Equations and inequations : lesson for 11th grade.

11th grade math lessons

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1 I. Solving the second degree equation :
- 1.1 1. Definition and vocabulary :
- 1.2 2. Solving the second degree equation :
  - 1.2.1 2.1. Writing f(x) in canonical form:
  - 1.2.2 2.2. Solving the equation ax²+bx+c=0 :
2 II. Factorisation and sign of the trinomial :
- 2.1 1. Factorisation of the trinomial :
- 2.2 2. Sign of the trinomial :
3 III. Graphical representations of trinomial functions :

The equations and inequations of the second degree in a math lesson for 11th grade where we will approach the resolution with the delta discriminant and the factorization of a polynomial of the second degree as well as the study of its sign. In this first lesson, we will study graphical interpretation.

In all this chapter, we will consider a a non-zero real.

I. Solving the second degree equation :

1. Definition and vocabulary :

Definition:

A second degree equation, with one unknown x, is an equation that can be written as , where $a,\,b,\,c$ are three real numbers given with $a\neq0$ .
Solving the equation $ax^2+bx+c\,=\,0$ , 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 $a\neq0$ .

2.1. Writing f(x) in canonical form:

Definition:

Since $a\neq0$ , $f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})$ or $x^2+\frac{b}{a}x=(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}$

so $f(x)=a%5B(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2}%5D$ .

The latter is called the canonical form of f.

2.2. Solving the equation ax²+bx+c=0 :

Ownership:

We pose $\Delta\,=b^2-4ac$ as follows $f(x)=a%5B(x+\frac{b}{2a})^2-\frac{\Delta\,}{4a^2}%5D$

Case 1:

If $\Delta\,<0$ then $\frac{\Delta,}{4a^2}<0$ .

The number in square brackets is strictly positive so the equation f(x)=0 has no solution.

Second case:

If $\Delta\,=0$ then $f(x)=a(x+\frac{b}{2a})^2$ .

Since $a\neq,0$ , the equation f(x)=0 has one and only one solution:

$x=-\frac{b}{2a}$ .

Third case:

If $\Delta\,>0$ then $\Delta\,=(\sqrt{\Delta\,})^2$ and :

$f(x)=a%5B(x+\frac{b}{2a})^2-\frac{\Delta\,}{4a^2}%5D\,\\f(x)=a%5B(x+\frac{b}{2a})^2-(\frac{\sqrt{\Delta}\,}{2a})^2%5D\,\\f(x)=a(x+\frac{b}{2a}+\frac{\sqrt{\Delta}\,}{2a})(x+\frac{b}{2a}-\frac{\sqrt{\Delta}\,}{2a})$

$f(x)=a(x-\frac{-b-\sqrt{\Delta}\,}{2a})(x-\frac{-b+\sqrt{\Delta}\,}{2a}-)$

If we pose :

$x_1=\frac{-b-\sqrt{\Delta}\,}{2a}$ and $x_2=\frac{-b+\sqrt{\Delta}\,}{2a}$ then .

So since $a\neq\,0$ , the equation f(x)=0 has two distinct solutions x_1 and x_2 .

Definition:

The number b^2-4ac is called the discriminant of the second degree equation or the trinomial .

It is noted $\Delta$ (read “delta”).

Theorem:

a. When $\Delta\,<0$ , the equation has no solution in $\mathbb{R}$ .

b. When $\Delta\,=0$ , the equation has a double root: $x_1=\frac{-b\,}{2a}$ .

c. When $\Delta\,>0$ , the equation has two solutions:

$x_1=\frac{-b-\sqrt{\Delta}\,}{2a}$ and $x_2=\frac{-b+\sqrt{\Delta}\,}{2a}$ .

II. Factorisation and sign of the trinomial :

1. Factorisation of the trinomial :

We have seen in the proof of Theorem 1 that if

$\Delta\,>0$ then .

Theorem 2 : factorization of the trinomial.

When the equation f(x)=ax²+bx+c=0 has two solutions x_1 and x_2 ( in the case $\Delta\,>0$ ) then,

2. Sign of the trinomial :

Theorem

When $\Delta\,<0$ , is always of the sign of a.
When $\Delta\,=0$ , has the sign of a
When $\Delta\,<0$ , 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 :

Definition:

The curve of the function $f\,:\,x\,\mapsto \,ax^2+bx+c$ is a parabola. This parabola faces upwards when a>0 and faces downwards when a<0 .

Synopsis:

Examples:

Solve

Solution:

$\Delta\,=-3$ since $\Delta\,<0$ , the trinomial has no root in $\mathbb{R}$ .

Moreover a=1 so a>0 so for any real and $S=\mathbb{R}$ .

Solve the second degree inequation $-x^2+3x-2\,\geq\,\,0$ .

We have $\Delta\,=3^2-4\times \,(-1)\times \,(-2)=9-8=1$ .

The equation has two roots which are t:

$x_1=\frac{-3+\sqrt{1}}{2\times \,(-1)}=\frac{-2}{-2}=1$ = 1 and $x_2=\frac{-3-\sqrt{1}}{2\times \,(-1)}=\frac{-4}{-2}=2$ .

We have so the solution set of the second degree inequation $-x^2+3x-2\,\geq\,\,0$ is the interval $%5B1\,;2%5D$ .

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Equations and inequations of the second degree: 11th grade math lesson

I. Solving the second degree equation :

1. Definition and vocabulary :

2. Solving the second degree equation :

2.1. Writing f(x) in canonical form:

2.2. Solving the equation ax²+bx+c=0 :

II. Factorisation and sign of the trinomial :

1. Factorisation of the trinomial :

2. Sign of the trinomial :

III. Graphical representations of trinomial functions :

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