Let’s talk about the quadratic formula today. In algebra , a quadratic equation (from Latin quadratus for ” square “) is any equation that can be rearranged as standard

ax^{2}+bx+c=0

where x denotes an unknown , and a , b , and c represent known numbers, where a 0 . If a = 0 , then the equation is linear , not quadratic, because there is no period. The numbers a , b , and c are the coefficients of the equation and can be distinguished by calling them the quadratic coefficient , the linear coefficient , and the constant or free term , respectively. ax2

The values ​​of x that satisfy the equation are called solutions of the equation, and the roots or zeros of the expression on its left hand side. A quadratic equation has at most two solutions. If there is no real solution, there are two complex solutions. If there is only one solution, one says that it is a double root . A quadratic equation always has two roots if complex roots are included and a double root is counted for two. A quadratic equation can be made into an equivalent equation depending on

{\displaystyle ax^{2}+bx+c=a(xr)(xs)=0}

where r and s are solutions of x . Putting the square on a quadratic equation in standard form gives the quadratic formula , which expresses the solutions in the form a , b and c . Solutions to problems that could be expressed as quadratic equations were known as early as 2000 BC.

Since only one unknown is involved in a quadratic equation, it is called ” one-sided “. The quadratic equation only contains powers of x that are non-negative integers, and so it is a polynomial equation . In particular, it is a second degree polynomial equation, since the greatest power is two.

A quadratic equation with real or complex coefficients has two solutions, which are called roots . These two solutions may or may not be different, and they may or may not be real.

### Factoring by Inspection

It may be possible to express a quadratic equation ax2 + bx + c = 0 as the product ( px + q )( rx + s ) = 0 . In some cases, by simple inspection, it is possible to determine the values ​​of p , q , r, and s that make the two forms equivalent. If the quadratic equation is written in another form, the “zero multiplicative property” states that the quadratic equation is satisfied if px + q = 0 or rx + s = 0., Solving these two linear equations gives the roots of the quadratic.

For most students, factoring by observation is the first way they are exposed to solving quadratic equations. [2] : 202-207 If one is given a quadratic equation of the form x2 + bx + c = 0 , then the sought factor is of the form ( x + q ) ( x + s ) , and any Two numbers have to be found. q and s that connect b and whose product is c (this is sometimes called “Vieta’s rule” [3] and is related to Sources of Vieta ). As an example, x2 + 5 x + 6 as the factor ( x + 3)( x + 2) . In the more general case where a is not equal to 1 , trial and error can require considerable effort in estimation and investigation, assuming that it may be included at all by inspection.

Except in special cases such as b = 0 or c = 0 , factoring by inspection only works for quadratic equations with rational roots. This means that the vast majority of quadratic equations that arise in practical applications cannot be solved by factoring by inspection.

### Completing The Square

The procedure for completing the square uses algebraic identities

x^{2}+2hx+h^{2}=(x+h)^{2},

which represents a well-defined algorithm that can be used to solve any quadratic equation. Starting with the quadratic equation in standard form, ax2 + bx + c = 0

1. Divide each side by the coefficient a of the square term .
2. Subtract the constant term c / a from both sides .
3. Add the square of half of b / a , the coefficient of x , to both sides. This “completes the square”, turning the left side into a perfect square.
4. Write the left side as a square and simplify the right side if necessary.
5. Create two linear equations by equating the left side square root with the right side positive and negative square root.
6. Solve each of the two linear equations.

We get 22 + 4 x – 4 = 0 . Let us describe the use of this algorithm by solving

1)~~~~~~~~~~~~~~~~~~ \ x ^ 2 + 2x-2 = 0\\

2) ~~~~~~~~~~~~~~~~~~~~~~~~~\ x ^ 2 + 2x = 2\\

3)~~~~~~~~~~~~ \ x ^ 2 + 2x + 1 = 2 + 1\\

4) ~~~~~~~~~~~~~~~~~~~~~~~~~\ \left(x+1 \right)^2=3\\

5) ~~~~~~~~~~~~~~~~~~~~~~~~\ x+1=\pm\sqrt{3}\\

6) ~~~~~~~~~~~~~~~~~~~~~~~~~\ x=-1\pm\sqrt{3}

The plus-minus sign “±” indicates that both x = -1 + 3 and x = -1 – 3 are solutions to the quadratic equation.

