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1. Quadratic form

2. Introduction


4.The Quadratic Formula Explained
5.The simplest way

6.Positive and negative quadratic equation
Quadratic form
In mathematics, a quadratic form is a homogeneous polynomial of degree two in a number of
variables. For example,

is a quadratic form in the variables x and y.

Quadratic forms occupy central place in various branches of mathematics: number theory, linear
algebra, group theory (orthogonal group), differential geometry (Riemannian metric), differential
topology (intersection forms of four-manifolds), and Lie theory (the Killing form).

Quadratic forms are homogeneous quadratic polynomials in n variables. In the cases of one, two,
and three variables they are called unary, binary, and ternary and have the following explicit


where a,…,f are the coefficients.[1] Note that quadratic functions, such as ax2+bx+c in the one
variable case, are not quadratic forms, as they are typically not homogeneous (unless b and c are
both 0).

The theory of quadratic forms and methods used in their study depend in a large measure on the
nature of the coefficients, which may be real or complex numbers, rational numbers, or integers.
In linear algebra, analytic geometry, and in the majority of applications of quadratic forms, the
coefficients are real or complex numbers. In the algebraic theory of quadratic forms, the
coefficients are elements of a certain field. In the arithmetic theory of quadratic forms, the
coefficients belong to a fixed commutative ring, frequently the integers Z or the p-adic integers
Zp.[2] Binary quadratic forms have been extensively studied in number theory, in particular, in
the theory of quadratic fields, continued fractions, and modular forms. The theory of integral
quadratic forms in n variables has important applications to algebraic topology.

Using homogeneous coordinates, a non-zero quadratic form in n variables defines an (n−2)-
dimensional quadric in the (n−1)-dimensional projective space. This is a basic construction in
projective geometry. In this way one may visualize 3-dimensional real quadratic forms as conic

A closely related notion with geometric overtones is a quadratic space, which is a pair (V,q),
with V a vector space over a field k, and q:V → k a quadratic form on V. An example is given by
the three-dimensional Euclidean space and the square of the Euclidean norm expressing the
distance between a point with coordinates (x,y,z) and the origin:

An n-ary quadratic form over a field K is a homogeneous polynomial of degree 2 in n variables
with coefficients in K:

This formula may be rewritten using matrices: let x be the column vector with components x1,
…, xn and A = (aij) be the n×n matrix over K whose entries are the coefficients of q. Then

       q(x) = xtAx.

Two n-ary quadratic forms φ and ψ over K are equivalent if there exists a nonsingular linear
transformation T ∈GLn(K) such that

Let us assume that the characteristic of K is different from 2.

(The theory of quadratic forms over a field of characteristic 2 has important differences and
many definitions and theorems have to be modified.) The coefficient matrix A of q may be
replaced by the symmetric matrix 1/2(A + At) with the same quadratic form, so it may be
assumed from the outset that A is symmetric. Moreover, a symmetric matrix A is uniquely
determined by the corresponding quadratic form. Under an equivalence T, the symmetric matrix
A of φ and the symmetric matrix B of ψ are related as follows:

        B = TtAT.

The associated bilinear form of a quadratic form q is defined by

Thus, bq is a symmetric bilinear form over K with matrix A. Conversely, any symmetric bilinear
form b defines a quadratic form

        q(x) = b(x,x)

and these two processes are the inverses of one another. As a consequence, over a field of
characteristic not equal to 2, the theories of symmetric bilinear forms and of quadratic forms in n
variables are essentially the same.

The Quadratic Formula Explained

This lesson covers many ways to solve quadratics, such as taking square roots, completing the square,
and using the Quadratic Formula. But we'll start with solving by factoring.

You should already know how to factor quadratics. (If not, review Factoring Quadratics.) The new thing
here is that the quadratic is part of an equation, and you're told to solve for the values of x that make the
equation true. Here's how it works:

       Solve (x   – 3)(x – 4) = 0.

        The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at
        least one of the factors must be zero, I'll set them each equal to zero:

                 x – 3 = 0 or x – 4 = 0
          This gives me simple linear equations, and they're easy to solve:

                    x = 3 or x = 4

          And this is the solution they're looking for:   x = 3, 4

Note that "x = 3, 4" means the same thing as "x = 3 or         x = 4"; the only difference is the formatting. The
"x = 3, 4" format is more-typically used.


