Quadratic equation

The quadratic equation , quadratic equation or equation known as of the “second degree” is form

$ax^2\; +\; bx\; +\; C\; =\; 0\; \backslash ,$ where $a$, $b$ and $c$ are real coefficients or complex, $a$ not no one. The unknown factor $x$ can be real or complex.

History

The equation S of the second degree were posed at the Babylonian (one then sought a positive solution using an algorithm), at the Égyptiens, even at the Greeks (Livre II of the Elements of Euclide), but no civilization of this time explicitly studied the equations.

The quadratic equations were the first solved equations, the mathematical equation is invented at the same time as the Algèbre by the Moslem Iranian scientist Al-Khwarizmi at the 9th century, which took again this tradition, increased Greek knowledge for the demonstration, in order to find a solution (real and positive). The equations were presented under one of the following forms because a number was supposed to be positive:

$ax^2\; =\; bx\; +\; c$ $ax^2\; +\; bx\; =\; c$ $ax^2\; +\; C\; =\; bx$

Until the Rebirth, the algebra used neither symbols nor letters, and was purely verbal.

Solution of a quadratic equation with real coefficients

The incomplete equations $ax^2\; +\; B\; =\; 0$ and $ax^2\; +\; bx\; =\; 0$ are solved by simple special methods.

An equation of the form $ax^2\; +bx\; +\; C\; =\; 0$, of which all the coefficients are nonnull, is known as “complete equation”.

To facilitate the writing, one poses the function then defined by $f\; (X)\; =\; ax^2\; +\; bx\; +\; c$, then one carries out his reduction to solve finally the equation $f\; (X)\; =\; 0$. Here, one proposes to factorize using the Discriminant Δ.

Canonical form and Discriminant Δ

One proposes to use for that the remarkable Identités:

One calls this form of writing the form Canonique of the trinomial one.

That is to say $\backslash \; Delta\; =\; b^2\; -\; 4ac$. $\backslash \; Delta$ (delta) is called the Discriminant of this trinomial.

If Δ > 0

If $\backslash \; Delta\; >\; 0$, one can factorize $f\; (X)$ using a remarkable identity: then the equation $f\; (X)\; =0$ has two distinct real roots $x\_1$ and $x\_2$:

• $x\_1\; =\; \backslash \; frac\; \{-\; B\; +\; \backslash \; sqrt\; \{\backslash \; Delta\}\}\; \{2a\}$
• $x\_2\; =\; \backslash \; frac\; \{-\; B\; -\; \backslash \; sqrt\; \{\backslash \; Delta\}\}\; \{2a\}$

The factorized form of $f\; (X)$ is finally:

$f\; (X)\; =\; has\; (X\; -\; x\_1)\; (X\; -\; x\_2)$.

If Δ 0

If $\backslash \; Delta\; =\; 0$, one can write, consequently method, that $f\; (X)\; =\; 0$ is equivalent saying that: $\backslash \; left\; (X\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; \backslash \; right)\; ^2\; =\; 0$

The equation has a real root then doubles $x\_0$:

$x\_0\; =\; -\; \backslash \; frac\; \{B\}\; \{2a\}$

One can then factorize the function $f$ as follows:

$f\; (X)\; =\; has\; (x-x\_0)\; ^2~$

However, this writing is a remarkable identity; thus, any remarkable identity of the form $(a-b)\; ^2$ has as a discriminant 0, and its double root can be easily found, without same calculating the discriminant.

If Δ < 0

It is pointed out that the canonical form of trinomial is: $f\; (X)\; =\; has\; \backslash \; left\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; \backslash \; right)\; ^2\; -\; \backslash \; frac\; \{\backslash \; Delta\}\; \{4a^2\}\; \backslash \; right$, thus to solve $f\; (X)\; =\; 0$ amounts solving:

$\backslash \; left\; (X\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; \backslash \; right)\; ^2\; -\; \backslash \; frac\; \{\backslash \; Delta\}\; \{4a^2\}\; =\; 0$

Resolution as a whole of realities

That is to say $\backslash \; alpha$ a reality, such as: $\backslash \; alpha\; =\; \backslash \; left\; (X\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; \backslash \; right)\; ^2$.

One knows that a square cannot be negative, therefore $\backslash \; alpha\; \backslash \; Ge\; 0$. But implies $-\; \backslash \; Delta\; >\; 0$, but the sum of two positive numbers of which one is strictly positive is not never equal to zero. Therefore, if , there exists no root real with the trinomial one.

