Fuzzy-sets based analysis often involves solution of interval variables, as demonstrated in papers by Dourado et al. [1] , Mo et al. [2] , Tian et al. [3] , and Liao et al. [4] . In this conjunction, the solution of interval quadratic equations is of paramount importance (see, for example, anonymous, 2022 [5] ). Alolyan [6] apparently was the first who investigated the fuzzy quadratic equations. In their book A History of Mathematics, Carl B. Boyer and Uta C. Merzbach [7] stated that “[Quadratic] equations have been handled effectively by the Babylonians in some of the oldest problem texts”, approximately around 400 AD. This problem has also been tackled by the Indian mathematician Brahmagupta, in Chapter 18 of Brahmasphuṭasiddhānta, composed in 628 CE, according to Kim Pfloker in The Mathematics of Egypt, Mesopotamia, China, India, and Islam by Victor J. Katz [8] . Indeed, Brahmagupta gave the solution to the general quadratic equation defined as follows

a x 2 + b x = c (1)

where a , b , c are coefficients, and x denotes an unknown. He stated his recipe as “Diminish by the middle [number] the square-root of the rupas (rupas refer to the constantc) multiplied by four times the square and increased by the square of the middle [number]; divide the remainder by twice the square. [The result is] the middle [number].” This leads us to the following expressions of the roots of the quadratic equation defined in Equation (1), which are:

x 1 , 2 = ± 4 a c + b 2 − b 2 a (2)

This is equivalent to the well-known modern formula that we use, since, in Equation (1), c is on the right side of the quadratic equation. Note that we set that x 1 < x 2 . This implies that we need to be cautious whether a < 0 or a > 0 , because a cannot be zero in Equation (2).

Indeed, when a > 0 , to respect x 1 < x 2 , we will have to consider that:

x 1 = − b − b 2 − 4 a c 2 a     and     x 2 = − b + b 2 − 4 a c 2 a

On the other hand, when a < 0 , to get x 1 < x 2 , we have to consider that:

x 1 = − b + b 2 − 4 a c 2 a     and     x 2 = − b − b 2 − 4 a c 2 a

The mathematical explanation of the previous statement will be provided in Appendix A.

In this paper, we treat the cases of quadratic equations in which two of the three coefficients a, b and c constitute intervals. This implies that finding the formulas of x₁ and x₂ becomes a much harder task for some of the methods developed in our previous studies. Indeed, the method used by Elishakoff and Daphnis [9] cannot lead to the establishment of formulas for those problems, while Ioakimidis [10] reports that “quantifier elimination cannot be performed in a reasonable CPU time [so] all the present results were confined to only one interval parameter”. We cannot consider these methods in the rest of this article. Elishakoff and Miglis [11] developed a special parametrization technique as a cure for the dependency problem suffered by the classical interval analysis, that they referred to as interval parametrization. This method is a computational tool, rather than being an analytical technique.

This paper aims then to solve that issue, by offering a method that yields analytical formulas for cases in which two or all three coefficients of the quadratic equation are defined as intervals.

Interval analysis is extremely instrumental in case of uncertainty analysis when probability density functions or membership functions are unknown. Still, if we know the lower and upper bounds, we can use either least favourable or most favourable designs.

This method will be called sign-variation method in the rest of the article and is based upon the partial derivatives of the expressions of the roots of a quadratic equation. Indeed, we will consider every possible sign variation of each partial derivative of a root of the quadratic equation. Afterwards, we will be able to know how the root’s value evolves in regard to the variable that has been used to partially derive the root’s expression. The method will be developed in the next section, and the results yielded by the sign-variation method will be compared to the results obtained via the interval parametrization technique developed by Elishakoff and Miglis [11] , but also to the direct approach with the Maple software. The direct approach comprises of the straightforward minimization and the maximization of expressions utilizing Maple’s Minimize and Maximize commands [12] . Analogous methodology was applied by Landowski [13] , whereas Sevastjanov and Dymova [14] developed a fuzzy solution of interval linear equation, while Mamehrashi [15] utilizing the context of fuzzy sets, too.

We first introduce an example of using this method for a particular interval quadratic equation, and will provide in the Appendices B and C the results obtained for every possible subcase. Let us consider a quadratic equation with two variables serving as intervals, a and b, and c being a crisp quantity. We also state here that a > 0 , b < 0 and c < 0 , and that the roots of this equation are real, so b 2 − 4 a c > 0 . Since a and b constitute intervals, we have:

a = [ a _ ; a ¯ ] (3)

and

b = [ b _ ; b ¯ ] (4)

In Equations (3) and (4), the lower bar designates the smallest value of the variable, whereas the upper bar denotes the greatest value of the variable, and those values will be separated by a semi column in the rest of this study.

We will start by considering the first root x₁, which is:

x 1 = − b − b 2 − 4 a c 2 a (5)

Since x₁ depends on a, b and c, we can consider the following partial derivatives:

p d a 1 ( a , b , c ) = ∂ x 1 ∂ a = c a b 2 − 4 a c − − b − b 2 − 4 a c 2 a 2 (6)

p d b 1 ( a , b , c ) = ∂ x 1 ∂ b = − 1 − b b 2 − 4 a c 2 a (7)

We will not consider p d c 1 ( a , b , c ) = ∂ x 1 ∂ c here because it is not useful for our

current example, as c is a crisp quantity. However, it will be used when c constitutes an interval quantity in the Appendix B and Appendix C.

Let us now draw the sign-variation tables of both p d a 1 and p d b 1 related to our problem, so considering that a > 0 , b < 0 and c < 0 .