One important issue should be mentioned at this point: Just as with linear equations, the solutions to
quadratic equations may be verified by plugging them back into the original equation, and making sure
that they work, that they result in a true statement. For the above example, we would do the following:

Checking x = 3 in (x –     3)(x – 4) = 0:

So       ([3] – 3)([3] – 4) ?=? 0
             (3 – 3)(3 – 4) ?=? 0
                    (0)(–1) ?=? 0
                          0 = 0

Checking x = 4 in (x –     3)(x – 4) = 0:

          So ([4] – 3)([4] – 4) ?=? 0
                 (4 – 3)(4 – 4) ?=? 0
                         (1)(0) ?=? 0
                              0 = 0

So both solutions "check" and are thus verified as being correct.

         Solve x     + 5x + 6 = 0.

          This equation is already in the form "(quadratic) equals (zero)" but, unlike the previous example,
          this isn't yet factored. The quadratic must first be factored, because it is only when we
          MULTIPLY and get zero that we can say anything about the factors and solutions. We can't
          conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6),
          because we can add lots of stuff that totals zero.

          So the first thing I have to do is factor:

                    x2 + 5x + 6 = (x + 2)(x + 3)

          Set this equal to zero:

                    (x + 2)(x + 3) = 0
        Solve each factor:

                  x + 2 = 0 or x + 3 = 0
                  x = –2 or x = – 3
        The solution to x        + 5x + 6 = 0 is x = –3, –2

Checking x = –3 and    x = –2 in x2 + 5x + 6 = 0:

        [–3]2 + 5[–3] + 6 ?=? 0
               9 – 15 + 6 ?=? 0
               9 + 6 – 15 ?=? 0
                  15 – 15 ?=? 0
                        0 = 0

        [–2]2 + 5[–2] + 6 ?=? 0
               4 – 10 + 6 ?=? 0
               4 + 6 – 10 ?=? 0
                  10 – 10 ?=? 0
                        0 = 0

So both solutions "check".

       Solve x     – 3 = 2x.

        This equation is not in "(quadratic) equals (zero)" form, so I can't try to solve it yet. The first thing I
        need to do is get all the terms over on one side, with zero on the other side. Only then can I factor
        and solve:

                  x2 – 3 = 2x
                  x2 – 2x – 3 = 0
                  (x – 3)(x + 1) = 0
                  x – 3 = 0 or x + 1 = 0
                  x = 3 or x = –1
        Then the solution to x          – 3 = 2x is x = –1, 3

       Solve (x    + 2)(x + 3) = 12.

        Instead, I first have to multiply out and simplify the left-hand side, then subtract the 12 over to the
        left-hand side, and re-factor. Only then can I solve.

                  (x + 2)(x + 3) = 12
                  x2 + 5x + 6 = 12
                  x2 + 5x – 6 = 0
                  (x + 6)(x – 1) = 0
                  x + 6 = 0 or x – 1 = 0
                  x = –6 or x = 1

        Then the solution to (x      + 2)(x + 3) = 12 is x = –6, 1

       Solve x(x    + 5) = 0.

        Even though we are used to variable factors having variables and numbers (like the other factor,
        x + 5), a factor can contain only a variable, so "x" is a perfectly valid factor. So set the factors
        equal to zero, and solve:

                  x(x + 5) = 0
                  x = 0 or x + 5 = 0
                  x = 0 or x = –5

        Then the solution to x(x      + 5) = 0 is x = 0, –5
       Solve x     – 5x = 0.

        This two-term quadratic is easier to factor than were the previous quadratics: I can factor an x out
        of both terms, taking the x out front. (Warning: Do not "divide the x off", do not make it magically
        "disappear", or you'll lose one of your solutions!)

                  x(x – 5) = 0
                  x = 0 or x – 5 = 0
                  x = 0 or x = 5
        Then the solution to x       – 5x = 0 is x = 0, 5

There is one other case of two-term quadratics that you can factor:

       Solve x     – 4 = 0.

        This equation is in "(quadratic) equals (zero)" form, so it's ready to solve. The quadratic itself is a
        difference of squares, so I'll apply the difference-of-squares formula:

                  x2 – 4 = 0
                  (x – 2)(x + 2) = 0
                  x – 2 = 0 or x + 2 = 0
                  x = 2 or x = –2

        Then the solution is x   = –2, 2

Note: This solution may also be formatted as    "x = ± 2".

There is another way to work this last problem, which leads us to the next section...
    Solve x     –4=0       in another way

        Previously, I'd solved this by factoring the difference of squares, and solving each factor; the
        solution was "x = ± 2". However—

        I can also try isolating the squared variable term, putting the number over on the other side, like

                  x2 – 4 = 0
                  x2 = 4

        I know that, when solving an equation, I can do whatever I like to that equation as long as I do the
        same thing to both sides of the equation. On the left-hand side of this particular equation, I have
             2                                   2
        an x , and I need a plain x. To turn an x into an x, I can take the square root of each side of the


        Then the solution is x   =±2

Why did I need the "±" ("plus-minus") sign on the 2 when I took the square root of the 4?