## Quadratic Formula and Its Derivation

Completing the square can be used to derive a general formula for solving quadratic equations , called the quadratic formula. [5] The mathematical proof will now be briefly summarized. [6] It can be easily seen by polynomial expansion that the following equation is equivalent to a quadratic equation:

\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}.

Taking the square root of both the sides and separating x, we get:

{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}

Some sources, especially older ones, use alternative parameters of the quadratic equation such as ax 2 + 2 bx + c = 0 or ax 2 – 2 bx + c =0 ,

where the magnitude of b is half the normal. Is. One, possibly with the opposite sign. These result in slightly different forms for the solution, but are otherwise equivalent.

Several alternative etymologies can be found in the literature . These proofs are simpler than the standard meeting the square method, represent interesting applications of other techniques frequently used in algebra, or provide insight into other areas of mathematics.

A lesser known quadratic formula, as used in Müller’s method , gives the same root through the equation

{\displaystyle x={\frac {2c}{-b\pm {\sqrt {b^{2}-4ac}}}}.}

This can be derived from the standard quadratic formula by Vieta’s formulas , which assert that the product of the roots is c / a .

One property of this form is that it produces a valid root when a = 0 , while the second root has division by zero, because when a = 0 , the quadratic equation becomes a linear equation, which has an origin. Conversely, in this case, the more general formula consists of division by zero for one root and an indefinite form 0/0 for the other root . On the other hand, when c = 0 , the more general formula gives two true roots while this form gives zero roots and an indefinite form 0/0 .

It is sometimes convenient to reduce a quadratic equation so that its leading coefficient is one. This is done by dividing both sides by a , which is always possible because a is non-zero. This produces the reduced quadratic equation :

x^{2}+px+q=0,

Where P = B / A and Q = C / A. _ This mnemonic equation has the same solutions as the root.

The quadratic formula for the solution of the reduced quadratic equation, its coefficients, is written as:

{\displaystyle x={\frac {1}{2}}\left(-p\pm {\sqrt {p^{2}-4q}}\right),}

or equivalent:

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

### Differential

In a quadratic formula, the expression below the square root is called the discriminant of the quadratic equation, and is often represented using an upper case D or an upper case Greek delta:

\Delta = b^2 - 4ac.

A quadratic equation with real coefficients can have one or two different real roots or two different complex roots. In this case the discriminant determines the number and nature of the roots. There are three cases:

• If the discriminant is positive, then there are two different roots.
\frac{-b + \sqrt {\Delta}}{2a} \quad\text{and}\quad \frac{-b - \sqrt {\Delta}}{2a},

which are both real numbers. For quadratic equations with rational coefficients, if the discriminant is a square number, the roots are rational—in other cases they can be quadratic irrational.

• If the discriminant is zero, then there is actually a real root
-\frac{b}{2a},

Sometimes called repeated or double root.

• If the discriminant is negative, there are no real roots. Rather, there are two distinct (unrealistic) complex roots
{\displaystyle -{\frac {b}{2a}}+i{\frac {\sqrt {-\Delta }}{2a}}\quad {\text{and}}\quad -{\frac {b}{2a}}-i{\frac {\sqrt {-\Delta }}{2a}},}

which are complex conjugates of each other. In these expressions I is the imaginary entity.

Thus the roots are distinct if and only if the discriminant is non-zero, and the roots are real if and only if the discriminant is non-negative.

### Geometric Interpretation

The function f ( x ) = ax2 + bx + c is a quadratic function. [12] The general shape of a graph of any quadratic function is called a parabola. The location and size of the parabola, and how it opens, depends on the values ​​of a , b , and c . As shown in Figure 1, if a > 0 , the parabola has a minimum point and opens upwards. If a < 0 , then the parabola has a maximum point and opens downwards. The extreme point of the parabola, whether minimum or maximum, corresponds to its vertex. top of x-coordinates , and the y- coordinate of the vertex can be found by substituting this x – value in the function. The Y dialog point is located at (0, c ).

