Quadratic equation

A quadratic equation is an equation that is in the form

${\displaystyle ax^{2}+bx+c=0\quad }$

with lets write. Here are coefficients ; is the unknown . If there is also , one speaks of a purely square equation . ${\displaystyle a\neq 0}$ ${\displaystyle a,b,c}$ ${\displaystyle x}$ ${\displaystyle b=0}$

Your solutions can be found using the formula

${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

determine. In the real-number domain, the quadratic equation can have zero, one or two solutions. If the expression under the root is negative, there is no solution; if it is zero, there is a solution; if it is positive there are two solutions.

The left side of this equation is the term of a quadratic function (in more general terms: a polynomial of the second degree ) ,; the function graph of this function in the Cartesian coordinate system is a parabola . Geometrically, the quadratic equation describes the zeros of this parabola. ${\displaystyle f(x)=ax^{2}+bx+c}$ ${\displaystyle f(x)=0}$

General form and normal form

The general form of the quadratic equation is

${\displaystyle ax^{2}+bx+c=0\qquad (a\neq 0).}$

In this context , the term quadratic term , linear term and constant term (or absolute term ) of the equation. ${\displaystyle ax^{2}}$ ${\displaystyle bx}$ ${\displaystyle c}$

The equation is in normal form if , that is, if the quadratic term has the coefficient 1. From the general form, the normal form can be obtained by equivalence transformations by dividing by. With the definition ${\displaystyle a=1}$ ${\displaystyle a\neq 0}$

${\displaystyle p={\frac {b}{a}}}$   and   ${\displaystyle \displaystyle q={\frac {c}{a}}}$

the normal form can thus be written as

${\displaystyle x^{2}+px+q=0\,.}$

In the following, quadratic equations with real numbers as coefficients , and or as and are considered first. ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$ ${\displaystyle p}$ ${\displaystyle q}$

Solutions of the quadratic equation with real coefficients

A solution to the quadratic equation is a number that, when substituted for , satisfies the equation . If complex numbers are allowed as solutions, every quadratic equation has exactly two (possibly coincident) solutions, also called the roots of the equation . If you only look at the real numbers, a quadratic equation has zero to two solutions. ${\displaystyle x}$

Number of real zeros

The number of solutions can be determined with the help of the so-called discriminant (from Latin “discriminare” = “differentiate”). In the general case , in the normalized case (for derivation see below): ${\displaystyle D}$ ${\displaystyle D=b^{2}-4ac}$ ${\displaystyle D=p^{2}-4q}$

Position of the parabolas and effects on the number of zeros

The graphic shows the relationship between the number of real zeros and the discriminant:

• (A) Discriminant positive: The parabola has two points of intersection with the -axis, so there are two different real zeros and . ${\displaystyle x}$ ${\displaystyle x_{1}}$ ${\displaystyle x_{2}}$
• (B) Discriminant zero: The parabola has exactly one point of contact with the -axis, namely its vertex . There is therefore exactly one (double) real solution. The quadratic equation can be reduced to the form . ${\displaystyle x}$ ${\displaystyle ax^{2}+bx+c=a\cdot (x-x_{1})^{2}=0}$
• (C) Negative discriminant: The parabola has no intersection with the -axis, there are no real solutions to the quadratic equation. If you allow complex numbers as the basic set for the solutions, you get two different complex solutions. These are conjugated to one another , that is, they have the same real part and their imaginary parts only differ in their sign . ${\displaystyle x}$

Simple special cases

If the coefficient is the linear term or the absolute term , the quadratic equation can be solved by simple equivalence transformations without the need for a general formula. ${\displaystyle b=0}$ ${\displaystyle c=0}$

Missing linear link

The all-square equation with is equivalent to ${\displaystyle ax^{2}+c=0}$ ${\displaystyle a\neq 0}$

${\displaystyle x^{2}=-{\frac {c}{a}}\,.}$

The solutions are

${\displaystyle x_{1,2}=\pm {\sqrt {-{\frac {c}{a}}}}\,.}$

In the real case there are no real solutions for. The complex solutions are then ${\displaystyle {\tfrac {c}{a}}>0}$