For p d a 1 ( a , b , c ) = ∂ x 1 ∂ a , considering b < 0 and c < 0 , the signs of the partial derivative are, for different regions of variation of the parameter a:

Using Table 1, we see that ∂ x 1 / ∂ a is positive when a > 0 . That means that, regarding the variable a, if a is increasing, then x₁ is also increasing. Consequently, if we want to minimize x₁ regarding a, we need to use a _ , thus using a ¯ will result in maximizing x₁.

We can process to doing the same thing with the variable b.

For p d b 1 ( a , b , c ) = ∂ x 1 ∂ b , considering a > 0 and c < 0 , Table 2 is obtained:

As shown in Table 2, ∂ x 1 / ∂ b is negative when b < 0 . That means that, regarding the variable b, if b is increasing, then x₂ is decreasing. So, if we want to minimize x₁ regarding b, we need to use b ¯ , so using b _ will result in maximizing x₁.

Now combining what we obtained above yields the following expressions:

x _ 1 = − b ¯ − b ¯ 2 − 4 a _ c 2 a _ (8)

x ¯ 1 = − b _ − b _ 2 − 4 a ¯ c 2 a ¯ (9)

Let us now conduct analogous analysis with the root x₂, which is:

x 2 = − b + b 2 − 4 a c 2 a (10)

The corresponding partial derivatives of x₂ in regards to a and b are:

p d a 2 ( a , b , c ) = ∂ x 2 ∂ a = − c a b 2 − 4 a c − − b + b 2 − 4 a c 2 a 2 (11)

p d b 2 ( a , b , c ) = ∂ x 2 ∂ b = − 1 + b b 2 − 4 a c 2 a (12)

Let us now draw the sign-variation tables of p d a 1 and p d b 1 related to our problem, so considering that a > 0 , b < 0 and c < 0 , as we did for x₁:

For p d a 2 ( a , b , c ) = ∂ x 2 ∂ a , considering b < 0 and c < 0 :

As seen in Table 3, ∂ x 2 / ∂ a is negative when a > 0 . That means that, regarding the variable a, if a is increasing, then x₂ is decreasing. Consequently, if we want to minimize x₂ regarding a, we need to use a _ . Using a ¯ will then maximize x₂.

For p d b 2 ( a , b , c ) = ∂ x 2 ∂ b , considering a > 0 and c < 0 :

Now, with the variable b, we see in Table 4 that ∂ x 2 / ∂ b is also negative when b < 0 . That means that, regarding the variable b, if b is increasing, then x₂ is also decreasing. So, if we want to minimize x₂ regarding b, we need to use b ¯ , thus using b _ will result in maximizing x₂.

We then obtain the following formulas for x₂:

x _ 2 = − b ¯ + b ¯ 2 − 4 a ¯ c 2 a ¯ (13)

x ¯ 2 = − b _ + b _ 2 − 4 a _ c 2 a _ (14)

We now obtained the formulas for both roots of a quadratic equation with two variables, namely a and b, serving as intervals, in the specific case in which a > 0 , b < 0 and c < 0 .

Let us now compare the results yielded by this method with the direct approach from minimization and maximization commands of the Maple software, by considering a numerical example of that interval quadratic equation.

As we need to satisfy a > 0 , b < 0 , c < 0 , and b 2 − 4 a c > 0 , we consider the following example:

a = [ a _ ; a ¯ ] = [ 5 ; 6 ] ,     b = [ b _ ; b ¯ ] = [ − 11 ; − 10 ] ,     c = − 1 (15)

The direct approach using Equation (15) yields:

x _ 1 = − 0.095445 ( at   a _ = 5   and   b ¯ = − 10 ) ,   x ¯ 1 = − 0.086799 ( at   a ¯ = 6   and   b _ = − 11 ) (16)

x ¯ 2 = 1.761294 ( at   a ¯ = 6   and   b ¯ = − 10 ) ,   x _ 2 = 2.287434 ( at   a _ = 5   and   b _ = − 11 ) (17)

The sign-variation formulas using Equations (9), (10), (13), (14) and (15) give the following results:

x _ 1 = − 0.095445 ,     x ¯ 1 = − 0.086799 (18)

x _ 2 = 1.761294 ,     x ¯ 2 = 2.287434 (19)

As is seen, we obtain the same results, with the same values for parameters a and b. We will extend this to every subcase, and provide Tables C1-C3 in Appendix C that summarize all results for every possible subcase, for the cases in which two or all the parameters constitute intervals. We will also provide multiple examples of different subcases to compare the results given by those formulas with the other methods, namely interval parametrization and direct approach from minimization and maximization commands of the Maple software.

One might argue that considering the case of a as an interval, then b also, and then assuming that combining the results from both separated cases leads to obtaining the formulas when both variables are considered as intervals might be questionable. This method actually works because the sign of both ∂ x n / ∂ a and ∂ x n / ∂ c , n ∈ { 1 ; 2 } is constant whatever the values of the other variables are. Consequently, when we consider ∂ x n / ∂ b , n ∈ { 1 ; 2 } , it ensures that the tables shown above will remain valid whatever the value a and c will take in each subcase.

One could also point out that some of the signs obtained in the sign-variation tables are not obvious to determine. We will then provide in the Appendix D, the proofs that these obtained signs are valid, namely using Mathematica commands, in order to be on the safe side and prevent an error.

