I. ax 2 =0 – incomplete quadratic equation (b=0, c=0 ). Solution: x=0. Answer: 0.
Solve equations.
2x·(x+3)=6x-x 2 .
Solution. Let's open the brackets by multiplying 2x for each term in brackets:
2x 2 +6x=6x-x 2 ; We move the terms from the right side to the left:
2x 2 +6x-6x+x 2 =0; Here are similar terms:
3x 2 =0, hence x=0.
Answer: 0.
II. ax 2 +bx=0 –incomplete quadratic equation (c=0 ). Solution: x (ax+b)=0 → x 1 =0 or ax+b=0 → x 2 =-b/a. Answer: 0; -b/a.
5x 2 -26x=0.
Solution. Let's take out the common factor X outside of brackets:
x(5x-26)=0; each factor can be equal to zero:
x=0 or 5x-26=0→ 5x=26, divide both sides of the equality by 5 and we get: x=5.2.
Answer: 0; 5,2.
Example 3. 64x+4x 2 =0.
Solution. Let's take out the common factor 4x outside of brackets:
4x(16+x)=0. We have three factors, 4≠0, therefore, or x=0 or 16+x=0. From the last equality we get x=-16.
Answer: -16; 0.
Example 4.(x-3) 2 +5x=9.
Solution. Applying the formula for the square of the difference of two expressions, we will open the brackets:
x 2 -6x+9+5x=9; transform to the form: x 2 -6x+9+5x-9=0; Let us present similar terms:
x 2 -x=0; we'll take it out X outside the brackets, we get: x (x-1)=0. From here or x=0 or x-1=0→ x=1.
Answer: 0; 1.
III. ax 2 +c=0 –incomplete quadratic equation (b=0 ); Solution: ax 2 =-c → x 2 =-c/a.
If (-c/a)<0 , then there are no real roots. If (-с/а)>0
Example 5. x 2 -49=0.
Solution.
x 2 =49, from here x=±7. Answer:-7; 7.
Example 6. 9x 2 -4=0.
Solution.
Often you need to find the sum of squares (x 1 2 +x 2 2) or the sum of cubes (x 1 3 +x 2 3) of the roots of a quadratic equation, less often - the sum of the reciprocal values of the squares of the roots or the sum of arithmetic square roots of the roots of a quadratic equation:
Vieta's theorem can help with this:
x 2 +px+q=0
x 1 + x 2 = -p; x 1 ∙x 2 =q.
Let's express through p And q:
1) sum of squares of the roots of the equation x 2 +px+q=0;
2) sum of cubes of the roots of the equation x 2 +px+q=0.
Solution.
1) Expression x 1 2 +x 2 2 obtained by squaring both sides of the equation x 1 + x 2 = -p;
(x 1 +x 2) 2 =(-p) 2 ; open the brackets: x 1 2 +2x 1 x 2 + x 2 2 =p 2 ; we express the required amount: x 1 2 +x 2 2 =p 2 -2x 1 x 2 =p 2 -2q. We got a useful equality: x 1 2 +x 2 2 =p 2 -2q.
2) Expression x 1 3 +x 2 3 Let us represent the sum of cubes using the formula:
(x 1 3 +x 2 3)=(x 1 +x 2)(x 1 2 -x 1 x 2 +x 2 2)=-p·(p 2 -2q-q)=-p·(p 2 -3q).
Another useful equation: x 1 3 +x 2 3 = -p·(p 2 -3q).
Examples.
3) x 2 -3x-4=0. Without solving the equation, calculate the value of the expression x 1 2 +x 2 2.
Solution.
x 1 +x 2 =-p=3, and the work x 1 ∙x 2 =q=in example 1) equality:
x 1 2 +x 2 2 =p 2 -2q. We have -p=x 1 +x 2 = 3 → p 2 =3 2 =9; q= x 1 x 2 = -4. Then x 1 2 +x 2 2 =9-2·(-4)=9+8=17.
Answer: x 1 2 +x 2 2 =17.
4) x 2 -2x-4=0. Calculate: x 1 3 +x 2 3 .
Solution.
By Vieta's theorem, the sum of the roots of this reduced quadratic equation is x 1 +x 2 =-p=2, and the work x 1 ∙x 2 =q=-4. Let's apply what we have received ( in example 2) equality: x 1 3 +x 2 3 =-p·(p 2 -3q)= 2·(2 2 -3·(-4))=2·(4+12)=2·16=32.
