User:GoreyCat/AlgebraSandbox

Numerical Axioms

It is possible to define a regular set of numbers in a formal fashion. The set of Peano axioms define the series of numbers known as the natural numbers. They are as follows:

1. There is a whole number 0.
2. Every natural number a has a successor, denoted by a + 1.
3. There is no natural number whose successor is 0.
4. Distinct natural numbers have distinct successors: if a <> b, then a + 1 <> b + 1.
5. If a property is possessed by 0 and also by the successor of every natural number it is possessed by, then it is possessed by all natural numbers.

Let us attempt to motivate these axioms. We want these axioms to eliminate any set which is not the natural numbers. E.g., any set fulfilling the above should at least be infinite.

The first two are obvious properties of the natural numbers (and of integers) as we know them. Note that some prefer to use 1 as the lowest number. The reason for choosing zero has root in [set theory], in which the first natural number is chosen as the empty set $\emptyset$.

The 3rd axiom prevents circularity. If this axiom was not included, defining $0+1=0$ would trivially fulfill the remaining axioms --- prove this for yourself by considering each remaining axiom!

The 4th prevents a partial loop. Consider a the set $\{0,1\}$ and set $0+1=1$ and $1+1=1$. This set fulfills every axiom but the 4th --- prove this for yourself.

The 5th is sometimes called the induction axiom. It ensures that the set is connected, i.e. that we can reach any number by using the 2nd axiom repeatedly on 0. An example of a set that fulfills every axiom but the 5th is $\left\{0,{\frac {1}{2}},1,{\frac {3}{2}},\dots \right\}$ with the usual meaning of +1.

From this we can deduce the existence of a series of quantities like this:

• 0
• 0 + 1
• 0 + 1 + 1
• 0 + 1 + 1 + 1
• 0 + 1 + 1 + 1 + 1
• 0 + 1 + 1 + 1 + 1 + 1
• and so on

where '0' is a constant and the first natural number and '1' is a constant natural number equivalent to the difference in value between two consecutive natural numbers.

This set is sufficient for counting. However, it is inconvenient to refer to a large natural number as '0' followed by the requisite large number of '+ 1' expressions. Due to this, each of the natural numbers is given a label, and to make the labelling easier another axiom is introduced:

'0 + 1' is equivalent to '1'.

Thus the series of natural numbers may be written so for some brevity:

• 0
• 1
• 1 + 1
• 1 + 1 + 1
• 1 + 1 + 1 + 1
• 1 + 1 + 1 + 1 + 1
• and so on

Once this is done, giving each quantity its own label is trivial. And so the series of natural numbers can then be written:

• 0
• 1
• 2
• 3
• 4
• 5
• and so on

What are Roots?

Roots are the inverse operation for exponents. An expression with roots is called a radical expression. It's easy, although perhaps tedious, to compute exponents given a root. For instance 7*7*7*7 = 49*49 = 2401. So, we know the fourth root of 2401 is 7, and the square root of 2401 is 49. What is the third root of 2401? This article gives a formula for determining the answer, while this article gives a detailed explanation of roots.

Finding the value for a particular root is difficult. This is because exponentiation is a different kind of function than addition, subtraction, multiplication, and division. When we graph functions we will see that expressions that use exponentiation use curves instead of lines. We will see using algebra that not all of these expression are functions, that knowing when an expression is a relation or a function can allow us to make certain types of assumptions, and we can use these assumptions to build mental models for topics that would otherwise be impossible to understand.

For now we will deal with roots by turning them back into exponents.

If a root is defined as the nth root of X, it is represented as ${\sqrt[{n}]{x}}=r$. We get rid of the root by raising our answer to the nth power, i.e. $r^{n}=x.\!\,$

Problems Using Roots

Square root

If you take the square root of a number, the result is a number which when squared gives the first number. This can be written symbolically as:

${\sqrt {x}}=y{\mbox{ if }}y^{2}=x\,$. In the series of real numbers, $y^{2}\geq 0$ regardless of the value of $y$. As such, we cannot define the ${\sqrt {x}}$ when $x<0$.