${\displaystyle x_{1,2}=\pm \mathrm {i} {\sqrt {\frac {c}{a}}}.}$

For example, the equation has the solutions . The equation has no real solutions, which are complex solutions . ${\displaystyle x^{2}-3=0}$ ${\displaystyle x_{1,2}=\pm {\sqrt {3}}}$ ${\displaystyle 2x^{2}+8=0}$ ${\displaystyle x_{1,2}=\pm 2\mathrm {i} }$

Missing constant term

From the equation , by factoring out , i. i.e., it must or apply. So the two solutions are ${\displaystyle ax^{2}+bx=0}$ ${\displaystyle x(ax+b)=0}$ ${\displaystyle x=0}$ ${\displaystyle ax+b=0}$

${\displaystyle x_{1}=0}$ and ${\displaystyle x_{2}=-{\tfrac {b}{a}}\,.}$

For example, the equation has the solutions and . ${\displaystyle 3x^{2}-2x=0}$ ${\displaystyle x_{1}=0}$ ${\displaystyle x_{2}={\tfrac {2}{3}}}$

Equation in vertex form

The vertex shape

${\displaystyle a(x-d)^{2}+e=0}$

is a variation of the all-square equation . Like this, it can be solved by "backward computing": First you subtract and divide by . this leads to ${\displaystyle ax^{2}+c=0}$ ${\displaystyle e}$ ${\displaystyle a}$

${\displaystyle (x-d)^{2}=-{\frac {e}{a}}}$.

For it follows ${\displaystyle -{\frac {e}{a}}\geq 0}$

${\displaystyle x-d={\sqrt {-{\frac {e}{a}}}}}$or . ${\displaystyle x-d=-{\sqrt {-{\frac {e}{a}}}}}$

The solutions are obtained by adding ${\displaystyle d}$

${\displaystyle x_{1}=d+{\sqrt {-{\frac {e}{a}}}}}$and . ${\displaystyle x_{2}=d-{\sqrt {-{\frac {e}{a}}}}}$

The two complex solutions are obtained for ${\displaystyle -{\frac {e}{a}}<0}$

${\displaystyle x_{1}=d+\mathrm {i} {\sqrt {\frac {e}{a}}}}$and . ${\displaystyle x_{2}=d-\mathrm {i} {\sqrt {\frac {e}{a}}}}$

Example:

${\displaystyle {\begin{aligned}3(x-2)^{2}-5&=0&&|+5;:3\\(x-2)^{2}&={\frac {5}{3}}&&|\pm {\sqrt {\ }}\\x-2&=\pm {\sqrt {\frac {5}{3}}}&&|+2\\x&=2\pm {\sqrt {\frac {5}{3}}}\end{aligned}}}$

Solve with a square extension

In solving with quadratic completion , the binomial formulas are used to convert a quadratic equation in general form or in normal form to the vertex form, which can then be easily solved.

The first or second binomial formula is used in the form . To do this, the quadratic equation is transformed so that the left side has the shape . Then add up on both sides . This is the "square addition". The left side now has the shape and can be transformed to using the binomial formula . Then the equation is in the easy-to-solve vertex form. ${\displaystyle (x\pm d)^{2}=x^{2}\pm 2dx+d^{2}}$ ${\displaystyle x^{2}\pm 2dx}$ ${\displaystyle d^{2}}$ ${\displaystyle x^{2}\pm 2dx+d^{2}}$ ${\displaystyle (x\pm d)^{2}}$

This is best explained using a specific numerical example. The quadratic equation is considered

${\displaystyle 3x^{2}-15x+18=0}$

First, the equation is normalized by dividing by the leading coefficient (here 3):

${\displaystyle x^{2}-5x+6=0}$

The constant term (here 6) is subtracted on both sides:

${\displaystyle x^{2}-5x=-6}$

Now the actual square addition follows: The left side must be completed so that a binomial formula (here the second) can be applied backwards. That from the above binomial formula is then , so both sides of the equation must be added: ${\displaystyle d}$ ${\displaystyle {\tfrac {5}{2}}}$ ${\displaystyle d^{2}=\left({\tfrac {5}{2}}\right)^{2}}$

