\frac{b^2}{4a}$. We then obtain the factorization
$$ax^2+bx+\frac{b^2}{4a}=ax^2+\Big(\frac{b}{2}+\frac{b}{2}\Big)x+\frac{b^2}{4a}=\Big(ax+\frac{b}{2}\Big)\Big(x+\frac{b}{2a}\Big).$$
That there are an infinite possibilities for $C$ is because there are numerous ways to decompose the middle coefficient $b$ (we used
$b=\frac{b}{2}+\frac{b}{2}$ above, but one can also use $\frac{b}{3}+\frac{2b}{3},~\frac{b}{4}+\frac{3b}{4},~\frac{b}{5}+\frac{4b}{5}, ~2b-b, 3b-2b$,
and so on, obtaining a separate factorization for each decomposition). In general, the decomposition of $b$ can be accomplished via
$b=kb+(1-k)b$, where $k$ is an integer or a rational number.
We can view the present problem as dealing with a product of sums
, while a previous one was about a sum of products
.
One thing to note about such quadratics is that they have a simple factorization character: they either factor as $a(x+2)^2$, or they do not factor at all -- over the integers.
Thus, when the leading coefficient and the constant term of a quadratic equation are interchanged, the roots
become inverted.
This also happens for a linear equation: if $a,b\neq 0$, then $ax+b=0$ and $bx+a=0$ have
reciprocal solutions $\frac{-b}{a}$ and $\frac{-a}{b}$, respectively.
That's just about it, for degrees $1$ and $2$.
In the case of a
cubic or higher degree polynomial, more than one interchange is needed.
Some strategic substitutions that simplify the process of solving quadratic and (some) higher degree polynomial equations will be explored in this section.
The main idea is that some polynomial equations in which certain coefficients are missing may be easier to solve than the ones in which all
coefficients are present.
In order to obtain the appropriate substitutions, one has to work backwards -- take a given polynomial
function $f(x)$, apply a horizontal transformation $f(x-h)$ to it; then, in the transformed polynomial function, equate the unwanted
coefficient to zero and solve for $h$ -- if possible. Now if $x$ in $f(x)$ is replaced by $x-h$, the unwanted term will be displaced.
Basically, what this reduction does is to move
the vertex of the parabola from wherever it was to a point on the $y$-axis.
Observe that the reduced quadratic is easier to solve
than the original one; however, this method of reducing a quadratic to a form where the middle term is missing is not a very efficient means
of solving a quadratic equation, though it works.
Thus, among quadratics, only perfect square trinomials can be reduced to their simplest forms under the above substitution. The vertex becomes translated to the origin, in a single move.
Both the original form and the reduced form have the same discriminant.
In a quadratic, the reduction strategy is always capable of eliminating the second term, but not the constant term.
In general,
the farther, the harder
. If the polynomial is of degree three or more, it becomes progressively difficult to eliminate terms beyond the
second term. For example, to find an appropriate substitution that will eliminate the third term, one will have to solve a quadratic equation; to
find a suitable substitution that will eliminate the fourth term, one will have to solve a cubic equation, and so on.
Beautiful, beautiful thing. The traditional method uses completing the square; the reduction method goes through a different route.
For a little bit of history regarding cubic equations, see
this article.
This reduction strategy is usually called
Cardano's method,
and as the article stated, the original idea came from Scipione del Ferro
and Tartaglia (independently of each other).
It follows that the general solution of such cubics are easier to write down.
This is the general situation.
This generalizes question 7 above.
Both directions are immediate. If such a reduction is possible, then the polynomial has only one root -- namely $0$ -- in the reduced state. In view of the original substitution, the starting polynomial will have one repeated root. Conversely, if the polynomial is a perfect $n$th power, write it in the form $(Ax+C)^{n}$, then find a substitution that makes it become $ax^n$.
As an application of the above reduction procedure, we derive some relationships among the coefficients of polynomials that have repeated factors -- repeated as many times as their degrees. Call them perfect power polynomials.