Examples:

• ${\sqrt {9}}=3$ since $3^{2}=3\cdot 3=9.$
• If $x=25$ then ${\sqrt {x-16}}={\sqrt {25-16}}={\sqrt {9}}=3$.
• If $x=7$ then ${\sqrt {x-16}}$ is undefined because ${\sqrt {x-16}}={\sqrt {7-16}}={\sqrt {-9}}$, but there is no number $y$ so that $y^{2}=-9$. Notice that the answer is not $-3$ since $-3\cdot -3=9$ and not $-9$.

You may notice or discover that there is a solution to square roots of negative numbers. This will be discussed in the future chapter of Complex Numbers, which will require learning intermediate concepts.

Cube roots

Roots do not have to be square. One can also take the cube root of a number (${\sqrt[{3}]{}}$ ). The cube root is the number which, when cubed (multiplied by itself and then multiplied by itself again), gives back the original number. For example, the cube root of 8 is 2 because $2\cdot 2\cdot 2=8$, or:

${\sqrt[{3}]{8}}=2.$

Other roots

There are an infinite number of possible roots all in the form of ${\sqrt[{n}]{a}}$ which corresponds to $a^{\frac {1}{n}}$, when expressed using exponents. If ${\sqrt[{n}]{a}}={b}$ then $b^{n}=a\,$.

The only exception is 0. ${\sqrt[{0}]{a}}$ is undefined, as it corresponds to $a^{\frac {1}{0}}$ , resulting in a division by zero. Even if you attempt to discover the 0th root of 1, you will not make progress as practically any number to the power of zero equals 1, leaving only an undefined result.

Irrational numbers

If you square root a whole number which is not itself the square of a rational number the answer will have an infinite number of decimal places. Such a number is described as irrational and is defined as a number which cannot be written as a rational number: ${\frac {a}{b}}$, where a and b are integers.

However, using a calculator you can approximate the square root of a non-square number:

${\sqrt {3}}\approx 1.73205080757$

The result of taking the square root is written with the approximately equal sign $\approx$ because the result is an irrational value which cannot be written in decimal notation exactly. Writing the square root of 3 or any other non-square number as ${\sqrt {3}}$ is the simplest way to represent the exact value.

Proof

Taking ${\sqrt {3}}$ as an example.

Suppose ${\sqrt {3}}$ is rational and ${\sqrt {3}}$ = ${\frac {a}{b}}$ where a and b are integers and relatively prime.

$3={\frac {a^{2}}{b^{2}}}$

$a^{2}=3b^{2}$

This implies that 3 is a factor of $a^{2}$. Since a is an integer and 3 is prime, 3 is a factor of a. Let a = 3k where k is an integer.

$a^{2}=3b^{2}$

$(3k)^{2}=3b^{2}$

$9k^{2}=3b^{2}$

$b^{2}=3k^{2}$

Similarly, 3 is a factor of b, which contradicts the first statement that a and b are relatively prime. Therefore ${\sqrt {3}}$ cannot be rational. Therefore ${\sqrt {3}}$ is irrational.

Irrational numbers also appear when attempting to take cube roots or other roots. However, they are not restricted to roots, and may also appear in other mathematical constants (e.g. π, e, φ, etc.).

With practice the concept of slope for linear functions becomes intuitive. It makes sense that the line that fits the equation $y=2x$ has a steeper ascent then the line that fits the equation $y=1/2x$. You only have to move horizontally one unit to change your vertical direction two for the former when you graph $y=2x$. How many blocks do you need to move horizontally to change your vertical direction by one for the line $y=1/2x$?

When we express concepts like $y=x^{2}$ the abstract behavior of what is being represented becomes a little harder to see.

A monomial of one variable, let's say x, is an algabraic expression of the form

$cx^{m}\$

where

• $c$ is a constant, and
• $m$ is a non-negative integer (e.g., 0, 1, 2, 3, ...).