Because it might have been a positive 2 or a negative 2 that was squared to get that 4 in the original

.The answer I got above, with the "±" sign, matches the solution I got when I solved this equation using
the difference-of-squares factoring formula. Thus, this result confirms the need to use the "±" sign when
solving by square-rooting. (In mathematics, we need to be able to get the same answer, no matter which
valid method you happen to have used in order to arrive at that answer.)

A benefit of this square-rooting process is that it allows us to solve some quadratics that we could not
have solved before. For instance:

       Solve x      – 50 = 0.

        This quadratic has a squared part and a number part. I'll start by adding the numerical term to the
        other side of the equaion (so the squared part is by itself), and then I'll square-root both sides. I'll
        need to remember to simplify the square root:

                  x2 – 50 = 0
                  x2 = 50
        Then the solution is

While we could have gotten the previous integer solution by factoring, we could never have gotten this
radical solution by factoring. Factoring is clearly useful, but additional techniques can allow us to find
solutions to additional sorts of equations.

       Solve (x   – 5)2 – 100 = 0.

        This quadratic has a squared part and a number part. I'll start by adding the number to the other
        side of the equation, so the squared binomial is by itself. Then I'll square-root both sides,
        remembering to simplify my results:

                (x – 5)2 – 100 = 0
                (x – 5)2 = 100

                x – 5 = ±10
                x = 5 ± 10
                x = 5 – 10 or x = 5 + 10
                x = –5 or x = 15

        Since the equation, after square-rooting, did not contain any square roots, I was able to simplify
        down to simple values. Copyright © 2006-2008 All Rights Reserved

                The solution is x     = –5, 15

This example points out the importance of remembering the "±" when we square-root both sides:

By the way, since the solution to the equation consisted of nice neat "whole" numbers, this quadratic
could also have been solved by multiplying out the square and simplifying to get "x – 10x – 75", which
then could have been factored as "(x – 15)(x + 5)".

       Solve (x   – 2)2 – 12 = 0

        This quadratic has a squared part and a number part. I'll add the number to the other side (so the
        squared part is by itself), and then I'll square-root both sides, remembering to simplify:

                (x – 2)2 – 12 = 0
                (x – 2)2 = 12

        Then the solution is
This quadratic equation, unlike the previous one, could not be solved by factoring. But how would I have
solved it, if they had not given me the quadratic already put into "(squared part) minus (a number)" form?
This concern lead ".


The simplest way
Often, the simplest way to solve "ax + bx + c = 0" for the value of x is to factor the quadratic, set each
factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't
factor at all, or you just don't feel like factoring. While factoring may not always be successful, the
Quadratic Formula can always find the solution.

The Quadratic Formula uses the "a", "b", and "c" from "ax + bx + c", where "a", "b", and "c" are just
numbers; they are the "numerical coefficients". The Formula is derived from the process of completing the
square, and is formally stated as:

         For ax     + bx + c = 0, the value of x is given by:

For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0".
Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root.
And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root
or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them
back" on your test, and you'll mess yourself up. Remember that "b " means "the square of ALL of b,
including its sign", so don't leave b being negative, even if b is negative, because the square of a
negative is a positive.

In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run.
Trust me on this!

Here are some examples of how the Quadratic Formula works:

       Solve   x2 + 3x – 4 = 0

        This quadratic happens to factor:

                  x2 + 3x – 4 = (x + 4)(x – 1) = 0

        ...so I already know that the solutions are x = –4 and x = 1. How would my solution look in the
        Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:
        Then, as expected, the solution is x   = –4, x = 1.

Suppose you have ax + bx + c = y, and you are told to plug zero in for y. The corresponding x-values
are the x-intercepts of the graph. So solving ax + bx + c = 0 for x means, among other things, that you
are trying to find x-intercepts. Since there were two solutions for x + 3x – 4 = 0, there must then be two
x-intercepts on the graph. Graphing, we get the curve below:

As you can see, the x-intercepts match the solutions, crossing the x-axis at x = –4 and x = 1. This shows
the connection between graphing and solving: When you are solving "(quadratic) = 0", you are finding the
x-intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the
Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make
sure that the displayed x-intercepts have the same decimal values as do the solutions provided by the
Quadratic Formula.