Recall that the reduced form of a quadratic, under the usual substitution, is $ax^2 + \frac{4ac-b^2}{4a}$. Moreover, in a perfect power polynomial, only the leading term survives after reduction, so we must have $\frac{4ac-b^2}{4a}=0$, whence $b^2=4ac$. The other direction is also easy.
If $a$ is eliminated between $b^2=3ac$ and $bc=9ad$, the result is $c^2=3bd$. Thus, an equivalent requirement is that $b^2=3ac$ and $c^2=3bd$. Stated differently, the sequences $a,~b,~3c$ and $b,~c,~3d$ are both geometric.
The third condition, $6b^2=16ac$, can be reduced to $3b^2=8ac$. We have chosen not to reduce it to its simplest form for a purpose. There will be similar scenario in what follows.
As our manner is, we have chosen not to reduce $10b^2=25ac$. This is to allow the inquisitive reader to notice what is going on.
Observe that the number of conditions to be satisfied among the coefficients is one less than the degree of the polynomial. Observe also that these conditions are independent of each other.
Beautiful, beautiful thing.
Each of these conditions -- expressible in terms of the leading coefficient $a_{n}$ and
the coefficient of the second term $a_{n-1}$ -- is to be satisfied by a perfect power polynomial of degree $n$.
Observe that if $a_{n-1}=0$,
then the above equation implies that all other subsequent coefficients are forced to be zero as well. Thus, in a perfect power polynomial of degree $n$,
no term must be missing -- or else the polynomial reduces to the form $a_{n}x^{n}$. So we know immediately that something like $x^2+9$ is not a
perfect square trinomial because the middle term is missing.
For simplicity, we work with a quadratic equation in the usual standard form:
$$ax^2+bx+c=0,\quad a\neq 0,\quad \textrm{with discriminant}\quad \Delta_{1} = b^2-4ac.$$
We also assume that the quadratic has roots $\alpha$ and $\beta$. We wish to construct new quadratic equations whose roots are
positive integral powers of $\alpha$ and $\beta$,
and see how the discriminants of the new quadratic equations relate to the original discriminant.
Unlike linear operations
on the roots of a quadratic (which leave the leading term untouched),
these exponential operations do affect the leading term, but more
significantly, that middle man, $b$.
One more thing: the exercises in this section rely heavily on factoring and expanding, and you
can be sure that they will stretch your factoring faculty. But do not worry.
Thus, the relationship between the discriminants in this case is $$\Delta_{2}=b^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.
Thus, the relationship between the discriminants in this case is $$\Delta_{3}=(b^2-ac)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.
Thus, the relationship between the discriminants in this case is $$\Delta_{4}=b^2(b^2-2ac)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.
Thus, the relationship between the discriminants in this case is $$\Delta_{5}=(b^4-3acb^2+a^2c^2)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.
Thus, the relationship between the discriminants in this case is $$\Delta_{6}=b^2(b^2-ac)^2(b^2-3ac)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.
Thus, the relationship between the discriminants in this case is $$\Delta_{7}=(b^6-5acb^4+6a^2c^2b^2-a^3c^3)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.
It is evident that things can escalate quite quickly beyond exponent $7$, especially for that middle man, the coefficient of $x$. Despite this, is there a pattern in the successive coefficients of $x$, or in the progressive discriminants? Something is definitely happening.
In words, if the sum
of three non-zero real numbers is zero, then the sum of the products of the numbers, taken two at a time,
cannot be zero.
In the case of two numbers, this is very evident -- both numbers have to be non-zero in order
to satisfy the specified condition,
so the only possible product, namely $\alpha\beta$, is also non-zero.
Here's an example (not a proof!) in the case of three numbers:
let $\alpha=2,\beta=0,\gamma=-2$. One has $$\alpha+\beta+\gamma=2+0+(-2)=0,$$ but $$\alpha\beta+\beta\gamma+\gamma\alpha=
2(0)+0(-2)+(-2)(2)=-4\neq 0.$$
What does the implication $\alpha+\beta+\gamma=0\implies\alpha\beta+\beta\gamma+\gamma\alpha\neq 0$
illustrate in relation to the roots of a cubic equation?