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


Solutions of the quadratic equation ax 2 + bx + c = 0 correspond to the roots of the function f ( x ) = ax 2 + bx + c , since they are values ​​of x for which f ( x ) = 0 . As shown in Figure 2, if a , b , and c are real numbers and the domain of f is the set of real numbers, then the roots of f are exactly the x -coordinate of the points where the graph touchesX axis. As shown in Figure 3, if the discriminant is positive, the graph touches the x-axis at two points; If zero, the graph touches at a point; And if negative, then the graph does not touch the x -axis.

Period

x – r

is a factor of the polynomial

ax^{2}+bx+c

If and only if r is a root of the quadratic equation

ax^{2}+bx+c=0.

It follows from the quadratic formula that

ax^2+bx+c = a \left( x - \frac{-b + \sqrt {b^2-4ac}}{2a} \right) \left( x - \frac{-b - \sqrt {b^2-4ac}}{2a} \right).

In the special case 2 = 4 ac where the quadratic has only one different root ( ie the discriminant is zero), the quadratic polynomial can be factored as

ax^2+bx+c = a \left( x + \frac{b}{2a} \right)^2.

### Graphical Solution

ax^2+bx+c=0

can be concluded from the quadratic function of the graph

y=ax^2+bx+c,

which is a parabola.

If the parabola intersects the x -axis at two points, there are two real roots, which are the x -coordinates of these two points (also called x – dialogs), If the parabola is tangent to the x -axis, there is a double origin, which is the x – coordinate of the point of contact between the graph and the parabola .

If the parabola does not intersect the x -axis, then there are two complex conjugate roots. Although these roots cannot be seen on the graph, they can have real and imaginary parts.

Let h and k be the x – coordinate and y- coordinate , respectively, of the vertex of the parabola (that is, the point with the maximum or minimum y- coordinate. The quadratic function can be rewritten as)

{\displaystyle y=a(x-h)^{2}+k.}

Let d be the distance between the y- coordinate 2 k on the axis of the parabola , and a point on the parabola with the same y- coordinate (see the figure; there are two points that give the same distance, due to the symmetry of the parabola). . Then the real part of the roots is h , and their imaginary part is ± d . ie the roots are

{\displaystyle h+id\quad {\text{and}}\quad x-id,}

or in the case of figure example

{\displaystyle 5+3i\quad {\text{and}}\quad 5-3i.}

### Avoiding Loss of Importance

Although the quadratic formula provides an exact solution, the result is not accurate if the real numbers are approximated during the computation, as is usual in numerical analysis, where the real numbers are referred to as floating point numbers (“real” in many programming languages). is) is estimated by . In this context, the quadratic formula is not completely constant.

This occurs when the roots have different orders of magnitude, or, equivalently, when 2 and 2 – 4 ac are close in magnitude. In this case, the subtraction of two nearly identical numbers would cause a loss of significance or a catastrophic cancellation of the smaller root. To avoid this, which root is smaller in magnitude, r can be calculated as where R is the root which is larger in magnitude.

{\displaystyle (c/a)/R}


Another form of cancellation can occur between the positions of the discriminator 2 and 4 ac , that is, when the two origins are very close. This can result in loss of up to half of the true significant figures in the roots.

#### Examples and Applications

The positive solution of the golden ratio quadratic equation is found as. x2 – x – 1=0

The equations of circles and other conic sections—ellipse, parabola, and hyperbola—are quadratic equations in two variables.

Given the cosine or sine of an angle, finding the cosine or sine of half the greater angle involves solving a quadratic equation.

The process of simplifying an expression involving the square root of an expression to include the square root of another expression involves finding two solutions to a quadratic equation.

Descartes’ theorem states that for every four kissing (mutual tangent) circles, their radii satisfy a particular quadratic equation.

The equation given by Fuse’s theorem, giving the relation between the radius of the inscribed circle of a bicentral quadrilateral, the radius of its circumscribed circle, and the distance between the centers of those circles, can be expressed as a quadratic equation for which two The distance between the centers of circles is a solution in terms of their radii. The second solution of the same equation in the case of the relevant radius gives the distance between the center of the circumscribed circle and the center of the extremity of the former tangent of the quadrilateral.

The critical points of a cubic function and the inflection point of a quartic function are found by solving a quadratic equation.