Note, however, that the calculator's display of the graph will probably have some pixel-related round-off
error, so you'd be checking to see that the computed and graphed values were reasonably close; don't
expect an exact match.

       Solve 2x     – 4x – 3 = 0. Round your answer to two decimal places, if necessary.
        There are no factors of (2)(–3) = –6 that add up to –4, so I know that this quadratic cannot be
        factored. I will apply the Quadratic Formula. In this case, a = 2, b = –4, and c = –3:

        Then the answer is x     = –0.58, x = 2.58, rounded to two decimal places.

Warning: The "solution" or "roots" or "zeroes" of a quadratic are usually required to be in the "exact" form
of the answer. In the example above, the exact form is the one with the square roots of ten in it. You'll
need to get a calculator approximation in order to graph the x-intercepts or to simplify the final answer in a
word problem. But unless you have a good reason to think that the answer is supposed to be a rounded
answer, always go with the exact form.

Compare the solutions of 2x      – 4x – 3 = 0 with the x-
intercepts of the graph:

Remember: The "solutions" of an equation are also the x-
intercepts of the corresponding graph.

Somebody (possibly in seventh-century India) was solving a lot of quadratic equations by completing the
square. At some point, he noticed that he was always doing the exact same steps in the exact same
order for every equation. Taking advantage of the one of the great powers and benefits of algebra
(namely, the ability to deal with abstractions, rather than having to muck about with the numbers every
single time), he made a formula out of what he'd been doing:
        The Quadratic Formula: For ax         + bx + c = 0, the value of x is given by

The nice thing about the Quadratic Formula is that the Quadratic Formula always works. There are some
quadratics (most of them, actually) that you can't solve by factoring. But the Quadratic Formula will
always spit out an answer, whether the quadratic was factorable or not.

I have a lesson on the Quadratic Formula, which gives examples and shows the connection between the
discriminant (the stuff inside the square root), the number and type of solutions of the quadratic equation,
and the graph of the related parabola. So I'll just do one example here. If you need further instruction,
study the lesson at the above hyperlink.

       Use the Quadratic Formula to solve x           – 4x – 8 = 0.

        Looking at the coefficients, I see that a = 1, b = –4, and c = –8. I'll plug them into the Formula,
        and simplify. I should get the same answer as before:

        Then the solution is

The nice thing about the Quadratic Formula (as compared to completing the square) is that we are just
plugging into a formula. There are no "steps" to remember, and there are fewer opportunities for
mistakes. Copyright © Elizabeth -2008 All Rights Reserved


       Solve x     – 8x + 15 = 0 by using the following graph.
          The graph is of the related quadratic, y = x – 8x + 15, with the x-intercepts being where y = 0.
          The point here is to look at the picture (hoping that the points really do cross at whole numbers,
          as it appears), and read the x-intercepts (and hence the solutions) from the picture.

                  The solution is x   = 3, 5 Copyright © -2008 All Rights Reserved
Since x    – 8x + 15 factors as (x – 3)(x – 5), we know that our answer is correct.
         Solve 0.3x     – 0.5x – 5/3 = 0 by using the following graph.

          For this picture, we are labelled a bunch of points. Partly, this was to be helpful, because the x-
          intercepts are messy ,but mostly this was in hopes of confusing us, in case I had forgotten that
          only the x-intercepts, not the vertices or y-intercepts, correspond to "solutions".

          The x-values of the two points where the graph crosses the x-axis are the solutions to the
            The solution is   x = –5/3, 10/3

   Find the solutions to the following quadratic:

   They haven't given me the quadratic equation, so I can't check my work algebraically. (And,
    technically, they haven't even given me a quadratic to solve; they have only given me the picture
    of a parabola from which I am supposed to approximate the x-intercepts, which really is a
    different question....)
        I ignore the vertex and the y-intercept, and pay attention only to the x-intercepts. The "solutions"
        are the x-values of the points where the pictured line crosses the x-axis:

                The solution is   x = –5.39, 2.76

"Solving" quadratics by graphing is silly in "real life", and requires that the solutions be the simple
factoring-type solutions such as "x = 3", rather than something like "x = –4 + sqrt(7)". In other words,
they either have to "give" you the answers or they have to ask you for solutions that you could have found
easily by factoring. About the only thing we can gain from this topic is reinforcing our understanding of the
connection between solutions and x-intercepts: the solutions to "(some polynomial) equals (zero)"
correspond to the x-intercepts of "y equals (that same polynomial)". If you come away with an
understanding of that concept, then you will know when best to use your graphing calculator or other
graphing software to help you solve general polynomials; namely, when they aren't factorable.