${\displaystyle x^{2}-5x+\left({\frac {5}{2}}\right)^{2}=\left({\frac {5}{2}}\right)^{2}-6}$

The left side is reshaped according to the binomial formula, the right side simplified:

${\displaystyle \left(x-{\frac {5}{2}}\right)^{2}={\frac {1}{4}}}$

this leads to

${\displaystyle x-{\frac {5}{2}}=\pm {\frac {1}{2}}}$,

so to the two solutions and . ${\displaystyle x_{1}={\frac {5}{2}}+{\frac {1}{2}}=3}$ ${\displaystyle x_{2}={\frac {5}{2}}-{\frac {1}{2}}=2}$

General solution formulas

One can also solve quadratic equations by using one of the general solution formulas derived with the help of quadratic completion.

Solution formula for the general quadratic equation ( abc formula)

The solutions to the general quadratic equation are: ${\displaystyle ax^{2}+bx+c=0}$

${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

In parts of Germany and Switzerland, the formula is colloquially referred to as the "midnight formula" because students should be able to recite it even if you wake them up at midnight and ask for the formula . In Austria, the expression large formula is used .

By expanding with the term , a form of the midnight formula is obtained, which can also be used for the linear case , but which can no longer provide the calculation of the solution because of division by zero. In both cases, the solution formula is not required anyway. However, for very small amounts , the alternative form is more robust to numerical cancellation. ${\displaystyle -b\mp {\sqrt {b^{2}-4ac}}}$ ${\displaystyle a=0}$ ${\displaystyle c=0}$ ${\displaystyle x={\frac {-b}{a}}}$ ${\displaystyle a}$

${\displaystyle x_{1,2}={\frac {\left(-b\pm {\sqrt {b^{2}-4ac}}\right)\cdot \left(-b\mp {\sqrt {b^{2}-4ac}}\right)}{2a\cdot \left(-b\mp {\sqrt {b^{2}-4ac}}\right)}}={\frac {2c}{-b\mp {\sqrt {b^{2}-4ac}}}}}$

If you put the equation in the form

${\displaystyle ax^{2}+2\beta x+c=0}$

indicates (i.e. with ), one obtains the somewhat simpler solution formula ${\displaystyle \beta =b/2}$

${\displaystyle x_{1,2}={\frac {-\beta \pm {\sqrt {\beta ^{2}-ac}}}{a}}}$

Solution of the abc formula in the case of a negative discriminant

If the discriminant introduced above is negative, the root of a negative number has to be calculated for the solutions. There are no solutions for this in the number range of real numbers. In the area of ​​complex numbers, the following applies . This term determines the imaginary part of the two conjugate solutions, one with a positive, one with a negative sign. The term before with becomes the constant real part of the two solutions: ${\displaystyle D=b^{2}-4ac}$ ${\displaystyle {\sqrt {D}}=i{\sqrt {-D}}}$ ${\displaystyle -{\tfrac {b}{2a}}}$

${\displaystyle x_{1,2}=-{\frac {b}{2a}}\pm i\cdot {\frac {\sqrt {4ac-b^{2}}}{2a}}}$ (complex case with negative discriminant).

Derivation of the abc formula

The general shape results from reshaping according to the method of square completion :

${\displaystyle {\begin{array}{rcll}ax^{2}+bx+c&=&0&|-c\\[1ex]ax^{2}+bx&=&-c&|{}\cdot 4a\\[1ex]4a^{2}x^{2}+4abx&=&-4ac&|+b^{2}{\text{ (quadratische Ergänzung)}}\\[1ex](2ax)^{2}+2\cdot 2ax\,b+b^{2}&=&b^{2}-4ac&|{\text{ Umformen mit binomischer Formel}}\\[1ex](2ax+b)^{2}&=&b^{2}-4ac&|\pm {\sqrt {\quad }}\\[1ex]2ax+b&=&\pm {\sqrt {b^{2}-4ac}}&|-b\\[1ex]2ax&=&-b\pm {\sqrt {b^{2}-4ac}}&|:(2a)\\[1ex]x&=&{\dfrac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}&\end{array}}}$

Solution formula for the normal form ( pq formula)

If the normal form is present, the solutions are according to the pq formula ${\displaystyle x^{2}+px+q=0}$

${\displaystyle x_{1,2}=-{\frac {p}{2}}\pm {\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}\,.}$

In Austria this formula is known as a small solution formula .