The sign-variation method also works for the case when only one parameter a, b or c is considered as an interval, but we need to consider every subcase possible. It is then less optimal to use than quantifier elimination method [10] for example, that has been the best way to yield analytical formulas when only one parameter a, b or c of quadratic equations constitutes an interval. We will, at the end of our article in the conclusion, draw Table C3 that summarizes the advantages and the drawbacks of each method.

As is seen the present methodology is based upon heavy use of sensitivity derivatives.

This method is advantageous over the parametrization technique by Elishakoff and Miglis [11] since the latter is a purely numerical technique, whereas the current methodology constitutes an analytical approach. It is also superior to the technique developed by Elishakoff and Daphnis [9] since the latter study is not able to cover all possible cases of interval equations.

Now, we consider the case in which all the coefficients of a quadratic equation are serving as intervals. When it comes to interval parametrization, as it needs to be computed in order to give results, we will not get analytical formulas that could directly lead to solving the problem. In order to get analytical formulas, we will then keep using the sign-variation method.

Let us first consider the case with all the parameters a, b and c constituting intervals, with the sign-variation method.

a = [ a _ ; a ¯ ] (120)

b = [ b _ ; b ¯ ] (121)

c = [ c _ ; c ¯ ] (122)

Let us consider here a > 0 , b < 0 and c < 0 for this example. We assume that a ≠ 0 and b 2 − 4 a c > 0 too.

We can apply the sign-variation analysis method in order to find the formulas of minimized and maximized values of both roots of the equation, the same way we did previously.

For x₁ first, as Table 1 shows, the value ∂ x 1 / ∂ a is positive when a > 0 . Consequently, if we want to minimize x₁ regarding a, we need to use a _ , thus using a ¯ will result in maximizing x₁.

The value ∂ x 1 / ∂ b is negative when b < 0 according to Table 2. So, if we want to minimize x₁ with respect to b, we need to use b ¯ , then we need to use b _ to maximize x₁.

The value ∂ x 1 / ∂ c is positive when c < 0 in Table B6 in Appendix B. To minimize x₁ with respect to c, we need to use c _ , using c ¯ will result in maximizing x₁. These results together yield the following formulas:

x _ 1 = − b ¯ − b ¯ 2 − 4 a _ c _ 2 a _ (123)

x ¯ 1 = − b _ − b _ 2 − 4 a ¯   c ¯ 2 a ¯ (124)

For x₂, Table 3 shows that the value ∂ x 2 / ∂ a is negative when a > 0 . Consequently, if we want to minimize x₂ regarding a, we need to use a ¯ , thus using a _ will result in maximizing x₂.

The derivative ∂ x 2 / ∂ b is negative when b < 0 as demonstrates Table 4. So, if we want to minimize x₂ regarding b, we need to use b ¯ , then we need to use b _ to maximize x₂.

In Table B16, the derivative ∂ x 2 / ∂ c is negative when c < 0 . To minimize x₂ regarding c, we need to use c ¯ , using c _ will result in maximizing x₂. These propositions yield the formulas:

x _ 2 = − b ¯ + b ¯ 2 − 4 a ¯   c ¯ 2 a ¯ (125)

x ¯ 2 = − b _ + b _ 2 − 4 a _ c _ 2 a _ (126)

Let us consider another case: a < 0 , b > 0 and c < 0 . We still assume that a ≠ 0 and b 2 − 4 a c > 0 too.

The derivative ∂ x 1 / ∂ a is positive when a < 0 as Table B3 shows. Consequently, if we want to minimize x₁ regarding a, we need to use a _ , thus using a ¯ will result in maximizing x₁.

The derivative ∂ x 1 / ∂ b is positive when b > 0 as seen in Table B9. So, if we want to minimize x₁ regarding b, we need to use b _ , then we need to use b ¯ to maximize x₁.

Table B5 shows that the value ∂ x 1 / ∂ c is positive when c < 0 . To minimize x₁ with respect to c, we need to use c _ , using c ¯ will result in maximizing x₁.Those results yield the following formulas:

x _ 1 = − b _ + b _ 2 − 4 a _ c _ 2 a _ (127)

x ¯ 1 = − b ¯ + b ¯ 2 − 4 a ¯   c ¯ 2 a ¯ (128)

For x₂, Table B13 demonstrates that the value ∂ x 2 / ∂ a is negative when a < 0 . Consequently, if we want to minimize x₂ regarding a, we need to use a ¯ , thus using a _ will result in maximizing x₂.

The derivative ∂ x 2 / ∂ b is negative when b > 0 in Table B20. So, if we want to minimize x₂ with respect to b, we need to use b ¯ , then we need to use b _ to maximize x₂.

In Table B15, we see that ∂ x 2 / ∂ c is also negative when c < 0 . To minimize x₂ regarding c, we need to use c ¯ , using c _ will result in maximizing x₂. This leads us to the following formulas:

x _ 2 = − b ¯ − b ¯ 2 − 4 a ¯   c ¯ 2 a ¯ (129)

x ¯ 2 = − b _ − b _ 2 − 4 a _ c _ 2 a _ (130)

We cannot consider every possible case, so we provide in the Appendix C Table C3 that contains the results for every possible subcase when all coefficients constitute intervals.

