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In all this chapter, we will consider a a non-zero real.
I. Solving the second degree equation :
1. Definition and vocabulary :
- A second degree equation, with one unknown x, is an equation that can be written as
, where
are three real numbers given with
.
- Solving the equation
, is to find all the numbers
such that
.
- Such a number p is called the solution or root of the equation.
2. Solving the second degree equation :
Let f(x) = ax²+bx+c with .
2.1. Writing f(x) in canonical form:
Since ,
or
so .
The latter is called the canonical form of f.
2.2. Solving the equation ax²+bx+c=0 :
We pose as follows
Case 1:
If then
.
The number in square brackets is strictly positive so the equation f(x)=0 has no solution.
Second case:
If then
.
Since , the equation f(x)=0 has one and only one solution:
.
Third case:
If then
and :
If we pose :
and
then
.
So since , the equation f(x)=0 has two distinct solutions
and
.
The number is called the discriminant of the second degree equation
or the trinomial
.
It is noted (read “delta”).
a. When , the equation has no solution in
.
b. When , the equation has a double root:
.
c. When , the equation has two solutions:
and
.
II. Factorisation and sign of the trinomial :
1. Factorisation of the trinomial :
We have seen in the proof of Theorem 1 that if
then
.
When the equation f(x)=ax²+bx+c=0 has two solutions and
( in the case
) then,
2. Sign of the trinomial :
- When
,
is always of the sign of a.
- When
,
has the sign of a
- When
,
has the sign of a, except when x is between the roots, in which case f(x) and a have opposite signs.
Application:
To solve a second degree inequation, we determine the sign of the associated trinomial.
III. Graphical representations of trinomial functions :
The curve of the function is a parabola. This parabola faces upwards when
and faces downwards when
.
Synopsis:
Examples:
Solve
Solution:
since
, the trinomial has no root in
.
Moreover a=1 so a>0 so for any real
and
.
Solve the second degree inequation .
We have .
The equation has two roots which are t:
= 1 and
.
We have so
the solution set of the second degree inequation
is the interval
.
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