The integer $m$ is called the degree of the monomial.

The idea of a monomial of degree zero appears a bit mystical since it always represents one, except when the value of the variable is set equal to zero when the result is undefined. This idea allows us preserve the value of the constant in the monomial. We know that $cx^{0}$ is always equal to $c$ because even though we have 0 x's (somethings) we still have a c. When x = 0 things are difficult because the value we started with, 0, represents nothing.

For a monomial of power 1 we are multiplying C by one instance of our variable. When $x=0$ we get $c*0=0$. When $x\neq 0$ we are multiplying c by 1 x. If x is less than 1 then c gets smaller, if x is more than 1 c gets bigger. When x is between 0 and -1 c gets smaller slower, when x is less than -1 c gets smaller faster.

A monomial with power two is one that "squares" the value of x. The reference to square is because using the multiplication operation once allows us to measure area. If you have something that is one unit on each side this is called a square unit. If you divide both sides of your square unit in half, you get 4 quarter units. We represent this with math by doing the multiplication $1/2*1/2=1/4$ Squaring something is a non-intuitive operation until you become comfortable with the graph of the function. We can see this with the story of the mathematician who was offered a reward by his king. The mathematician said he wanted a single grain of wheat, squared every day for 30 days. For the first seven days the king's servants delivered 1, 2, 4, 16, 256, 65,536 grains of wheat to the mathematician. On the seventh day the value was 4,294,967,296 (4 gig in computer terms)... Sometimes the story ends with the king re-negotiating, sometimes the story ends with the king executing the mathematician to preserve his kingdom, and sometimes the king is astute enough not to take the deal.

A monomial with power three is one that "cubes" the value of x. This is because we use the operation x*x*x to measure the volume that a given area of x*x takes up. If you have a cube that is 1 unit on each side and cut each side in half you will find that you have created 8 cubes. If the mathematician had asked to have the single grain of wheat cubed than the servants would have delivered 1, 8, 512, $134\times 10^{6}$, $242\times 10^{22}$ grains of wheat and the kings deal would have needed to be re-negotiated two days earlier.

Polynomials

A polynomial of one variable, x, is an algebraic expression that is a sum of one or more monomials. The degree of the polynomial is the highest degree of the monomials in the sum. An polynomial $P(x)$ can generically be expressed in the form

$P(x)=a_{n}x^{n}+...+a_{i}x^{i}+...a_{2}x^{2}+a_{1}x+a_{0}\$

or

$\ P(x)=\sum _{i=0}^{n}a_{i}x^{i}$

The constants a_i are called the coefficients of the polynomial.

Each of the individual monomials in the above sum, whose coefficient a_i ≠ 0, is called a term of the polynomial. When i = 0, xⁱ = 1 and the corresponding term simply equals the constant a_i. Also when i = 1, the corresponding term equals a_i x.

A polynomial having two terms is called a binomial. A polynomial having three terms is called a trinomial.

Polynomial Equations

We refer to all functions with one independent variable as $P(x)$. Each instance of $P(x)$ can be represented by an equation (either a monomial or a polynomial) which may have one or more places where the dependent variable is equal to zero. These places are called roots and they represent the number(s) whose value(s) for x make the function $P(x)=0$ true. These roots are called the zeroes of the polynomial (singular is zero).

A polynomial of degree 1, will always look like a line when you graph it, and always has 1 real zero. A polynomial of degree 2, a quadratic function, can have 0, 1, or 2 real zeroes. A polynomial of degree 3 (a cubic function) can have 1 or 3 real zeroes. A polynomial of degree 4 can have 0, 2, or 4 real zeroes. Complex (unreal) zeroes, when present, always come in pairs. In general, a polynomial of degree n, where n is odd, can have from 1 to n real zeroes. A polynomial of degree n, where n is even, can have from 0 to n real zeroes.