Let us now consider the same problem defined above, with the parameters a, b and c still constituting the intervals defined in Equations (120), (121) and (122):

We introduce the average value of a denoted a_ave and the deviation value of a denoted a_dev. We proceed analogously for b and c, so the average values of b and c denoted respectively b_ave and c_ave and the deviation values of b and c denoted b_dev and c_dev, as follows:

a a v e = a ¯ + a _ 2 (131)

a d e v = a ¯ − a _ 2 (132)

b a v e = b ¯ + b _ 2 (133)

b d e v = b ¯ − b _ 2 (134)

c a v e = c ¯ + c _ 2 (135)

c d e v = c ¯ − c _ 2 (136)

a, band c become now:

a = a a v e + a d e v t 1 (137)

b = b a v e + b d e v t 2 (138)

c = c a v e + c d e v t 3 (139)

using

t 1 = [ − 1 ; 1 ] (140)

t 2 = [ − 1 ; 1 ] (141)

t 3 = [ − 1 ; 1 ] (142)

The roots of this equation can now be written, using Equations (137), (138) and (139), with a > 0 :

x 1 = − ( b a v e + b d e v t 2 ) − ( b a v e + b d e v t 2 ) 2 − 4 ( a a v e + a d e v t 1 ) ( c a v e + c d e v t 3 ) 2 ( a a v e + a d e v t 1 ) (143)

x 2 = − ( b a v e + b d e v t 2 ) + ( b a v e + b d e v t 2 ) 2 − 4 ( a a v e + a d e v t 1 ) ( c a v e + c d e v t 3 ) 2 ( a a v e + a d e v t 1 ) (144)

The roots of this equation are, when a < 0 , using Equations (137), (138) and (139):

x 1 = − ( b a v e + b d e v t 2 ) + ( b a v e + b d e v t 2 ) 2 − 4 ( a a v e + a d e v t 1 ) ( c a v e + c d e v t 3 ) 2 ( a a v e + a d e v t 1 ) (145)

x 2 = − ( b a v e + b d e v t 2 ) − ( b a v e + b d e v t 2 ) 2 − 4 ( a a v e + a d e v t 1 ) ( c a v e + c d e v t 3 ) 2 ( a a v e + a d e v t 1 ) (146)

We defined the expressions for both methods with a > 0 , b < 0 and c < 0 , we can now proceed to evaluate their results numerically. Let us consider the following example:

a = [ a _ ; a ¯ ] = [ 3 ; 4 ] ,     b = [ b _ ; b ¯ ] = [ − 11 ; − 10 ] ,     c = [ c _ ; c ¯ ] = [ − 2 ; − 1 ] (147)

The direct approach gives the following results using Equation (147):

x _ 1 = − 0.189255 ( at   a _ = 3 , b ¯ = − 10   and   c _ = − 2 ) , x ¯ 1 = − 0.088087 ( at   a ¯ = 4 , b _ = − 11   and   c ¯ = − 1 ) (148)

x _ 2 = 2.596291 ( at   a ¯ = 4 , b ¯ = − 10   and   c ¯ = − 1 ) , x ¯ 2 = 3.840266 ( at   a _ = 3 , b _ = − 11   and   c _ = − 2 ) (149)

The sign-variation method formulas yield, using Equations (123)-(126) and (147):

x _ 1 = − 0.189255 ,     x ¯ 1 = − 0.088087 (150)

x _ 2 = 2.596291 ,     x ¯ 2 = 3.840266 (151)

Let us now use the interval parametrization formulas with Equations (143), (144) and (147):

x _ 1 = − 0.189255 ( at   t 1 = − 1 , t 2 = 1   and   t 3 = − 1 ) , x ¯ 1 = − 0.088087 ( at   t 1 = 1 , t 2 = − 1   and   t 3 = 1 ) (152)

x _ 2 = 2.596291 ( at   t 1 = 1 , t 2 = 1   and   t 3 = 1 ) , x ¯ 2 = 3.522588 ( at   t 1 = − 1 , t 2 = − 1   and   t 3 = − 1 ) (153)

These results match with to those found using both other methods.

Let us try with the other subcase detailed above, a < 0 , b > 0 and c < 0 . We define:

a = [ a _ ; a ¯ ] = [ − 4 , − 3 ] ,     b = [ b _ ; b ¯ ] = [ 10 , 11 ] ,     c = [ c _ ; c ¯ ] = [ − 2 , − 1 ] (154)

The direct approach gives the following results using Equation (154):

x _ 1 = 0.093282 ( at   a ¯ = − 3 , b ¯ = 11   and   c ¯ = − 1 ) , x ¯ 1 = 0.219223 ( at   a _ = − 4 , b _ = 10   and   c _ = − 2 ) (155)

x _ 2 = 2.280776 ( at   a _ = − 4 , b _ = 10   and   c _ = − 2 ) , x ¯ 2 = 3.573384 ( at   a ¯ = − 3 , b ¯ = 11   and   c ¯ = − 1 ) (156)

The sign-variation method formulas yield using Equations (127)-(130) and (154):

x _ 1 = 0.093282 ,     x ¯ 1 = 0.219223 (157)

x _ 2 = 2.280776 ,     x ¯ 2 = 3.573384 (158)

Let us use the interval parametrization formulas with Equations (145), (146) and (154), to get:

x _ 1 = 0.093282 ( at   t 1 = 1 , t 2 = 1   and   t 3 = 1 ) , x ¯ 1 = 0.219223 ( at   t 1 = − 1 , t 2 = − 1   and   t 3 = − 1 ) (159)

x _ 2 = 2.280776 ( at   t 1 = − 1 , t 2 = − 1   and   t 3 = − 1 ) , x ¯ 2 = 3.573384 ( at   t 1 = 1 , t 2 = 1   and   t 3 = 1 ) (160)

We again get corresponding results between all methods.