Well, it says that both the sum of roots
and the sum of
products of the roots, taken two at a time
, cannot be zero simultaneously, provided ALL the roots
are real.
What of a cubic equation like $x^3-1=0$, in which both ($\alpha+\beta+\gamma$ and $\alpha\beta+\beta\gamma+\gamma\alpha$) are zero, one might ask.
What's happened here is that not all the roots are
real; in fact, by explicit calculation, the roots are found to be $1,~\frac{-1\pm i\sqrt{3}}{2}$.
Similar to the previous exercise. Aside: if $\alpha\beta+\beta\gamma+\gamma\alpha=0$, then there are just two possibilities: either ALL THREE ($\alpha,\beta,\gamma$) are non-zero or EXACTLY TWO of them are zero (remember that we don't want all three to be zero at the same time).
Part of the underlying reason is that in any choice of the $\alpha_{i}$'s, at least two of them must be non-zero in order to satisfy
the original hypothesis. Then, taking two of them at a time will eventually include the two non-zero member numbers at some point, which guarantees
that at least one of the overall two-products
will be non-zero, since the $\alpha_{i}$'s come from what is known as a field, or, more generally,
an integral domain. Something else guarantees that the overall sum-of-two-products is non-zero.
What's the contrapositive of the above statement? If $b^2-2ac< 0$, then not all the roots of the cubic are real.
There you have it -- a condition for complex roots: $b^2-2ac< 0$. This works for any polynomial, and it saves us the stress of having to compute the discriminant
-- however, there are drawbacks to this approach, so the usual discriminant is still dominant.
For a proof, use appropriate Vieta's formulas and a generalization of the identity $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$.
There's an implicit assumption above, namely $a_{n},a_{n-1},a_{n-2}\neq 0$, that is, all the FIRST THREE terms
are present.
Recall that we've already seen this result for quadratics, where the proof was accomplished by a direct computation of the quadratic
discriminant.
In the more general case, our one-sided discriminant will be do, showing that the presence (or absence) of complex roots is hugely
influenced by the FIRST THREE coefficients.
Again we see how the FIRST THREE highest degree terms affect the nature of the roots of a polynomial equation.
A quadratic example: $x^2+9=0$. Compared with the standard form $ax^2+bx+c=0$, we have $a=1,b=0,c=9$.
The second term is missing, and $a,c$ have the same signs.
Our one-sided discriminant, in this case, is $b^2-2ac=0^2-2(1)(9)=-18 < 0$.
A cubic example: $x^3+2x-4=0$. This cubic contains a pair of complex roots due to the absence of the $x^2$ term and the fact that the $x^3$ term
and the $x$ term have the same signs.
And so on.
The assumption that $d\neq 0$ removes the possiblity of any of the roots $\alpha,\beta,\gamma$ being zero. Thus, we won't run into the problem of having to divide by zero in the reciprocals $\frac{1}{\alpha},~\frac{1}{\beta},~\frac{1}{\gamma}$.
As in the case of a quadratic, translating just the roots of a cubic is equivalent to a horizontal translation of the entire cubic function. Similarly, scaling the roots in the form $\frac{\alpha}{k},~\frac{\beta}{k},~\frac{\gamma}{k}, ~k\neq 0$, is equivalent to the horizontal stretch or compression $f(kx)$.
If we remove the requirement that $k$ is an integer, then the conclusion will be restated: linear operations on the roots of a cubic have linear effect on the leading coefficients.
Observe that when $h=0$, one reobtains the original quartic, as expected.
Isn't this familiar from Taylor series? Almost.
One application of this formula is that we can use it to quickly
write down the polynomial equation that results from translating the roots of the original polynomial $f(x)$ horizontally by $h$ units.
It was assumed implicitly in the above exercise that the given quartics have real roots. The result still holds in the case of complex roots.
This may be taken as a characterization of horizontal reflection of even degree polynomials.