When we graph polynomials each zero is a place where the polynomial crosses the x axis. A polynomial of degree one can be generically written as $P(x)=Mx+C$ where M and C can be any real number. We will see that quadratic functions are curves. The curve can bend before it ever touches the X axis in which case it has no zeroes, It can bend just as it touches the X axis, in which case it can have just one zero, or it can open up above or below the X axis in which case it will have two zeroes. If you think about this you will see that polynomials with an odd degree (1,3,5, ...) have to be positive and negative, so they have to cross the X axis at least once. Polynomials with an even degree (2,4,6,....) might always be positive or negative and never have a zero.

Normally we represent a function in the form $P(x)=y$, but when we are looking for the roots of the function we want y to be equal to zero so we solve for the equation of $P(x)$ where $P(x)=0$

Order	Name	Number of bumps	Where found
1	linear	no bumps - straight line	straight line equations
2	quadratic	one bump	equations involving area and vibrations
3	cubic	two bumps	equations involving volumes
4	quartic	three bumps	some physics equations (melting ice)
n (5+)		n-1 bumps	very rare

Solving Polynomial Equations

Some polynomial equations can be solved by factoring, and all equations of degrees 1-4 can be solved completely by formulae. Above degree 4, there are no formulae for solving completely, and you must rely on numerical analysis or factoring. This means that for polynomials of degree greater than 4 it is often impossible to find exact solutions.

Rational roots of polynomial equations

Often we are interested in the rational roots of polynomials. A root is much like a factor of a number. For instance all even numbers have a factor of two. This means you can write the even numbers as two times another number. That is the numbers 2, 4, 6, 8 ... can be written as 2*1, 2*2, 2*3, 2*4 ... . This fact is helpful when you have a fraction of two even numbers. Given a fraction of two even numbers called N and M $N/M$ you could reduce the fraction by re-writing it as $2n/2m$. By keeping fractions in lowest terms it's easier to know when you can add or subtract them without looking for a common denominator.

An example of a use of a polynomial equation

There is a story that in grade school the mathematician Gauss was asked to add the numbers 1 to 100 sequentially. He is said to have intuited the sum could be expressed with the formula n(n+1)/2 and quickly gave the answer 5050. The basis of this formula is that the numbers 1 through 49 added to the numbers 99 through 51 each yield 100. It is interesting to look at how this formula works for the values 9 and 10. For 10 we add the numbers 1+9, 2+ 8, 3+ 7, 4+ 6 to get 40 and we add the two remaining terms 5 and 10 to get 55. For 9 we add the terms 1 + 8, 2 + 7, 3 + 6, 4+ 5 to get 4*9 = 36 + 9 = 45. In the first case the n + 1 is the odd number and represents adding the 10 and the middle number, the 5. In the second case the n is the odd number and the n+1 represents the sum for the preceding terms in the formula. You may or may not find stories like this intriguing based on how your personality reacts to what is known as the foundational crisis of mathematics. Learning mathematics is a lot like learning a foreign language. Some people seem more adept at learning languages than others, but with hard work learning a new language is something we can all do.

Multiplying polynomials together

When we multiply polynomials together we rely heavily on the distributive property.

For instance when we multiply 67 by 5 we can divide the equation into (60 + 7)*5 = (300 + 35) = 335. Additionally we can apply the commutative property to multiply multidigit numbers. 67*25 = (60 + 7)(20 + 5) = ((60 + 7)*20) + ((60 + 7) *5) = (60*20) + (7*20) + (60*5) + (7*5) = 1200 + 140 + 300 + 35 = 1675. These properties are the foundation for the different forms of the mechanical calculating tool the abacus.