This article offers a method that leads to obtaining analytical formulas for the roots of quadratic equations in which two or all three coefficients serve as intervals. As seen, the results yielded by the sign-variation method using numerical examples keep matching with those obtained using a direct approach that is based on Minimize and Maximize commands of the Maple software, also with those obtained using the interval parametrization method introduced by Elishakoff and Miglis [11] . Let us now draw a table that summarizes the advantages and the drawbacks of the classic interval analysis method used by Elishakoff and Daphnis (2015), the quantifier elimination used by Ioakimidis [10] , the interval parametrization method by Elishakoff and Miglis [11] , and the sign-variation method presented in this article.

Table 5 aims to show which method is the most adapted when solving a problem involving quadratic equations in which one, two or even all three of the coefficients could serve as intervals.

Obtaining analytical formulas seems remarkable to those problems because then they could be used by hand for any similar problem. Thus, quantifier elimination method seems to be the most adapted method to solve problems dealing with quadratic equations in which one coefficient serves as an interval, whereas sign-variation method appears as the only analytical way to deal with quadratic equations in which two or three of the coefficients constitute intervals.

This work was conducted when Mr. Nicolas Yvain of SIGMA Clermont, Aubière, France, was on internship at the Florida Atlantic University. The work reported herein was not associated with funding. Authors declare no conflict of interest. Authors appreciate the constructive and helpful comments of the reviewers. We are also grateful to Ms. Nancy Ho for editorial assistance.

The authors declare no conflicts of interest regarding the publication of this paper.

Yvain, N. and Elishakoff, I. (2023) Novel Method to Deal with Interval Quadratic Equations via Sign-Variation Analysis. Journal of Applied Mathematics and Physics, 11, 3212-3250. https://doi.org/10.4236/jamp.2023.1110208

As stated in the Introduction part, we have to adapt which of the two root’s expression is x₁ and which is x₂ depending on the sign of a, in order to respect x 1 < x 2 .

Indeed, let us consider both expressions of the root of a quadratic equation, which are:

r 1 = − b − b 2 − 4 a c 2 a       and       r 2 = − b + b 2 − 4 a c 2 a (189)

In order to better understand when r 1 < r 2 , we need to study their respective signs.

No matter what is the sign of a, we always have − b − b 2 − 4 a c < − b + b 2 − 4 a c , since b 2 − 4 a c is a positive quantity.

We shall now consider to take the sign of a into account. As said before, a cannot be zero otherwise r₁ and r₂ do not exist.

If a > 0 :

− b − b 2 − 4 a c < − b + b 2 − 4 a c becomes − b − b 2 − 4 a c / 2 a < − b + b 2 − 4 a c / 2 a

We now consider that a < 0 :

− b − b 2 − 4 a c < − b + b 2 − 4 a c turns into − b + b 2 − 4 a c / 2 a < − b − b 2 − 4 a c / 2 a

Thus, if a > 0 then, to respect x 1 < x 2 , we shall write the roots as follows:

x 1 = − b − b 2 − 4 a c 2 a     and     x 2 = − b + b 2 − 4 a c 2 a (190)

While, if we have a < 0 , the formulas of the roots should be written as:

x 1 = − b + b 2 − 4 a c 2 a     and     x 2 = − b − b 2 − 4 a c 2 a (191)

Let us proceed to establish the sign-variation tables for the root x₁, defined in Equation (5), for the subcases that were not studied in the article. We need to introduce the last partial derivative of x₁ that hasn’t been considered yet, in regard to the variable c, as follows:

p d c 1 ( a , b , c ) = ∂ x 1 ∂ c = 1 b 2 − 4 a c (192)

For p d a 1 ( a , b , c ) = ∂ x 1 ∂ a , considering b > 0 and c > 0 :

For p d a 1 ( a , b , c ) = ∂ x 1 ∂ a , considering b < 0 and c > 0 :

For p d a 1 ( a , b , c ) = ∂ x 1 ∂ a , considering b > 0 and c < 0 :

For p d c 1 ( a , b , c ) = ∂ x 1 ∂ c , considering a > 0 and b > 0 :

For p d c 1 ( a , b , c ) = ∂ x 1 ∂ c , considering a < 0 and b > 0 :

For p d c 1 ( a , b , c ) = ∂ x 1 ∂ c , considering a > 0 and b < 0 :

For p d c 1 ( a , b , c ) = ∂ x 1 ∂ c , considering a < 0 and b < 0 :

As we can see from Table 1 and Tables B1-B7, for both ∂ x 1 / ∂ a and ∂ x 1 / ∂ c , when they are defined, their sign remains constantly positive whatever the other variables could be.

Let us now establish the tables for ∂ x 1 / ∂ b , with the missing subcases:

For p d b 1 ( a , b , c ) = ∂ x 1 ∂ b , considering a > 0 and c > 0 :

For p d b 1 ( a , b , c ) = ∂ x 1 ∂ b , considering a < 0 and c > 0 :

For p d b 1 ( a , b , c ) = ∂ x 1 ∂ b , considering a < 0 and c < 0 :

Table B10. Sign variation of p d b 1 ( a , b , c ) = ∂ x 1 ∂ b with a < 0 and c < 0 .

When it comes to ∂ x 1 / ∂ b , we see in Table 2 and Tables B7-B9 that the variations are more complex than for the other variables in Table 1 and Tables B1-B7.

Let us now proceed to do the same thing for the other root, x₂, that is defined in Equation (10). Since x₂ also depends on a, b and c, we can consider this last partial derivative:

p d c 2 ( a , b , c ) = ∂ x 2 ∂ c = − 1 b 2 − 4 a c (193)

As we did for the previous root, we will proceed to creating the sign-variation tables of each of those functions, considering the missing subcases that were not treated in the article.