Resolution as a whole of the complex

However, there exist two complex roots $z\_1$ and $z\_2$. Knowing that $\backslash \; Delta\; =\; b^2\; -\; 4ac$ and that $\backslash \; Delta\; =\; -\; \backslash \; Delta\; .i^2$, pose $\backslash \; delta\; =\; \backslash \; sqrt\; \{-\; \backslash \; Delta\}\; =\; \backslash \; sqrt\; \{4ac\; -\; b^2\}$. Thus, $\backslash \; Delta\; =\; \backslash \; delta^2.\; i^2$.

By taking again factorization already used if $\backslash \; Delta\; >\; 0$, one finds:

$z\_1\; =\; \backslash \; frac\; \{-\; B\; -\; I\; \backslash \; delta\}\; \{2a\}$ and $z\_2\; =\; \backslash \; frac\; \{-\; B\; +\; I\; \backslash \; delta\}\; \{2a\}$.

The function is factorized then:

$az^2+bz+c=a\; (z-z\_1)\; (z-z\_2)\; ~$

Let us notice that in all the cases, a polynomial of the second degree has two roots: either two distinct real roots, or two confused real roots (i.e. a root doubles), or two complex roots (consequence of the Théorème of Alembert-Gauss).

Examples

• $x^2\; +\; 3\; X\; +\; 3\; =\; 0$ does not have a solution in the whole of realities because . However, in the whole of the complexes, she admits two solutions $z\_1$ and $z\_2$ such as $z\_1\; =\; \backslash \; frac\; \{-\; 3\; -\; \backslash \; sqrt\; \{3\}\; I\}\; \{2\}$ and $z\_2\; =\; \backslash \; frac\; \{-\; 3\; +\; \backslash \; sqrt\; \{3\}\; I\}\; \{2\}$
• $x^2\; -2x\; +\; 1\; =\; 0$ thus has a discriminant null Δ has as a double solution $x\_0=-\; \backslash \; tfrac\; \{-\; 2\}\; \{2\}\; =1$
• $7x\; +\; 15\; -\; 2x^2\; =\; 0$ has a discriminant $\backslash \; Delta\; =\; 169$ strictly positive thus admits two solutions: $x\_1=\; \backslash \; frac\; \{-\; 7\; \backslash \; sqrt\; \{169\}\}\; \{2\; \backslash \; times\; (-\; 2)\}=\; \backslash \; frac\; \{-\; 7-13\}\; \{-\; 4\}\; =\; \backslash \; frac\; \{20\}\; \{4\}\; =\; 5$ and $x\_2=\; \backslash \; frac\; \{-\; 7+\; \backslash \; sqrt\; \{169\}\}\; \{2\; \backslash \; times\; (-\; 2)\}\; =\; \backslash \; frac\; \{-\; 7+13\}\; \{-\; 4\}\; =\; \backslash \; frac\; \{6\}\; \{-\; 4\}\; =\; \backslash \; frac\; \{3\}\; \{2\}$.

By using the obvious roots

The roots of a polynomial of the second degree have several interesting properties - called relations of Viète - and which can simplify their research. Either $S$ the sum of the roots, one has $S\; =\; x\_1\; +\; x\_2\; =\; \backslash \; frac\; \{-\; B\; +\; \backslash \; sqrt\; \{\backslash \; Delta\}\}\; \{2a\}\; +\; \backslash \; frac\; \{-\; B\; -\; \backslash \; sqrt\; \{\backslash \; Delta\}\}\; \{2a\}\; =\; \backslash \; frac\; \{-\; 2b\}\; \{2a\}\; =\; -\; \backslash \; frac\; \{B\}\; \{has\}$

Either $P$ the product of the roots, one has

It is thus very easy to calculate these two values. And as soon as one found one of the two roots of a polynomial (by doing a little mental Calcul and by testing values simple to calculate like 0,1,2,-1…), the second root becomes obvious: $x\_2\; =\; S\; -\; x\_1\; =\; -\; \backslash \; tfrac\; \{B\}\; \{has\}\; -\; x\_1$ or $x\_2\; =\; \backslash \; tfrac\; \{P\}\; \{x\_1\}\; =\; \backslash \; tfrac\; \{C\}\; \{ax\_1\}$. Thus, with trinomial the $x^2\; +\; 3x\; -\; 4$, one finds like first $x\_1\; root\; =\; 1$ and as $\backslash \; tfrac\; \{C\}\; \{has\}\; =\; -4$, one does not need even more to calculate to find the second $x\_2\; root\; =\; -4$. Finally, the use of obvious roots and the properties of roots of a polynomial makes it possible to largely accelerate the search for these roots.