When multiplying polynomials together we do similar operations. We use the commutative property to divide the multiplier into its component parts and multiply the multiplicant by each of these parts. For instance to multiply $x^{2}+x$ by $x+1$ we first write the multiplicand and multiplier in terms of powers of x. This gives us $x^{2}+x+0x^{0}$ and $x+1x^{0}$ The terms raised to the zero power represent constant integer terms in our equations. Next we apply the commutative property to rewrite the equations as $[(x^{2}+x+0x^{0})*1x^{1}]+[(x^{2}+x+0x^{0})*1x^{0}]$. We simplify these equations to be $[x^{3}+x^{2}+0x^{1}]+[x^{2}+x+0x^{0}]$ (notice how our integer term drops out). Finally we combine like terms to get the answer x^3 + 2x^2 + x +0x^0. Let's repeat that in the more familiar columnar format of multiplication:

         1x^2 + 1x^1 + 0x^0
*               1x^1 + 1x^0
--------------------------
         1x^2 + 1x^1 + 0x^0
+ 1x^3 + 1x^2 + 0x^1
--------------------------
= 1x^3 + 2x^2 + 1x^1 + 0x^0 = x^3 + 2x^2 + x

By breaking a polynomial into its terms

If we have a polynomial P(x)

$P(x)=a_{n}x^{n}+...+a_{i}x^{i}+...a_{2}x^{2}+a_{1}x+a_{0}\$

The only possible rational roots (roots of the form p/q) are in the form

${\frac {p}{q}}={\frac {{\mbox{factor}}\ {\mbox{of}}\ a_{0}}{{\mbox{factor}}\ {\mbox{of}}\ a_{n}}}$ (Also know as the rational root theorem, or RRT)

Binomials

binomial is a sum or difference of two monomials. These can also be called polynomials, but to specify, these are binomials.

Examples

2x + 2

2y - 7

How to factor

To factor binomials, find the greatest common factor between the terms and factor.

Example

4x + 2

The greatest common factor between these terms is 2 because both of the terms can be divided by it and the coefficient and constant is still an integer. The example factored would become:

2(2x+1)

External links

• Online interactive exercises on polynomials.

Quadratic equations

Up to now you have only dealt with equations and expressions involving just x; in this section we'll move onto solving things which have $x^{2}$ in them.

All quadratic equations can be arranged in the form $ax^{2}+bx+c=0\ (a\neq 0)$, and a,b,c are all constants. Now let's look at some examples:

Examples: Rearrange the following equations in the form $ax^{2}+bx+c=0\qquad$:

$(1)\qquad x(x-3)=3-5x;\qquad$

Solution for (1):

$x^{2}-3x=3-5x\qquad$

$x^{2}+2x=3\qquad$

$x^{2}+2x-3=0\qquad$

Note that in the first step you distributed the x on the left side of the equation. The second step was obtained by adding a 5x to both sides of the equation and subsequently subtracting a 3 from both sides of the equation.

$(2)\qquad 2x+1=(x^{2}+2){\sqrt {3}}$

Solution for (2):

${\begin{matrix}2x+1&=&x^{2}{\sqrt {3}}+2{\sqrt {3}}\\-x^{2}{\sqrt {3}}+2x+1-2{\sqrt {3}}&=&0\\x^{2}{\sqrt {3}}-2x-1+2{\sqrt {3}}&=&0\end{matrix}}$

Note that in the last step, both sides are multiplied by -1, to make the term $-x^{2}{\sqrt {3}}$ positive, so that the solving of the equation would be easier.

Factorization

Factorization is the most common way to solve quadratic equations. Let us consider again the first example above: $x(x-3)=3-5x$ We have already simplified the equation into

$x^{2}+2x-3=0$

Now, we want to factorize the equation - that is to say, get it into a form such as:

$(x+''{\text{something}}'')(x+''{\text{something else}}'')=0$

Look at the number term c. In this example, it is -3. Now, if we are lucky, the numbers "something" and "something else" will turn out to be nice whole numbers, so let's think of two numbers that will multiply together to give -3. Either 3 and -1, or -3 and 1. But we also need to get the x term correct (here, b=2). In fact, we need our two factors of c to add together to make b. And (3)+(-1)=2. So, we have found our 'somethings': they are 3 and -1. Let's fill them in.