For p d a 2 ( a , b , c ) = ∂ x 2 ∂ a , considering b > 0 and c > 0 :

Table B11. Sign variation of p d a 2 ( a , b , c ) = ∂ x 2 ∂ a with b > 0 and c > 0 .

For p d a 2 ( a , b , c ) = ∂ x 2 ∂ a , considering b < 0 and c > 0 :

Table B12. Sign variation of p d a 2 ( a , b , c ) = ∂ x 2 ∂ a with b < 0 and c > 0 .

For p d a 2 ( a , b , c ) = ∂ x 2 ∂ a , considering b > 0 and c < 0 :

Table B13. Sign variation of p d a 2 ( a , b , c ) = ∂ x 2 ∂ a with b > 0 and c < 0 .

For p d c 2 ( a , b , c ) = ∂ x 2 ∂ c , considering a > 0 and b > 0 :

Table B14. Sign variation of p d c 2 ( a , b , c ) = ∂ x 2 ∂ c with a > 0 and b > 0 .

For p d c 2 ( a , b , c ) = ∂ x 2 ∂ c , considering a < 0 and b > 0 :

Table B15. Sign variation of p d c 2 ( a , b , c ) = ∂ x 2 ∂ c with a < 0 and b > 0 .

For p d c 2 ( a , b , c ) = ∂ x 2 ∂ c , considering a > 0 and b < 0 :

Table B16. Sign variation of p d c 2 ( a , b , c ) = ∂ x 2 ∂ c with a > 0 and b < 0 .

For p d c 2 ( a , b , c ) = ∂ x 2 ∂ c , considering a < 0 and b < 0 :

Table B17. Sign variation of p d c 2 ( a , b , c ) = ∂ x 2 ∂ c with a < 0 and b < 0 .

As we can see from Table 3 and Tables B11-B17 for both ∂ x 2 / ∂ a and ∂ x 2 / ∂ c , when they are defined, their sign remains constantly negative whatever the other variables could be.

Let us now build the tables for ∂ x 2 / ∂ b , with the missing subcases:

For p d b 2 ( a , b , c ) = ∂ x 2 ∂ b , considering a > 0 and c > 0 :

Table B18. Sign variation of p d b 2 ( a , b , c ) = ∂ x 2 ∂ b with a > 0 and c > 0 .

For p d b 2 ( a , b , c ) = ∂ x 2 ∂ b , considering a < 0 and c > 0 :

Table B19. Sign variation of p d b 2 ( a , b , c ) = ∂ x 2 ∂ b with a < 0 and c > 0 .

For p d b 2 ( a , b , c ) = ∂ x 2 ∂ b , considering a < 0 and c < 0 :

Table B20. Sign variation of p d b 2 ( a , b , c ) = ∂ x 2 ∂ b with a < 0 and c < 0 .

We can analogously denote that, for ∂ x 2 / ∂ b , we get in Table 4 and Tables B18-B20 variations that are more complex than for the other variables in Table 3 and Tables B11-B17.

Now that we have built all the sign-variations tables of every partial derivative of both roots of the quadratic equations, we can proceed to establishing the analytic formulas.

Since there is a lot of subcases, and this implies a lot of formulas, we had to summarize the results in tables, that will be established for every subcase possible, when two but also when all the coefficients of quadratic equations constitute intervals.These formulas always require that a ≠ 0 and b 2 − 4 a c > 0 too, as we are only dealing with the real roots of interval quadratic equations. As seen in Section 4.3, the formulas established for a quadratic equation in which a and c constitute intervals are the same whatever the signs of a, b and c are. The following tables that will be presented are then used when a and b, or when b and c, or when a, b and c serve as intervals.

Please note that the expressions of x₁ and x₂ should change whether a is a positive or negative quantity, as stated in the Introduction part of this study.

We will start in Table C1 by considering that a and b constitute intervals, and that c is a crisp quantity.

Let us give an example on how to read this table. We consider each subcase possible for the values of a, b and c, and we give for each root which value of the parameters will lead to obtaining the minimum values and maximum values of the root. For example, when a > 0 , b > 0 and c > 0 , the minimum value of the first root x₁ will be obtained using b ¯ and a _ , as stated in the cell marked with the star sign * in Table C1.

This leads us to the following expression for the lowest value of the first root x _ 1 :

x _ 1 = − b ¯ − b ¯ 2 − 4 a _ c 2 a _

Let us now establish Table C2, in which b and c are now the coefficients of quadratic equations that serve as intervals, as a is a deterministic quantity here. This Table should be read analogously to Table C1.

We can finally consider the case when all coefficients of a quadratic equation constitute intervals in Table C3, that should again be read analogously to Table C1.

One could argue that some of the signs of the partial derivatives obtained in Tables 1-4, Tables B1-B19 are not obvious and should deserve a deeper explanation. We will then provide justifications in this last part of the Appendices to make sure that these signs are actually valid.

In order to justify the signs obtained, we will use a computerized method, with the software Mathematica, with the Minimize and Maximize commands.

As stated in the Document Center of Mathematica [12] , those formulas should be used as follows:

Minimize[f, cons, x] and Maximize[f, cons, x] (c1)

where f corresponds to the function that needs to be minimized/maximized under the constraints stated in cons, in regard to the variable x. We will use the traditional form defined as //tf at the end of each command to get clearer results, as follows:

tf = Traditional Form; (c2)

We will check the signs for p d a 1 ( a , b , c ) = ∂ x 1 ∂ a , p d b 1 ( a , b , c ) = ∂ x 1 ∂ b , p d a 2 ( a , b , c ) = ∂ x 2 ∂ a and p d b 2 ( a , b , c ) = ∂ x 2 ∂ b , since finding the signs of p d c 1 ( a , b , c ) = ∂ x 1 ∂ c and p d c 2 ( a , b , c ) = ∂ x 2 ∂ c does not yield any particular issue when looking at their formulas.

