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

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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 :

  1. A second degree equation, with one unknown x, is an equation that can be written as ax^2+bx+c=0, where a,\,b,\,c are three real numbers given with a\neq0.
  2. Solving the equation ax^2+bx+c\,=\,0, is to find all the numbers p such that ap^2+bp+c=0.
  3. 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:


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 :


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:


Third case:

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



If we pose :

x_1=\frac{-b-\sqrt{\Delta}\,}{2a} and x_2=\frac{-b+\sqrt{\Delta}\,}{2a} then f(x)=a(x-x_1)(x-x_2).

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


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

It is noted \Delta (read “delta”).


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 f(x)=a(x-x_1)(x-x_2).

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 :

  1. When \Delta\,<0, f(x) is always of the sign of a.
  2. When \Delta\,=0, f(x) has the sign of a
  3. When \Delta\,<0, f(x) has the sign of a, except when x is between the roots, in which case f(x) and a have opposite signs.


To solve a second degree inequation, we determine the sign of the associated trinomial.

III. Graphical representations of trinomial functions :


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


synthesis trinomial equation second degree


Solve x^2+3x+3>0


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

Moreover a=1 so a>0 so x^2+3x+3>0 for any real x 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 -x^2+3x-2=0 has two roots which are t:

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

We have a=-1 so a<0 the solution set of the second degree inequation -x^2+3x-2\,\geq\,\,0 is the interval %5B1\,;2%5D.

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