Solution of the pq formula in the case of a negative discriminant

As with the abc formula , when is negative, there are no solutions in the real number range. The complex solutions then result in: ${\displaystyle {\tfrac {1}{4}}D=\left({\tfrac {p}{2}}\right)^{2}-q}$

${\displaystyle x_{1,2}=-{\frac {p}{2}}\pm i\cdot {\sqrt {q-\left({\frac {p}{2}}\right)^{2}}}}$

Derivation of the pq formula

The formula results from the normal form of the quadratic equation by adding the square :

${\displaystyle {\begin{array}{rcll}x^{2}+px+q&=&0&|-q\\[1ex]x^{2}+px&=&-q&|+\left({\dfrac {p}{2}}\right)^{2}{\text{ (quadratische Ergänzung)}}\\[1ex]x^{2}+2\cdot {\dfrac {p}{2}}\ x+\left({\dfrac {p}{2}}\right)^{2}&=&\left({\dfrac {p}{2}}\right)^{2}-q&|{\text{ binomische Formel}}\\[1ex]\left(x+{\dfrac {p}{2}}\right)^{2}&=&\left({\dfrac {p}{2}}\right)^{2}-q&|\pm {\sqrt {\ }}\\[1ex]x+{\dfrac {p}{2}}&=&\pm {\sqrt {\left({\dfrac {p}{2}}\right)^{2}-q}}&|-{\dfrac {p}{2}}\\[1ex]x&=&-{\dfrac {p}{2}}\pm {\sqrt {\left({\dfrac {p}{2}}\right)^{2}-q}}\end{array}}}$

Another possibility to derive the formula, is that one in the abc- formula , and sets the denominator and 2 draws in the root. ${\displaystyle a=1}$ ${\displaystyle b=p}$ ${\displaystyle c=q}$

Decomposition into linear factors

With the solutions, the quadratic normalized polynomial can be broken down into linear factors:

${\displaystyle x^{2}+px+q=(x-x_{1})\cdot (x-x_{2})}$

and the non-standardized in

${\displaystyle ax^{2}+bx+c=a\cdot (x-x_{1})\cdot (x-x_{2})}$

Theorem of Vieta

If the quadratic equation is in normal form and has the solutions and , then applies ${\displaystyle x_{1}}$ ${\displaystyle x_{2}}$

${\displaystyle 0=x^{2}+px+q=(x-x_{1})\cdot (x-x_{2})=x^{2}-(x_{1}+x_{2})x+x_{1}x_{2}}$.

By comparing the coefficients , one obtains Vieta's theorem

${\displaystyle \,x_{1}+x_{2}=-p}$   and   . ${\displaystyle x_{1}\cdot x_{2}=q}$

In particular, if and are whole numbers , the solutions can often be found quickly with a little practice by trying out whether the divisor pairs of result as a sum . For example, one gets for the solutions and through the decomposition with . ${\displaystyle p}$ ${\displaystyle q}$ ${\displaystyle q}$ ${\displaystyle -p}$ ${\displaystyle x^{2}+4x+3=0}$ ${\displaystyle x_{1}=-1}$ ${\displaystyle x_{2}=-3}$ ${\displaystyle 3=(-1)(-3)}$ ${\displaystyle (-1)+(-3)=-4}$

Numerical calculation

If the solutions are determined numerically and differ by orders of magnitude, the problem of cancellation can be avoided by varying the above formulas as follows:

${\displaystyle x_{1}=-{\frac {p}{2}}-\operatorname {sgn}(p)\cdot {\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}}$