We define the functions to study as follows in the Mathematica software:

p d a 1 = c / ( a ∗ ( b ∗ b − 4 ∗ a ∗ c ) 0.5 ) − ( ( − b − ( b ∗ b − 4 ∗ a ∗ c ) 0.5 ) / ( 2 ∗ ( a ∗ a ) ) ) ; p d a 1 / / t f (c3)

p d b 1 = ( − 1 − ( b ) / ( ( b ∗ b − 4 ∗ a ∗ c ) 0.5 ) ) / ( 2 ∗ a ) ; p d b 1 / / t f (c4)

p d a 2 = − c / ( a ∗ ( b ∗ b − 4 ∗ a ∗ c ) 0.5 ) − ( ( − b + ( b ∗ b − 4 ∗ a ∗ c ) 0.5 ) / ( 2 ∗ ( a ∗ a ) ) ) ; p d a 2 / / t f (c5)

p d b 2 = ( − 1 + ( b ) / ( ( b ∗ b − 4 ∗ a ∗ c ) 0.5 ) ) / ( 2 ∗ a ) ; p d b 2 / / t f (c6)

We can now proceed to look for the minimum or maximum values of these functions, considering multiple subcases for the variables a, b and c. We will consider the minimum value of the function when the subcase in the corresponding table depicts that the sign of the function is positive and will look for the maximum value of the function when the subcase in the corresponding table depicts that the sign of the function is negative. Indeed, if we obtain that the minimum value of the function is 0, then it shows that the function is positive, whereas if we get that the maximum value of the function is 0, then we can argue that the function is negative.

Let us now do the whole process for two different cases as examples.

We will first consider p d a 1 ( a , b , c ) = ∂ x 1 ∂ a , and, as seen in Table 1, Table B1, Table B7 and Table B3, p d a 1 is always positive when defined (namely when a ≠ 0 ). Let us for example verify Table 1.

We need to set the adapted constraints for our command, they will be b < 0 and c < 0 (subcase studied in Table 1), a ≠ 0 (so that p d a 1 is defined), and b 2 − 4 a c > 0 (since we are only considering real roots). Since we want to verify that p d a 1 is positive, we will use the Minimize command, and we will use it with respect to the variable a because we considered p d a 1 .

This leads us to entering the following command in Mathematica:

t 1 = M i n i m i z e [ p d a 1 , a ≠ 0 ∧ c < 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c7)

which outputs the following results:

{ 0. c < 0 ∧ b < 0 ∞ True , { a → Indeterminate } }

As we can see, when respecting b < 0 and c < 0 , we get that the minimum value of p d a 1 ( a , b , c ) = ∂ x 1 ∂ a is 0. We can then argue that p d a 1 is positive when b < 0 and c < 0 .

We can proceed identically for the rest of the Tables related to p d a 1 , namely Tables B5-B7.

We will enter the same commands, and just change one of the constraints, which will become b > 0 and c > 0 for Table 5, b > 0 and c > 0 for Table B1 and b > 0 and c > 0 for Table B2.

As we want to prove that p d a 1 is positive when b > 0 and c > 0 , b > 0 and c > 0 and b > 0 and c > 0 , we use the following commands 8, 9 and 10:

t 2 = M i n i m i z e [ p d a 1 , a ≠ 0 ∧ c > 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c8)

which returns

{ 0. c > 0 ∧ b < 0 ∞ True , { a → Indeterminate } }

t 3 = M i n i m i z e [ p d a 1 , a ≠ 0 ∧ c < 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c9)

which outputs

{ 0. c < 0 ∧ b > 0 ∞ True , { a → Indeterminate } }

t 4 = M i n i m i z e [ p d a 1 , a ≠ 0 ∧ c > 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c10)

which leads to

{ 0. c > 0 ∧ b > 0 ∞ True , { a → Indeterminate } }

These commands (8), (9) and (10) justify the signs found in Table 1, Tables B1-B3.

We can process analogously for p d a 2 ( a , b , c ) = ∂ x 2 ∂ a , which is always a negative quantity when defined (namely when a ≠ 0 ). We will just input the same formulas that have been used for p d a 1 , but with Maximize commands because we want to show that p d a 2 is negative in Table 3, Tables B11-B13. This leads us to using the following commands (11)-(14) in Mathematica:

t 5 = M a x i m i z e [ p d a 2 , a ≠ 0 ∧ c > 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c11)

which returns

{ 0. c > 0 ∧ b > 0 − ∞ True , { a → Indeterminate } }

t 6 = M a x i m i z e [ p d a 2 , a ≠ 0 ∧ c < 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c12)

which outputs

{ 0. c < 0 ∧ b > 0 − ∞ True , { a → Indeterminate } }

t 7 = M a x i m i z e [ p d a 2 , a ≠ 0 ∧ c > 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c13)

which leads to

{ 0. c > 0 ∧ b < 0 − ∞ True , { a → Indeterminate } }

t 8 = M a x i m i z e [ p d a 2 , a ≠ 0 ∧ c < 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , a ] / / t f (c14)

which yields the following results

{ 0. c < 0 ∧ b < 0 − ∞ True , { a → Indeterminate } }

Commands (11), (12), (13) and (14) justify the signs of p d a 2 shown in Table 3, Tables B10-B12.