Note:

• If $c$ is null, 0 is obvious root of the polynomial.
• If $a\; +\; B\; +\; C\; =\; 0$, 1 is obvious root of the polynomial.
• If $a\; -\; B\; +\; C\; =\; 0$, -1 is obvious root of the polynomial.
• If $4a\; +\; 2b\; +\; C\; =\; 0$, 2 is obvious root of the polynomial.
• If $4a\; -\; 2b\; +\; C\; =\; 0$, -2 is obvious root of the polynomial

Profit of precision in the numerical resolution

When $\backslash \; Delta\; >\; 0$, if $b$ is positive, the expression of $x\_1$ results in calculating the difference of the two numbers $\backslash \; sqrt\; \{\backslash \; Delta\}$ and $b$. If this calculation is made numerically, by a method of Floating decimal point on a Ordinateur for example, that involves a loss of precision, all the more serious as $\backslash \; sqrt\; \{\backslash \; Delta\}$ is very close to $b$, or that $4ac$ is small compared to $b^2$.

Using the properties of the roots, one calculates $x\_1$ without loss of precision:

$x\_1\; =\; \backslash \; frac\; C\; \{has\; x\_2\}$

If $b$ is negative, one calculates $x\_2$ such as:

$x\_1\; =\; \backslash \; frac\; \{-\; B\; +\; \backslash \; sqrt\; \{\backslash \; Delta\}\}\; \{2a\}$ $x\_2\; =\; \backslash \; frac\; C\; \{has\; x\_1\}$

Reduced discriminant

If $b$ is even, one can use the reduced discriminant.

One poses $b\text{'}\; =\; \backslash \; tfrac\; \{B\}\; \{2\}$

Reduced discriminant: $\backslash \; Delta\text{'}\; =\; b\text{'}^2\; -\; ac$

If $\backslash \; Delta\text{'}\; >\; 0$, the solutions is $x\_1$ and $x\_2$:

$x\_1\; =\; \backslash \; frac\; \{-\; b\text{'}\; +\; \backslash \; sqrt\; \{\backslash \; Delta\text{'}\}\}\; \{has\}$ $x\_2\; =\; \backslash \; frac\; \{-\; b\text{'}\; -\; \backslash \; sqrt\; \{\backslash \; Delta\text{'}\}\}\; \{has\}$

If $\backslash \; Delta\text{'}\; =\; 0$, there is a root $x\_0\; =\; \backslash \; tfrac\; \{doubles\; -\; b\text{'}\}\; \{has\}$

Solution of a quadratic equation with complex coefficients

One writes

One distinguishes two cases according to whether the discriminant $\backslash \; Delta=b^2\; -\; 4ac$ is null or not.

If $\backslash \; Delta\; \backslash \; neq\; 0$ then one can pose $\backslash \; delta^2=\; \backslash \; Delta$ and one obtains factorization then:

$f\; (X)\; =\; has\; \backslash \; left\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; -\; \backslash \; frac\; \{\backslash \; delta\}\; \{2a\}\; \backslash \; right)\; \backslash \; left\; (X\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; +\; \backslash \; frac\; \{\backslash \; delta\}\; \{2a\}\; \backslash \; right)\; \backslash \; right$

One from of deduced that the equation admits two solutions:

$\backslash \; frac\; \{-\; b+\; \backslash \; delta\}\; \{2a\}$ and $\backslash \; frac\; \{-\; B\; \backslash \; delta\}\; \{2a\}$

If $\backslash \; Delta=0$ then

$f\; (X)\; =\; has\; \backslash \; left\; (X\; +\; \backslash \; frac\; \{B\}\; \{2a\}\; \backslash \; right)\; ^2$

The equation admits a single solution

$\backslash \; frac\; \{-\; B\}\; \{2a\}$

Remark : The solutions of a quadratic equation with complex coefficients are in general two complex numbers which are not combined, contrary to the case of a quadratic equation with real coefficients whose discriminant is strictly negative.

Internal bonds

• Function of the second degree
• Polynomial
• polynomial Equation

Bonds

Simple: Quadratic equation

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