${\displaystyle x_{2}={\frac {q}{x_{1}}}}$

Here has the value for and otherwise has the value . The second formula is based on Vieta's theorem . ${\displaystyle \operatorname {sgn}(p)}$ ${\displaystyle -1}$ ${\displaystyle p<0}$ ${\displaystyle 1}$

example

For the equation

${\displaystyle 4x^{2}-12x-40=0}$

result as solutions according to the abc formula

${\displaystyle x_{1,2}={\frac {-(-12)\pm {\sqrt {(-12)^{2}-4\cdot 4\cdot (-40)}}}{2\cdot 4}},}$

so and . ${\displaystyle x_{1}=-2}$ ${\displaystyle x_{2}=5}$

To use the pq formula, the general form is first converted into the normal form by dividing the equation by 4:

${\displaystyle x^{2}-3x-10=0}$

The solutions result from the pq formula

${\displaystyle x_{1,2}=-{\frac {-3}{2}}\pm {\sqrt {\left({\frac {-3}{2}}\right)^{2}-(-10)}},}$

thus also and . ${\displaystyle x_{1}=-2}$ ${\displaystyle x_{2}=5}$

With the help of the decompositions and one obtains the same solutions with Vieta's theorem. ${\displaystyle -10=(-2)\cdot 5}$ ${\displaystyle 5-2=3}$

Further examples

${\displaystyle x^{2}+2x-35=0\ }$	For the discriminant applies: . The two real solutions and result ${\displaystyle D}$ ${\displaystyle D>0}$ ${\displaystyle x_{1}=-7}$ ${\displaystyle x_{2}=5}$
${\displaystyle x^{2}-4x+4=0\ }$	The discriminant is . The (double) real solution is . ${\displaystyle D=0}$ ${\displaystyle x=2}$
${\displaystyle x^{2}+12x+37=0\ }$	There are no real solutions because the discriminant is negative. The complex solutions result from and . ${\displaystyle x_{1}=-6+i}$ ${\displaystyle x_{2}=-6-i}$

Generalizations

Complex coefficients

The quadratic equation

${\displaystyle az^{2}+bz+c=0}$

with complex coefficients , always has two complex solutions that coincide if and only if the discriminant is zero. ${\displaystyle a,b,c\in \mathbb {C} }$ ${\displaystyle a\neq 0}$ ${\displaystyle z_{1},z_{2}\in \mathbb {C} }$ ${\displaystyle b^{2}-4ac}$

As in the real case, the solutions can be calculated by adding the square or using the solution formulas given above. In general, however, a square root of a complex number must be calculated.

example

For the quadratic equation

${\displaystyle z^{2}-(\mathrm {i} +1)z+\mathrm {i} =0\ }$

the discriminant has the value . The two solutions and result . ${\displaystyle D=-2\mathrm {i} =(\mathrm {i} -1)^{2}}$ ${\displaystyle z_{1}=1}$ ${\displaystyle z_{2}=\mathrm {i} }$

Quadratic equations in general rings

In general, in abstract algebra, one calls an equation of form

${\displaystyle x^{2}+px+q=0}$

with elements p , q of a body or ring a quadratic equation . In solids and, more generally, in areas of integrity , it has at most two solutions, in arbitrary rings it can have more than two solutions.

If solutions exist, then one obtains them in commutative rings also with the pq- formula, if the characteristic of the ring is not equal to 2. Here, however, all possible square roots of the discriminant must be taken into account. For a finite field of characteristic 2 making the approach and passes by means of a system of linear equations for n coefficients a _i from . ${\displaystyle \mathbb {F} _{2^{n}}\cong \mathbb {F} _{2}(\varrho )}$ ${\displaystyle x=\textstyle \sum _{i=0}^{n-1}a_{i}\varrho ^{i}}$ ${\displaystyle x^{2}=\textstyle \sum _{i=0}^{n-1}a_{i}\varrho ^{2i}}$ ${\displaystyle \mathbb {F} _{2}}$

example

The quadratic equation

${\displaystyle x^{2}-1=0}$

has the four solutions 1, 3, 5 and 7 in the remainder class ring . ${\displaystyle \mathbb {Z} /8\mathbb {Z} }$

history

As early as 4000 years ago in the Old Babylonian Empire , quadratic equations were solved, for example in the following way: The quadratic equation is equivalent to the system of equations and . The approach or is now made for x . For the product results ${\displaystyle x^{2}+p=sx}$ ${\displaystyle xy=p}$ ${\displaystyle x+y=s}$ ${\displaystyle x={\tfrac {s}{2}}+e}$ ${\displaystyle y={\tfrac {s}{2}}-e}$ ${\displaystyle p}$

${\displaystyle p=xy=\left({\frac {s}{2}}+e\right)\cdot \left({\frac {s}{2}}-e\right)=\left({\frac {s}{2}}\right)^{2}-e^{2}}$.