Let us now study p d b 1 and p d b 2 . We will proceed analogously to the method we used for p d a 1 and p d a 2 .However, since the sign of p d b 1 and p d b 2 can change whether b is a negative or a positive quantity, as in Table B8, Table B10, Table B18, and Table B20, we will need to adjust the constraints and add one in regard to the sign of b. We also do not need to set a ≠ 0 anymore since we will set either a > 0 or a < 0 . We will start with Table 2, Table 4, Table B9 and Table B19 because the sign of the functions in those Tables does not change with the value of b, and will finish by considering Table B8, Table B10, Table B18 and Table B20 that need more commands to be justified.

Let us start with the sign of p d b 1 in Table 2, that is negative. We input the following command:

t 9 = M a x i m i z e [ p d b 1 , a > 0 ∧ c < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c15)

Command (15) yields the following results:

{ 0. c < 0 ∧ a > 0 − ∞ True , { b → Indeterminate } }

In Table 4, p d b 2 is negative. We then use command (16) in Mathematica:

t 10 = M a x i m i z e [ p d b 2 , a > 0 ∧ c < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c16)

which outputs:

{ 0. c < 0 ∧ a > 0 − ∞ True , { b → Indeterminate } }

Table B9 demonstrates that p d b 1 is a positive quantity. We use the following command (17):

t 11 = M i n i m i z e [ p d b 1 , a < 0 ∧ c > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c17)

Command (17) leads to the following results:

{ 0. c > 0 ∧ a < 0 ∞ True , { b → Indeterminate } }

Table B18 also points out that p d b 2 is a positive quantity. We can input in Mathematica:

t 12 = M i n i m i z e [ p d b 2 , a < 0 ∧ c > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c18)

which yields the following results:

{ 0. c > 0 ∧ a < 0 ∞ True , { b → Indeterminate } }

Now that we have treated the simpler cases, let us draw our attention on Table B8, Table B10, Table B18, and Table B20, that will each need two commands. Indeed, we need to consider that the sign of b has an influence on the sign of the functions, which was not the case before.

Table B8 shows that p d b 1 is positive when b is negative, and p d b 1 is negative when b becomes positive. We then need to enter commands (19) and (20) in Mathematica:

t 13 = M i n i m i z e [ p d b 1 , a > 0 ∧ c > 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c19)

Command (19) results in:

{ 0. c > 0 ∧ a > 0 ∞ True , { b → Indeterminate } }

Command (20) is the following:

t 14 = M a x i m i z e [ p d b 1 , a > 0 ∧ c > 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c20)

that yields:

{ − 1 a c > 0 ∧ a > 0 − ∞ True , { b → Indeterminate } }

−1/a being a negative quantity here, and also being the maximum of p d b 1 when a > 0 , b > 0 and c > 0 , we can then conclude that p d b 1 is indeed negative here.

In Table B10, p d b 1 is negative when b is negative, and p d b 1 is positive when b gets positive. We will enter the following commands (21) and (22) in Mathematica:

t 15 = M a x i m i z e [ p d b 1 , a < 0 ∧ c < 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c21)

which outputs:

{ 0. c < 0 ∧ a < 0 − ∞ True , { b → Indeterminate } }

Command (22) being:

t 16 = M i n i m i z e [ p d b 1 , a < 0 ∧ c < 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c22)

Command (22) gives the following results:

{ − 1 a c < 0 ∧ a < 0 ∞ True , { b → Indeterminate } }

−1/a is here a positive number. Since it is the maximum of p d b 1 when a < 0 , b > 0 and c < 0 , we can then conclude that p d b 1 is positive here.

In Table B18, we denote that p d b 2 is a negative quantity when b is negative, whereas it becomes a positive one when b is also positive. We can then input the following command (23):

t 17 = M a x i m i z e [ p d b 2 , a > 0 ∧ c > 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c23)

which outputs:

{ − 1 a c > 0 ∧ a > 0 ∞ True , { b → Indeterminate } }

−1/a, here, is a positive quantity since a > 0 . Since it is the maximum of p d b 2 , p d b 2 is indeed negative here.

We now type command (24), which is:

t 18 = M i n i m i z e [ p d b 2 , a > 0 ∧ c > 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c24)

Command (24) gives the following results:

{ 0. c > 0 ∧ a > 0 ∞ True , { b → Indeterminate } }

To conclude, our last Table to justify is Table B20. We observe that p d b 2 is a positive quantity when b is negative, and it becomes negative when b gets positive. This leads to command (25):

t 19 = M i n i m i z e [ p d b 2 , a < 0 ∧ c < 0 ∧ b < 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c25)

which results in:

{ − 1 a c < 0 ∧ a < 0 ∞ True , { b → Indeterminate } }

−1/a is here positive. Being the maximum of p d b 1 , the latter is indeed positive here.

We then type the following command (26):

t 20 = M a x i m i z e [ p d b 2 , a < 0 ∧ c < 0 ∧ b > 0 ∧ ( b ∗ b − 4 ∗ a ∗ c ) > 0 , b ] / / t f (c26)

Command (26) yields the following results:

{ 0. c < 0 ∧ a < 0 − ∞ True , { b → Indeterminate } }

Using all the provided commands above in the Mathematica software, we justified the signs obtained in Tables 1-4, Tables B1-B3, Tables B8-B13 and Tables B18-B20 since some of them are not immediate results.