As in a previous exercise,
we want the leading coefficients of both polynomials to have the same signs; this is to prevent the case where one of the polynomials
is reflected in both the $y$ and $x$ axes, with the result still holding.
The notion of a discriminant extends beyond quadratics to cubics and higher degree polynomials, but the computations
become progessively tedious as the degrees increase. For example, the discriminant of the cubic $ax^3+bx^2+cx+d$ is given by
$$\Delta_{3}=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd;$$ it contains five terms, each homogeneous of degree $4$.
The exercises in this section are mostly
a test of factoring and expanding ability. Assume throughout that the real numbers $a,b,c,d$ come from $ax^3+bx^2+cx+d$, and that
$\Delta_{3}=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$.
Such a cubic will have one pair of complex roots. In fact, it factors as $a(x+1)(x^2+1)$.
As we saw before, this also happens in the quadratic case. Does it seem to hold in general?
The discriminant is a perfect square in this case; it can be made zero if we impose an extra condition.
Compare this with the result of the previous question. The discriminant is always negative in this case, so the cubic contains a pair of complex roots -- in fact, it factors as $(ax+b)(x^2+1)$.
The discriminant is a multiple of a perfect square in this case. It can be made zero if $b=0$ or $c=0$ or $b=c$. However, $b=0$ is not allowed, in view of the fact that $a=-b$ from the onset. Why shouldn't $b$ be zero?
By re-writing this discriminant as $(c+d)(c-3d)(c-3d)^2$, we see that it is again a multiple of a perfect square. It can be made zero with additional assumptions.
In this case, the discriminant could be negative or positive, depending on whether $a$ and $c$ have the same or opposite signs.
This is expected, in view of the fact that $b^2=3ac$ and $bc=9ad$ are the conditions for the cubic to be a perfect cube quadrinomial (see the section on perfect power polynomials).
Recall that this was the case with quadratics as well. Does it seem to hold in general?
Just compute the discriminant of the reduced cubic.
In passing, this result is expected, and is the reason why conclusions
drawn from the roots of the reduced/depressed cubic also apply to the roots of the original cubic equation.
Let's consider a few numerical examples as instances of the above result (be aware that this DOESN'T prove the result).
In the cubic $x^3+2x^2+3x+4$, the coefficients $1,2,3,4$ form an arithmetic sequence with a common difference of $1$. The discriminant is:
$$2^2(3^2)-4(1)(3^3)-4(2^3)(4)-27(1^2)(4^2)+18(1)(2)(3)(4)=-200=-25\times 8.$$
The cubic $x^3-x-2$ has coefficients $1,0,-1,-2$ that form an arithmetic sequence with a common difference of $-1$. Its discriminant is:
$$0^2(-1)^2-4(1)(-1)^3-4(0)^3(-2)-27(1)^2(-2)^2+18(1)(0)(-1)(-2)=-104=-13\times 8.$$
The cubic $2x^3-x^2-4x-7$ has coefficients $2,-1,-4,-7$ that form an arithmetic sequence with a common difference of $-3$. Its discriminant is:
$$(-1)^2(-4)^2-4(2)(-4)^3-4(-1)^3(-7)-27(2)^2(-7)^2+18(2)(-1)(-4)(-7)=-5800=-725\times 8.$$
To prove the result, use the definition of the cubic discriminant and an enumeration of a general arithmetic sequence (e.g., $t,t+c,t+2c,t+3c$, where
$c$ is the common difference)
as the coefficients of the cubic.
To prove this, use the definition of the cubic discriminant and an enumeration of a general geometric sequence (e.g. $t,tr,tr^2,tr^3$, where $r$ is the common ratio) as the coefficients of the cubic.
Check that the cubic discriminant simplifies in this case, and is negative.
Check that the cubic discriminant is negative under the condition that $b$ and $d$ have the same signs.
If $b$ and $d$ have opposite signs, it may still happen that the discriminant is negative, but this will depend on the actual magnitudes
of the other coefficients.