## History

Babylonian mathematicians, in 2000 BC (demonstrated on Old Babylonian clay tablets) could solve problems related to areas and sides of rectangles. There is evidence dating this algorithm as far back as the Third Dynasty of Ur. In modern notation, problems usually involve solving a pair of simultaneous equations of the form:

{\displaystyle x+y=p,\ \ xy=q,}

which is equivalent to the statement that x and y are roots of the equation:

{\displaystyle z^{2}+q=pz.}

The steps given by the Babylonian scribes to solve the above rectangle problem in terms of x and y were as follows:

1. Calculate half of p .
2. Square the result.
3. Why decrease ?
4. Find the (positive) square root using a table of squares.
5. Combine the results of steps (1) and (4) to get x .

In modern notation this means to calculate , which is equivalent to the modern quadratic formula for large real roots (if any) with a = 1 , b = -p , and c = q .

{\displaystyle x=\left({\frac {p}{2}}\right)+{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}}{\displaystyle x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}}

Geometric methods were used to solve quadratic equations in Babylonia, Egypt, Greece, China and India. The Egyptian Berlin Papyrus, dating back to the Middle Kingdom (2050 BC to 1650 BC), contains the solution of a two-period quadratic equation. [17] Babylonian mathematicians from about 400 BC and Chinese mathematicians around 200 BC used geometric methods of dissection to solve quadratic equations with positive roots. [18] [ 19] The laws of quadratic equations were given in nine chapters on the mathematical art, a Chinese treatise on mathematics. [19] [20]These early geometric methods had no general formula. Euclid, the Greek mathematician, created a more abstract geometric method around 300 BC. With a purely geometric approach, Pythagoras and Euclid created a general procedure for finding solutions to a quadratic equation. In his work Arithmetic , the Greek mathematician Diophantus solved a quadratic equation, but gave only one root, even if both roots were positive. [21]

In 628 AD, an Indian mathematician Brahmagupta gave the first explicit (though still not completely general) solution to the quadratic equation ax 2 + bx = c : “The whole number is multiplied by four times the [coefficient] square. , add the square of the [coefficient] of the middle term; the square root of the same, less the [coefficient] of the middle term, is the value of dividing the square by twice the [coefficient].”It is equal to:

x={\frac {{\sqrt {4ac+b^{2}}}-b}{2a}}.

The Bakhshali manuscript , written in 7th century India, contained an algebraic formula for solving quadratic equations, as well as quadratic indeterminate equations (originally the type of ax / c = y [ clarification needed : this is linear, not quadratic ] ) . Muhammad ibn Musa al-Khwarizmi (Persia, 9th century), inspired by Brahmagupta, original research? ] developed a set of working formulas for positive solutions. Al-Khwarizmi goes on to provide complete solutions to the general quadratic equation, accepting one or two numerical answers for each quadratic equation while providing geometric proofs in the process. [22]

He also described the method of completing the square and accepted that the discriminant must be positive, which was proved by his contemporary ‘Abd al-Hamid ibn Turk (Central Asia, 9th century). ) which are given geometric figures to prove that if the discriminant is negative, then the quadratic equation has no solution.  While al-Khwarizmi himself did not accept negative solutions, later his successor Islamic mathematicians accepted negative solutions, as well as accepting irrational numbers as solutions. [24]Abu Kamil Shuja ibn Aslam (Egypt, 10th century) was the first to specifically accept irrational numbers (often as square roots, cube roots or fourth roots) as solutions to quadratic equations or as coefficients in an equation. The 9th century Indian mathematician Sridhar wrote rules for solving quadratic equations.

The Jewish mathematician Abraham Bar Hia Ha-Nasi (12th century, Spain) wrote the first European book to include complete solutions to the general quadratic equation. [27] His solution was based largely on the work of al-Khwarizmi. [22] The writings of the Chinese mathematician Yang Hui (1238–1298 AD) are the first known quadratic equations with negative coefficients of ‘x’, although he attributes this to the earlier Liu Yi. [28] By 1545 Gerolamo Cardano had compiled works relating to quadratic equations. The quadratic formula covering all cases was first derived by Simon Stevin in 1594. [29] In 1637 René Descartes published La Geometry which contained the quadratic formula we know today.