$(x+3)(x-1)=0$

Just to check, we can multiply out the brackets to check we have what we started with:

${\begin{matrix}(x+3)(x-1)=0\\x^{2}+3x-1x-3=0\\x^{2}+2x-3=0\end{matrix}}$

Now, we know that in an equation the left side is always equal to the right side. And in this case the right side of the equation is 0, so from that we can conclude the term $(x+3)(x-1)$ must equal to zero as well. And that means that either $(x+3)$ or $(x-1)$ must equal zero. (Not convinced? Remember (x+3) and (x-1) are just numbers. Can you find two non-zero numbers which multiply to make zero?)

Let's write that algebraically:

${\begin{matrix}x^{2}+2x-3=0\\x+3=0\qquad {\mbox{or}}\qquad x-1=0\\x=-3\qquad {\mbox{or}}\qquad x=1\end{matrix}}$

Thus, there are two different solutions to the same equation! This is the case for all quadratic equations. We say that this quadratic equation has two distinct and real roots.

With practice, you will often be able to write down the equation in factorised form almost immediately. Here is another example, in this case the x easily factorises out:

${\begin{matrix}2x^{2}&=9x\\2x^{2}-9x&=0\\x(2x-9)&=0\\x=0&\qquad {\mbox{or}}\qquad x={\frac {9}{2}}\end{matrix}}$

Completing the square

Sometimes the roots (solutions) of a quadratic equation cannot be easily obtained by factorisation. In such cases, we have to solve the equation by completing the square, or using the quadratic formula (see below).

In order to complete the square, we need to rewrite the given equation in the form $(x+a)^{2}=b$. Now here is an example:

${\begin{matrix}x^{2}+8x+9&=&0\\x^{2}+8x&=&-9\\x^{2}+8x+4^{2}&=&4^{2}-9\\(x+4)^{2}&=&7\\&x+4={\sqrt {7}}&\quad {\mbox{or}}\quad x+4=-{\sqrt {7}}\\&x=-4+{\sqrt {7}}&\quad {\mbox{or}}\quad x=-4-{\sqrt {7}}\\&x=-1.35&\quad {\mbox{or}}\quad x=-6.65\end{matrix}}$

In general, we get

$x^{2}+kx+\left({k \over 2}\right)^{2}=\left(x+{k \over 2}\right)^{2}$

Note that when we reach the stage of taking the square root of both sides of the equation, we might have a negative left-hand side. In this case, the roots will be complex. If you have not yet learned about complex numbers, it is possible to simply state that the equation "has no real roots".

Quadratic Formula

The quadratic formula is a special generalization of completing the square that allows the two roots of a quadratic equation to be obtained by simple substitution. It can be used to solve any quadratic equation and is very quick to work out on a calculator.

$ax^{2}+bx=-c\Rightarrow x^{2}+{\frac {b}{a}}x={\frac {-c}{a}}$

Complete the square:

$x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}={\frac {-c}{a}}+\left({\frac {b}{2a}}\right)^{2}$

Simplify:

Which equals 4y to the 19th power.

$\left(x+{\frac {b}{2a}}\right)^{2}={\frac {-c}{a}}+\left({\frac {b}{2a}}\right)^{2}$

$x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}+{\frac {-c}{a}}}}$

$x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}+{\frac {-4ac}{4a^{2}}}}}$

$x={\frac {-b}{2a}}\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}$

$x={\frac {-b}{2a}}\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}$

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

which is the desired form of the quadratic formula.

Hence, given that a quadratic is in the form $ax^{2}+bx+c=0$, the two roots are:

$x={-b\pm {\sqrt {b^{2}-4ac}} \over 2a}$

The quantity ${b^{2}-4ac}$ in the equation, known as the discriminant, is an indication of the solubility and nature of the roots:

• discriminant is positive—soluble over R, real roots
• discriminant is zero—soluble over R, real repeated (single) roots
• discriminant is negative—insoluble over R (yet soluble over C), no real roots

Weda's Theorem

If the quadratic equation $ax^{2}+bx+c=0\ (a\neq 0)$ has two real roots $x_{1}$ and $x_{2}$, then

$\left\{{\begin{matrix}x_{1}+x_{2}&=&-{\cfrac {b}{a}}\\x_{1}x_{2}&=&{\cfrac {c}{a}}\end{matrix}}\right.$

This is because $x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$ and $x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$. By simply adding or multiplying the two roots we will get the above two equations. This is called Weda's Theorem.