Check that the cubic discriminant is negative under the given conditions.
Alternatively, our one-sided discriminant
$b^2-2ac$ may be used in this case; in fact, we have $b^2-2ac=0^2-2ac=-2ac< 0$,
because $a$ and $c$ having the same signs guarantees that the product $ac$ is positive.
At first sight, it seems not right, that a polynomial should possess two degrees. But we'll clarify this shortly.
Let's mention that the domain of the polynomials under consideration is the set of integers (for simplicity, we restrict to the positive integers), so that in a sense we're dealing with
discrete functions that have internal polynomial character.
Suppose you are asked to find an equation that models the data presented in the table below:
n | f(n) |
---|---|
1 | 1 |
2 | 4 |
3 | 9 |
n | f(n) |
---|---|
1 | 1 |
2 | 4 |
3 | 9 |
4 | 16 |
n | f(n) |
---|---|
1 | r |
2 | s |
For the linear case, one obtains $$f(n)=(s-r)n+(2r-s);$$ for the quadratic case, $$f(n)=\frac{1}{2}\Big((s-r)n^2+(r-s)n+2r\Big).$$
ALWAYS ALWAYS ALWAYS remember that the instantaneous or temporary degree is limited to the data under consideration.
Let's also use this opportunity to mention that if the instantaneous degree is $2$, then the $n$-steps (or $x$-steps if you're more accustomed to $x$ and $y$)
doesn't matter. In fact, if we change the $n$-step in the table above to two:
n | f(n) |
---|---|
1 | r |
3 | s |
n | f(n) |
---|---|
1 | r |
2 | s |
3 | t |
You should obtain $$f(n)=\frac{1}{2}\Big((t+r-2s)n^2+(8s-5r-3t)n+(6r-6s+2t)\Big)$$ for the quadratic representation, and
$$f(n)=\frac{1}{6}\Big((t+r-2s)n^3+(6s-3t-3r)n^2+(2s+2t-4r)n+(12r-6s)\Big)$$ for the cubic model.
We want you to use the idea of FINITE DIFFERENCES to derive these expressions, rather than substituting directly into the given expressions.
This way, the purpose of this exercise will be realized.
At this point we should acknowledge that the polynomial having the instantaneous degree is often much more complicated
in appearance
-- and unnecessarily so -- than the actual, inherent polynomial. But that's the whole point of this exercise: we're mortgaging computational simplicity
for our own conceptual interest and conceived objective. We want you to hone your skills in working with FRACTIONS, FACTORIALS,
FINITE DIFFERENCES, and LINEAR SYSTEMS.
In the next set of exercises, we'll use instantaneous polynomials to describe some familiar number patterns.
n | f(n) |
---|---|
1 | 2 |
2 | 3 |
3 | 5 |
4 | 7 |
The CUBIC model is $$f(n)=-\frac{1}{6}\Big(n^3-9n^2+14n-18\Big);$$ and the QUARTIC model is $$f(n)=-\frac{1}{24}\Big(n^4-6n^3-n^2+6n-48\Big).$$
As always, our goal is that you derive these expressions using FINITE DIFFERENCES, rather than verifying them by substitution.
Something somewhat interesting is that we can extend this procedure to obtain a polynomial formula for any number of consecutive prime numbers,
but the HUGE (as in H-U-G-E) downside is that the computations are not practically implementable. Who is willing to derive a polynomial of degree $10$, for example?
n | f(n) |
---|---|
1 | 6 |
2 | 28 |
3 | 496 |
You should obtain $f(n)=223n^2-647n+430$.
Note that a CUBIC model is also possible for the above data.
NOTICE that the coefficients $223,-647,430$ add up to $6$, the first number.
This is a good way of checking if our model is correct.
n | f(n) |
---|---|
1 | 6 |
2 | 28 |
3 | 496 |
4 | 8128 |
In this case we have $$f(n)=\frac{1}{6}\Big(6718n^3-38970n^2+70016n-37728\Big)$$ for the cubic model. We leave you to derive the QUARTIC counterpart.