### Alternative Methods of Route Calculations

#### Vieta’s Formula

Vieta’s formulas give a simple relationship between the roots of a polynomial and its coefficients. Reverse the quadratic polynomial of the root

x_{1},x_{2}{\displaystyle P(x)=ax^{2}+bx+c}
{\displaystyle x_{1}+x_{2}=-{\frac {b}{a}},\quad x_{1}x_{2}={\frac {c}{a}}.}

These results immediately follow from the relation:

{\displaystyle \left(x-x_{1}\right)\left(x-x_{2}\right)=x^{2}-\left(x_{1}+x_{2}\right)x+x_{1}x_{2}=0,}

which can be compared with term to term

x^2 + (b/a)x +c/a = 0.


The first formula above gives a convenient expression when graphing a quadratic function. Since the graph is symmetric with respect to a vertical line through the vertex, when there are two real roots the x – coordinate of the vertex lies on the median of the roots (or the intersection). Thus the x – coordinate of the vertex is given by the expression

x_V = \frac {x_1 + x_2} {2} = -\frac{b}{2a}.

Given the y -coordinate, the above result in the quadratic equation can be obtained by substituting

y_V = - \frac{b^2}{4a} + c = - \frac{ b^2 - 4ac} {4a}.

As a practical matter, Vieta’s formulas provide a useful method for finding the roots of a quadratic in the case where one root is much smaller than the other. if | 2 |  , 1 | , then 1 + 2 x 1  , and we have the conjecture :

x_1 \approx -\frac{b}{a} .

The second Vieta’s formula then provides:

{\displaystyle x_{2}={\frac {c}{ax_{1}}}\approx -{\frac {c}{b}}.}

These formulas are much easier to evaluate than the quadratic formula under the condition of a major and a minor root, because the quadratic formula evaluates the smaller root as the difference of two very similar numbers (in the case of large b ), which causes rounding-off error in numerical evaluation.

Figure 5 (i) a direct evaluation using the quadratic formula (accurate when the roots are near each other in value) and (ii) an evaluation based on the above approximation of Vieta’s formulas (precise when the roots are widely spaced) ) As the linear coefficient b increases, the quadratic formula is initially accurate, and the approximate formula improves in accuracy, allowing b There is a small difference between the methods when increasing,

However, at some point the quadratic formula begins to lose accuracy due to round off error, while the approximation method continues to improve. As a result, the difference between methods begins to widen as the quadratic formula gets worse and worse.

This situation commonly arises in amplifier design, where widely separated roots are desired to ensure a stable operation (see phase response).

### Trigonometric Solution

In the days before calculators, people used mathematical tables—lists of numbers showing the results of calculations with different arguments—to simplify and speed up calculations. Tables of logarithms and trigonometric functions were common in math and science textbooks. Specific tables were published for applications such as astronomy, celestial navigation and statistics. Methods of numerical approximation existed, called prosthepheresis, that offered shortcuts around time-consuming tasks such as multiplication and taking powers and roots. [30] Astronomers, in particular, were concerned with ways that could speed up the long series of calculations involved in celestial mechanics calculations.

In this context we can understand the development of means of solving quadratic equations with the help of trigonometric substitution. Consider the following alternative form of the quadratic equation,

[1]~~~~ax^2 + bx \pm c = 0 ,\\

where the sign of the ± sign is chosen so that both a and c are positive. by substituting