But in practice, given a quadratic to solve, you should not start by drawing a graph. Which begs the
question: For any given quadratic, which method should you use?

Positive and negative quadratic equation

There is a simple, if slightly "dumb", way to remember the difference between right-side-up parabolas and
upside-down parabolas:

          positive quadratic y   = x2     negative quadratic y = –x
This can be useful information: If, for instance, you have an equation where a is negative, but you're
somehow coming up with plot points that make it look like the quadratic is right-side-up, then you will
know that you need to go back and check your work, because something is wrong.

Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point,
where the parabola changes direction, is called the "vertex".

If the quadratic is written in the form y = a(x – h) + k, then the vertex is the point (h, k). This makes
sense, if you think about it. The squared part is always positive (for a right-side-up parabola), unless it's
zero. So you'll always have that fixed value k, and then you'll always be adding something to it to make y
bigger, unless of course the squared part is zero. So the smallest y can possibly be is y = k, and this
smallest value will happen when the squared part, x – h, equals zero. And the squared part is zero when
x – h = 0, or when x = h. The same reasoning works, with k being the largest value and the squared part
always subtracting from it, for upside-down parabolas.

(Note: The "a" in the vertex form "y = a(x – h) + k" of the quadratic is the same as the "a" in the
common form of the quadratic equation, "y = ax + bx + c".)

Since the vertex is a useful point, and since you can "read off" the coordinates for the vertex from the
vertex form of the quadratic, you can see where the vertex form of the quadratic can be helpful, especially
if the vertex isn't one of your T-chart values. However, quadratics are not usually written in vertex form.
You can complete the square to convert ax + bx + c to vertex form, but, for finding the vertex, it's
simpler to just use a formula. (The vertex formula is derived from the completing-the-square process, just
as is the Quadratic Formula. In each case, memorization is probably simpler than completing the square.)

                             2                                                         –b
For a given quadratic y = ax + bx + c, the vertex (h, k) is found by computing h = /2a, and then
evaluating y at h to find k. If you've already learned the Quadratic Formula, you may find it easy to
memorize the formula for k, since it is related to both the formula for   h and the discriminant in the
Quadratic Formula: k = (4ac – b ) / 4a.

Now, you may think you can do this:But this doesn't make any sense! You already have two numbers that
square to 1; namely –1 and +1. And i already squares to –1. So it's not reasonable that i would also
square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the
ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and
convenient rules you used to have when dealing with square roots). In particular, YOU MUST ALWAYS

In your computations, you will deal with i just as you would with x, except for the fact that x2 is just x2, but
i2 is –1:

       Simplify 2i   + 3i.

        2i + 3i = (2 + 3)i = 5i

       Simplify 16i   – 5i.

        16i – 5i = (16 – 5)i = 11i

       Multiply and simplify (3i)(4i).

        (3i)(4i) = (3·4)(i·i) = (12)(i2) = (12)(–1) = –12

       Multiply and simplify (i)(2i)(–3i).

        (i)(2i)(–3i) = (2 · –3)(i · i · i) = (–6)(i2 · i)

                 =(–6)(–1 · i) = (–6)(–i) = 6i

Note this last problem. Within it, you can see that         , because i2 = –1. Continuing, we get:

This pattern of powers, signs, 1's, and i's is a cycle:

In other words, to calculate any high power of i, you can convert it to a lower power by taking the closest
multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent. For
example, a common trick question on tests is something along the lines of "Simplify i99", the idea being
that you'll try to multiply i ninety-nine times and you'll run out of time, and the teachers will get a good
giggle at your expense in the faculty lounge. Here's how the shortcut works:

        i99 = i96+3 = i(4×24)+3 = i3 = –i
That is, i99 = i3, because you can just lop off the i96. (Ninety-six is a multiple of four, so i96 is just 1, which
you can ignore.) In other words, you can divide the exponent by 4 (using long division), discard the
answer, and use only the remainder. This will give you the part of the exponent that you care above. Here
are a few more examples:

       Simplify i .

        i17 = i16 + 1 = i4 · 4 + 1 = i1 = i
       Simplify i         .

        i120 = i4 · 30 = i4· 30 + 0 = i0 = 1
       Simplify i             .

        i64,002 = i64,000 + 2 = i4 · 16,000 + 2 = i2 = –1

Now you've seen how imaginaries work; it's time to move on to complex numbers. "Complex" numbers
have two parts, a "real" part (being any "real" number that you're used to dealing with) and an "imaginary"
part (being any number with an "i" in it). The "standard" format for complex numbers is "a + bi"; that is,
real-part first and i-part last.

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