NOTICE that the coefficients add up to $6$, as expected: $$\frac{1}{6}\Big(6718-38970+70016-37728\Big)=\frac{1}{6}(36)=6.$$ Useful for
checking the correctness of our model.
n | f(n) |
---|---|
1 | 0 |
2 | 1 |
3 | 1 |
4 | 2 |
The Fibonacci example doesn't bode well for us, but it is good to be considered in view of our primary objective for this section.
In a nutshell it reveals the unnecessary complications that can arise with our approach. The entire Fibonacci sequence already has a simple formula
that describes it, whereas our polynomial model of the first four terms is already cubic. NOT so nice.
Useful for checking if our polynomial model is right.
Useful for checking if our polynomial model is right.
In particular, the last finite difference for the exponential function $f(n)=2^n$ is $2$.
For an exponential function of the form $f(n)=b^n$, the constant ratio between consecutive $f(n)$ values is always equal to $b$. Use this fact, and the preceding exercise.
Equality in the left member
of this inequality holds if and only if the triangle is isosceles. VERIFY this!
In passing, the
point of the above exercise is that the ratio of the length of an altitude to the length
of a right bisector
-- suitably constructed -- is always greater than or equal to $1$, but strictly less than $2$.
There you have it -- beautiful, simple thing.
Especially for those who don't like going through the rigmarole of finding
the equation of a right/perpendicular bisector of a line segment. You're welcome to use the above formula.
Minor reductions may be necessary when using this formula; for example, one may
have to divide by a common factor, or may have to re-write the equation in slope-intercept form, depending on
the requirement of the question in question. You read that correctly, question in question.
Quote this formula if you don't like the traditional procedure.
However, it is best
to follow the step-by-step procedure, rather than to rely on closed formulas (which are not always possible, or
if possible, may not be convenient).
NOTICE that one reobtains the right bisector equation via the
substitutions $x_{1}=\frac{x_{2}+x_{3}}{2},~y_{1}=\frac{y_{2}+y_{3}}{2}$. What does this say about the relationship
between an altitude and a right bisector?
This exercise gives us a way of constructing two lines that intersect at $45^{\circ}$
without using a protractor: Choose
the slope of the first line in such a way that it is not equal to $1$, then use the equation above to determine
the slope of the second line.
HINT: The (acute) angle $\theta$ between two lines with
slopes $m_{1}$ and $m_{2}$ satisfies $\tan\theta=|\frac{m_{2}-m_{1}}{1+m_{2}m_{1}}|$, where $||$ denotes the usual
absolute value.
The presence of the irrational number
$\sqrt{3}$ in the third coordinate says two things.
One, it supports a well-known fact: it is impossible to form an equilateral
triangle in which all the coordinates are rational numbers. That's the price
an equilateral triangle pays for having
equal sides:
sometimes, you won't have everything go your way, even when you try
many angles like a triangle.
The second thing it says is that every irrational number that
appears in the specification of the coordinates of an equilateral triangle must be bound to $\sqrt{3}$ somehow. For
example, the points $(-\sqrt{5},0),~(\sqrt{5},0),~(0,\sqrt{15})$ form an equilateral triangle. On the surface, it
appears that the coordinates are completely free of $\sqrt{3}$, but the fact that $\sqrt{15}=\sqrt{5}\times
\sqrt{3}$ shows that $\sqrt{3}$ was embedded in one of the coordinates.
Requiring that $x\neq 1$ is enough for the rational expression, but not for the logarithm.
As above, restricting $y\neq 1$ is enough for the rational expression, but not for the logarithm.
The case when $a=2,~x=1,~y=1$ is a special case.
The case of $a=2,~x=2,~y=1$ is a special case.
Aside: It has been suggested that freedom from boredom can be obtained by doing something
different from the norm, something different from your curriculum
.
This is another reason behind these exercises: If you feel bored to the core, feel free to take these exercises as
a chore; they may be able to bring you a cure -- for boredom.
What's going on here? Nothing yet.