Esericizi OFA di Analisi matematica 1

SECOND DEGREE EQUATIONS

An equation of the 2nd degree is said to be written in normal or canonical form if it is in the form ax + bx + c = 0 with real a, b and c and a ≠ 0.

It is called a discriminant of an equation of 2nd degree, and is denoted by Δ, the number b - 4ac.

The solutions are derived from the formula ±Δ = x1,2 = (-b ± √Δ) / 2a

EQUATIONS:

• If Δ > 0 the solutions are 2 and distinct - S = {(-b + √Δ) / 2a, (-b - √Δ) / 2a}
• If Δ = 0 the solutions are 2 coincident - S = {-b / 2a}
• If Δ < 0 the solutions do not exist - S = {Ø}

EQUATIONS SECOND DEGREE EQUATIONS - SPECIAL CASES:

Case I:

Given the equation ax = b, if a ≠ 0, then x = b/a

Case II:

Given the equation (x - a) ∙ (x - b) = 0 for the law of cancellation of the product you have x - a = 0 → x = a and x - b = 0 → x = b

Solutions are the numbers in the set {a, b}

Consider sets A = {0, 2, 4}, B = {1, 2, 3}, C = {x: x is odd}

(A U B) ∩ B = ?

The number 308 can be

EQUIVALENCE

Second principle of equivalence:

By multiplying or dividing the two members of an inequality by the same literal algebraic expression (or the same number other than 0) we obtain an inequality equivalent to that date, provided that we change the direction of the inequality in the event that that quantity is negative.

Example: 4x + 5 > 17 has x>3 as solution

Multiply by 2 both members

2(4x + 5) > 2(17 )

Get 8x + 10 > 34 which still has x > 3 as its solution

INEQUALITIES

You can graphically represent the set of solutions of an inequality on an oriented line. We adopt the convention of using an empty dot in case the limit value is not part of the solution set.

If the limit value is part of the solution set we use a full dot.

INEQUALITIES OF 1st DEGREE

In any inequality, a term may be transported from one member to another as long as it is changed in sign (Law of Carriage).

To solve a first-degree inequality to an unknown, the following steps must be developed:

a) full-form

1. Inequality is reduced;
2. Any parentheses are eliminated, performing the operations indicated;
3. The terms containing the unknown are moved to the first member and the terms known to the second member;
4. Inequality in normal form is reduced;
5. If the coefficient of x is not positive, it is made positive by multiplying by -1 all the inequality (thus changing both all the signs and the direction of the inequality);
6. You will find the set of solutions;
7. INEQUALITIES OF 1st DEGREE
   A first-degree inequality always admits infinite solutions real numbers
   GRADE II INEQUALITIES
   In mathematics, a second-degree or quadratic inequality is an algebraic inequality with a single unknown x that appears with a maximum degree of 2, and whose expression can be traced back to the form: ax^2+bx+c>0 ax^2+bx+c<0 a≠0, a,b,c real numbers
   GRADE II INEQUALITIES
   ax^2+bx+c>0 (ax^2+bx+c<0)
   RESOLUTION
   Step 1: check the sign of the coefficient a. In the event that a < 0, change the sign to all

The terms of the inequality and change the verse of the inequality.

Step 2: determine the real roots x1 and x2 (if they exist) of the equation of II degree associated ax^2+bx+c=0 with the formula ± √(b^2-4ac) = x1,2 / 2a

GRADE II INEQUALITIES

CASE I

Δ > 0, a > 0

Two real and distinct solutions

CASE II

Δ = 0, a > 0

Two real and coincident solutions

If the two solutions coincide:

CASE III

Δ < 0, a > 0

No real solution

FRACTIONAL INEQUALITIES

An inequality is called fractional if the unknown appears in the denominator. To solve the inequalities fratte it is not enough to multiply both members by the common denominator as in the case of equations because it is necessary to take into account the sign of the same.