Solving the binomial formula yields

${\displaystyle e={\sqrt {{\frac {1}{4}}s^{2}-p}}}$.

With this, the solution of the quadratic equation is also determined. The equation is discussed as an example . This is equivalent to the system of equations and . The above approach delivers ${\displaystyle e}$ ${\displaystyle x}$ ${\displaystyle x^{2}+210=29x}$ ${\displaystyle p=xy=210}$ ${\displaystyle s=x+y=29}$

${\displaystyle e={\sqrt {{\frac {1}{4}}s^{2}-210}}={\frac {1}{2}}.}$

For the solution of the quadratic equation results

${\displaystyle x={\frac {s}{2}}+e={\frac {29}{2}}+{\frac {1}{2}}=15}$.

The Greeks did not know negative numbers and had to make several case distinctions for the quadratic equation. Equations of the kind

${\displaystyle x^{2}+ax=a^{2}}$

are solved geometrically in Euclid ( II 11 ); the forms

${\displaystyle ax^{2}+c=bx}$ or. ${\displaystyle ax^{2}+bx=c}$

in Euclid (VI 28) or (VI 29).

geometric solution of the Brahmagupta equation . The area not hatched corresponds to . ${\displaystyle x^{2}+px=q}$ ${\displaystyle q}$

In Aryabhata and Brahmagupta the solution of the equation is found around 628 AD

${\displaystyle x^{2}+px=q}$

described in words. As you can see from the picture (left), the following division of the square applies:

${\displaystyle x^{2}+4x{\frac {p}{4}}+4\left({\frac {p}{4}}\right)^{2}=q+\left({\frac {p}{2}}\right)^{2}=\left(x+{\frac {p}{2}}\right)^{2}}$.

This immediately provides the solution in today's notation as

${\displaystyle x={\sqrt {\left({\frac {p}{2}}\right)^{2}+q}}-{\frac {p}{2}}}$.

Al-Chwarizmi was the first to put six different types of in the book of al-Kitāb al-muḫtaṣar fī ḥisāb al-ğabr wa-ʾl-muqābala ("The concise book on the calculation methods by supplementing and balancing") around 825 AD quadratic equations. The need for different types arises from ignorance of negative numbers and zero. Using a numerical example, he gave a geometric solution method for all types, which meant that only positive solutions were possible.

He presented the six types as text. The root means the solution sought and the capacity the square of the solution : ${\displaystyle x}$ ${\displaystyle x^{2}}$

• As for the fortunes that are equal to the roots (today:) , ${\displaystyle ax^{2}=bx}$
• As for the fortunes that are equal to the number (today:) , ${\displaystyle ax^{2}=c}$
• As for the roots, which are equal to a number (today:) , ${\displaystyle bx=c}$
• As for the fortunes and the roots that are equal to the number (today:) , ${\displaystyle ax^{2}+bx=c}$
• As for the fortune and the number that are equal to the roots (today:) and ${\displaystyle ax^{2}+c=bx}$
• As for the roots and the number, which are equal to wealth (today:) . ${\displaystyle ax^{2}=bx+c}$

A, b and c stand for nonnegative coefficients and x for the solution sought.