[2] ~~~  x = \sqrt{c/a} \tan\theta

and then multiplying through by cos2θ, we obtain

~~~~~~~~~~~~~~~~~~~~~~~~[३]   \sin^2\theta + \frac{b}{\sqrt {ac}} \sin\theta \cos\theta \pm \cos^2\theta = 0 .

Introducing functions of 2θ and rearranging, we obtain

[4] ~~~  \tan 2 \theta_n = + 2 \frac{\sqrt{ac}}{b} ,
[5]   ~~~~~~\sin 2 \theta_p = - 2 \frac{\sqrt{ac}}{b} ,

where the subscripts n and p correspond to the use of the negative or positive sign in Equation [1] , respectively. Substituting the two values ​​of n or p found from equations [4] or [5] in [2] gives the necessary roots [1] . Complex roots are in solution based on equation [5] if the absolute value of sin 2 p is greater than unity The amount of effort involved in solving quadratic equations using this mixed trigonometric and logarithmic table look-up strategy was two-thirds the effort using logarithmic tables alone. [31]A different trigonometric form would need to be used for computing complex roots. [32]For example, let’s say we had seven-local logarithmic and trigonometric tables available, and we wanted to solve the following to six-significant-figure accuracy:

4.16130x^2 + 9.15933x - 11.4207 = 0
1. A seven-space lookup table can only have 100,000 entries, and computing intermediate results at seven locations will typically require interpolation between adjacent entries.
2~~~~~~~~~~\log a = 0.6192290,  \log b = 0.9618637, \log c  = 1.0576927\\

3~~~~~~~~2 \sqrt{ac}/b = 2 \times 10^{(0.6192290 + 1.0576927)/2 - 0.9618637} = 1.505314 \\
4~~~~~~~~~~~~~~\theta = (\tan^{-1}1.505314) / 2 = 28.20169^{\circ} \text{ or } -61.79831^{\circ} \\
5~~~~~~~~~\log | \tan \theta | = -0.2706462 \text{ or } 0.2706462~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
6~~~~~~~~~~~~~~ \log\sqrt{c/a} = (1.0576927 - 0.6192290) / 2 = 0.2192318\\
7~~~~~~x_1 = 10^{0.2192318 - 0.2706462} = 0.888353 (rounded~ to ~six ~significant ~figures)\\
x_2 = -10^{0.2192318 + 0.2706462} = -3.08943


#### Solving Complex Roots in Polar Coordinates

If a quadratic equation has two complex roots with real coefficients—the situation where a and c need to have the same sign as each other—then the solutions to the roots can be expressed in polar form as

ax^2+bx+c=0b^{2}-4ac<0,
x_{1},\,x_{2}=r(\cos \theta \pm i\sin \theta ),

where is and

r={\sqrt {{\tfrac {c}{a}}}}\theta =\cos ^{{-1}}\left({\tfrac {-b}{2{\sqrt {ac}}}}\right).


##### Geometric Solution

A quadratic equation can be solved geometrically in a number of ways. One way is through Lil’s Method. Three coefficients a , b , c are drawn with right angles between them as in SA, AB and BC in Fig. 6. A circle is drawn taking the start and end point SC as diameter. The equation has a solution if it intersects the midline AB of the three, and the solution is the first coefficient a or SA divided by the negative of the distance from A along this line. So there is a 1 coefficient can be read straight off. Thus the solutions in the diagram are −AX1/SA and −AX2/SA.

The Carlyle circle, named after Thomas Carlyle, has the property that the solutions of the quadratic equation are the horizontal coordinates of the intersections of the circle with the horizontal axis. [35] Carlyle has used circles to develop the ruler and compass constructions of regular polygons.

The coefficients of the formula and its derivation remain true if a , b and c are complex numbers with any members, or more generally the field whose characteristic is not 2 . (In the field of characteristic 2, element 2a is zero and is impossible to divide by. )

symbol

\pm \sqrt {b^2-4ac}

In the formula “to be understood as any one of two elements whose square is 2 – 4 ac , if such elements exist”. In some fields, some elements do not have square roots and some have two; There is only one square root of zero except in areas with characteristic 2 . Even if a field does not contain the square root of a number, there is always a quadratic expansion field, so the quadratic formula always makes sense as a formula in that expansion field.

#### Feature 2

In the field of characteristic 2 , the quadratic formula, which relies on 2 being a unit, does not hold. Consider the Monic Quadratic Polynomial

x^{2} + bx + c

In the area of ​​specialty 2 . If b = 0 , the solution is reduced to find a square root, so the solution is

x = \sqrt{c}

and since then there’s only one root

-\sqrt{c} = -\sqrt{c} + 2\sqrt{c} = \sqrt{c}.
##### Summary,
\displaystyle x^{2} + c = (x + \sqrt{c})^{2}.

\frac{b}{a}R\left(\frac{ac}{b^2}\right)
\frac{b}{a}\left(R\left(\frac{ac}{b^2}\right)+1\right).