The procedure is as follows:

1. Establish the conditions of existence of the inequality, placing the denominators non-zero.
2. Bring the inequality in normal form. We then move all the terms to the left of the inequality sign, leaving zero

1. We compare the demand for inequality (that is, the verse of the inequality symbol) with what is obtained in the sign scheme, and determine the set of solutions of the inequality.

DISEQUAZIONI FRAZIONARIE

The set of solutions of the inequality is:

S: { x<-2 x>4} (-∞, -2) U (4, ∞)

ANALYTIC GEOMETRY

CARTESIAN PLANE

Given by a pair of orthogonal oriental straights:

• horizontal (x): abscis axis
• vertical (y): ordinate axis

They meet at a point that we call the origin of the axes and denoble with 0.

We fix a unit of measurement on both axes: each point of the abscis

axis and also each point of the ordinate axis determines and is determined by a real number.

The point P is identified with the ordered pair (xp,yp). The number xp is called abscissa of P, while the number yp is called ordinate of P.

The two coordinate lines divide the plane into four quadrants. The first is the one consisting of the points with positive absciss and positive ordinates, the rest are numbered following the counterclockwise direction.

ANALYTIC GEOMETRY

SEGMENTS P=(x1,y1) e Q=(x2,y2)

Midpoint

Segment length PQ

PH = | x2 - x1 |

HQ = | y2 – y1 |

BARYCENTRE

STRAIGHT LINE – implicit equation

A straight line in the Cartesian plane is the place of the points (x,y) that satisfy an equation of the type ax+by+c=0, for certain values of a,b,c R, with a and b not both equal to zero, i.e. (a,b)≠(0,0).

For certain particular values of a,b,c the equation line ax+by+c=0 verifies viewable geometric properties:

- // to the abscis axis

- // to the ordinate axis

- goes through the

originSTRAIGHT LINE – implicit equation

EQUATION OF THE LINE THROUGH TWO POINTS

In order to draw a straight line, it is sufficient to find two points that belong to it: in fact, given two points there is a single line that passes through them.

Given two points of the plane there is a single straight line passing through them. If the coordinates of two points P=(x1,y1) and Q=(x2,y2) are given, the equation of the only straight line through P and Q is given by (y−y1)(x2−x1)=(y2−y1)(x−x1)

From the example: P=(0,1/2) Q=(1,0) (y−1/2)(1−0)=(0−1/2)(x−0) y-1/2=-1/2x+ x + 2y -1 = 0

STRAIGHT LINE – implicit equation

PARALLELISM AND PERPENDICULARITY

Given a pair of lines it is possible to establish from their equation whether they are parallel or perpendicular.

Two lines are called PARALLEL if they do not have points in common (distinct parallels) or if they coincide (coincident parallels). Given two lines a1x+b1y+c1=0 and a2x+b2y+c2=0 they are

parallel if and only if: (a1,b1)=(ra2,rb2) for an appropriate real number ra1 b2 = b1 a2.

Two straights are called PERPENDICULAR when they are accidents and the angle formed by them is right, or 90°.

Two straights a1x+b1y+c1=0 and a2x+b2y+c2=0 are perpendicular if and only if a1 a2 + b1 b2=0.

STRAIGHT LINE – implicit equation

MUTUAL PROPOSITIONS

In general, given two straight lines r:a1x+b1y+c1=0 and s:a2x+b2y+c2=0 they may be in the following reciprocal positions:

1. coincident parallels, i.e. r=s;
2. distinct parallels, i.e. r || s, r≠s;
3. incident, if they are not coincident or parallel, and therefore meet in only one place;
4. perpendicular, or r ⊥ s: they are accidents and form an angle of 90°.

In algebraic terms, we consider the system of two equations in the two unknowns x,y

If the system does not allow solutions, the lines are parallel and not coincident.

If it admits only one solution they are accidents (and are perpendicular if a1a2+b1b2=0). In this case, the system