Using Weda's Theorem we can find the second root of a given quadratic equation without solving the equation.

Example: Given that one of the real roots of the equation $4x^{2}-13x+10=0$ is 2, find the other root without solving the equation.

Solution:

${\begin{matrix}x_{1}\cdot x_{2}&=&{\cfrac {c}{a}}\\\Rightarrow x_{1}\cdot 2&=&{\cfrac {10}{4}}\\x_{1}&=&{\cfrac {5}{4}}\end{matrix}}$

We can also determine the signs of two roots by applying the following rules:

1. the equation has two positive roots if $\Delta \geq 0,\ {\frac {c}{a}}>0,\ {\mbox{and}}\ {\frac {b}{a}}<0$;
2. the equation has two negative roots if $\Delta \geq 0,\ {\frac {c}{a}}>0,\ {\mbox{and}}\ {\frac {b}{a}}>0$;
3. the equation has two roots with different signs if $\ {\frac {c}{a}}<0$

($\Delta$ represents the discriminant of the equation.)

Another problem involving Weda's Theorem:

Example: For the equation $2x^{2}+ax+1=0$, given that the sum of squares of roots is $7{\frac {1}{4}}$, find the value of $a$.

Solution:

${\begin{matrix}(x_{1}+x_{2})^{2}-2x_{1}x_{2}&=&x_{1}^{2}+x_{2}^{2}\\\left(-{\cfrac {a}{2}}\right)^{2}-2\cdot {\cfrac {1}{2}}&=&7{\cfrac {1}{4}}\\{\cfrac {a^{2}}{4}}&=&8{\cfrac {1}{4}}\\a^{2}&=&33\\a&=&\pm {\sqrt {33}}\end{matrix}}$

Pythagorean Theorem ____________________

a^+b^=c^

Solving simultaneous linear and nonlinear equations

In previous chapters you have already learned how to solve simultaneous linear equations. Now we will learn how to solve a system of simultaneous linear and non-linear equations with two unknowns. It is usually done by substitution method.

Example: Solve the following simultaneous equations:

$\left\{{\begin{matrix}2x^{2}+y^{2}-5xy&=8&{\mbox{------ (1)}}\\y-x&=2&{\mbox{------ (2)}}\end{matrix}}\right.$

Solution:

${\begin{matrix}{\mbox{Rearrange (2):}}&y=x+2&{\mbox{------ (3)}}\\{\mbox{Substitute(3) into (1):}}&2x^{2}+(x+2)^{2}-5x(x+2)=8\\&2x^{2}+x^{2}+4x+4-5x^{2}-10x-8=0\\&x^{2}+3x+2=0\\&(x+2)(x+1)=0\\&x=-2\quad {\mbox{or}}\quad -1\\{\mbox{Substitute x=-2 back into (3):}}&y=-2+2=0\\{\mbox{Substitute x=-1 back into (3):}}&y=-1+2=1\end{matrix}}$

∴ x=-1 and y=1, or x=-2 and y=0.

Stemplots

The next thing we need to look at to properly read the plot is the key. Lets look at our example again.

Scores of Our Last Math Test

6|446
7|02668
8|022244888
9|026

9|2 = 92%

The key in this case is:

9|2 = 92%

What does the key tell us? It tells us, that in this case, the data we are looking at are percentages. It is always important to know what you are looking at, and not just be a robot and read numbers off of a plot. Now that we understand the title and the key, we know that by reading the data off of our plot, we are reading off grades in percentages.