geometric solution of Euclid's equation ${\displaystyle x^{2}+10x=39}$

As an example, the equation as it occurs in al-Khwarizmi ,

${\displaystyle x^{2}+10x=39}$

as a special case of having solved geometrically (see figure). To this end, the left side of the equation is understood as a square EFIH of the side length (and thus the area ) and two rectangles DEHG and BCFE with the sides and (and thus the area in each case ). The square and the two rectangles are combined to form a gnomon with the corner points BCIGDE as shown in the picture . This gnomon has an area of . If you add the square ABED of the side length (and thus the area ) to the square ACIG, then this has the area . On the other hand, however, according to the construction, this square ACIG has the side length and thus the area . Because of one closes and thus . The quadratic equation is thus "quadratically supplemented" with the (positive) solution . Note that this geometric method does not give the negative solution . ${\displaystyle x^{2}+bx=c}$ ${\displaystyle b,c>0}$ ${\displaystyle x}$ ${\displaystyle x^{2}}$ ${\displaystyle 5}$ ${\displaystyle x}$ ${\displaystyle 5x}$ ${\displaystyle x^{2}+10x=39}$ ${\displaystyle 5}$ ${\displaystyle 25}$ ${\displaystyle 39+25=64}$ ${\displaystyle 5+x}$ ${\displaystyle (5+x)^{2}}$ ${\displaystyle 64=8^{2}}$ ${\displaystyle 5+x=8}$ ${\displaystyle x=3}$ ${\displaystyle (x+5)^{2}=64}$ ${\displaystyle x=3}$ ${\displaystyle x=-13}$

With Heron of Alexandria and also with al-Khwarizmi the solution is given by

${\displaystyle ax^{2}+bx=c}$

described verbally; in today's notation as

${\displaystyle x={\frac {{\sqrt {ac+\left({\frac {b}{2}}\right)^{2}}}-{\frac {b}{2}}}{a}}}$.

However, Heron adds the Euclidean way as a geometric justification.

Around 1145 Robert von Chester and a little later Gerhard von Cremona translated the writings of al-Chwarizmi into Latin.

This brought the classification and the geometric solution methods to Europe.

Michael Stiefel wrote the book "Arithmetica integra" in 1544 AD, which is based on Christoph Rudolff's book "Behend vnnd Hubsch calculation through the artful rules of algebre so commonly called Coss" . By using negative numbers, the author succeeds in avoiding the case distinction for quadratic equations. But he does not yet allow negative numbers as solutions because he finds them absurd.

A new approach to solving a quadratic equation was offered by Vieta's theorem of roots , which was published posthumously in 1615 in his work De Aequationem Recognitione et Emendatione Tractatus duo .

In 1637, René Descartes described a method for solving quadratic equations with compasses and ruler in his work La Géométrie . He further showed that equations of a higher degree, in general, cannot be solved solely with a compass and ruler.

See also

• Linear equation
• Cubic equation
• Quartic equation

literature

• Bartel Leendert van der Waerden: Awakening Science . Volume 1: Egyptian, Babylonian and Greek Mathematics . 2nd Edition. Birkhäuser 1966.

Web links

Commons : Quadratic equation  - collection of images, videos, and audio files

• Video: From solving quadratic equations - pq formula and midnight formula and Vieta's theorem . Jakob Günter Lauth (SciFox) 2013, made available by the Technical Information Library (TIB), doi : 10.5446 / 17884 .

Individual evidence

1. ↑ Guido Walz: Equations and inequalities: plain text for non-mathematicians . Springer, 2018, ISBN 9783658216696 , p. 14
2. ^ ^{A b} Franz Embacher: Quadrartic equations . Script on mathe-online.at
3. ↑ Hans Wußing : 6000 Years of Mathematics - Volume 1. Springer Verlag, 2008, ISBN 978-3-540-77189-0 , pp. 237-241, doi: 10.1007 / 978-3-540-77192-0
4. ^ Helmuth Gericke : Mathematics in Antiquity, Orient and Occident . 7th edition. Fourier Verlag, 2003, ISBN 3-925037-64-0 , p. 198.
5. ↑ Hans Wußing : 6000 Years of Mathematics - Volume 1. Springer Verlag, 2008, ISBN 978-3-540-77189-0 , p. 278, doi: 10.1007 / 978-3-540-77192-0
6. ↑ Hans Wußing : 6000 Years of Mathematics - Volume 1. Springer Verlag, 2008, ISBN 978-3-540-77189-0 , pp. 341–342, doi: 10.1007 / 978-3-540-77192-0