Answer: x 1 3 +x 2 3 =32.
Question: what if we are given an unreduced quadratic equation? Answer: it can always be “reduced” by dividing term by term by the first coefficient.
5) 2x 2 -5x-7=0. Without deciding, calculate: x 1 2 +x 2 2.
Solution. We are given a complete quadratic equation. Divide both sides of the equality by 2 (the first coefficient) and obtain the following quadratic equation: x 2 -2.5x-3.5=0.
According to Vieta's theorem, the sum of the roots is equal to 2,5 ; the product of the roots is equal to -3,5 .
We solve it in the same way as the example 3) using the equality: x 1 2 +x 2 2 =p 2 -2q.
x 1 2 +x 2 2 =p 2 -2q= 2,5 2 -2∙(-3,5)=6,25+7=13,25.
Answer: x 1 2 + x 2 2 = 13,25.
6) x 2 -5x-2=0. Find:
Let us transform this equality and, using Vieta’s theorem, replace the sum of roots through -p, and the product of the roots through q, we get another useful formula. When deriving the formula, we used equality 1): x 1 2 +x 2 2 =p 2 -2q.
In our example x 1 +x 2 =-p=5; x 1 ∙x 2 =q=-2. We substitute these values into the resulting formula:
7) x 2 -13x+36=0. Find:
Let's transform this sum and get a formula that can be used to find the sum of arithmetic square roots from the roots of a quadratic equation.
We have x 1 +x 2 =-p=13; x 1 ∙x 2 =q=36. We substitute these values into the resulting formula:
Advice : always check the possibility of finding the roots of a quadratic equation using a suitable method, because 4 reviewed useful formulas allow you to quickly complete a task, especially in cases where the discriminant is an “inconvenient” number. In all simple cases, find the roots and operate on them. For example, in the last example we select the roots using Vieta’s theorem: the sum of the roots should be equal to 13 , and the product of the roots 36 . What are these numbers? Certainly, 4 and 9. Now calculate the sum of the square roots of these numbers: 2+3=5. That's it!
I. Vieta's theorem for the reduced quadratic equation.
Sum of roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term:
x 1 + x 2 = -p; x 1 ∙x 2 =q.
Find the roots of the given quadratic equation using Vieta's theorem.
Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), second coefficient p=-1, and the free member q=-30. First, let's make sure that this equation has roots, and that the roots (if any) will be expressed in integers. To do this, it is enough that the discriminant be a perfect square of an integer.
Finding the discriminant D=b 2 — 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .
Now, according to Vieta’s theorem, the sum of the roots must be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:
x 1 +x 2 =1; x 1 ∙x 2 =-30. We need to choose two numbers such that their product is equal to -30 , and the amount is unit. These are numbers -5 And 6 . Answer: -5; 6.
Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Let's make sure that there are integer roots. Let's find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us select the roots using Vieta’s theorem: the sum of the roots is equal to –р=-6, and the product of the roots is equal to q=8. These are numbers -4 And -2 .
In fact: -4-2=-6=-р; -4∙(-2)=8=q. Answer: -4; -2.
Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free member q=-4. Let's find the discriminant D 1, since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, so we do conclusion: The roots of this equation are not integers and cannot be found using Vieta’s theorem. This means that we solve this equation, as usual, using the formulas (in this case, using the formulas). We get:
Example 4). Write a quadratic equation using its roots if x 1 =-7, x 2 =4.
Solution. The required equation will be written in the form: x 2 +px+q=0, and, based on Vieta’s theorem –p=x 1 +x 2=-7+4=-3 → p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x 2 +3x-28=0.
Example 5). Write a quadratic equation using its roots if:
II. Vieta's theorem for a complete quadratic equation ax 2 +bx+c=0.
The sum of the roots is minus b, divided by A, the product of the roots is equal to With, divided by A:
x 1 + x 2 = -b/a; x 1 ∙x 2 =c/a.
Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11=0.
Solution.
We make sure that this equation will have roots. To do this, it is enough to create an expression for the discriminant, and, without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 . Now let's use theorem Vieta for complete quadratic equations.
x 1 +x 2 =-b:a=- (-7):2=3,5.
Example 7). Find the product of the roots of a quadratic equation 3x 2 +8x-21=0.
Solution.
Let's find the discriminant D 1, since the second coefficient ( 8 ) is an even number. D 1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to Vieta’s theorem, the product of roots x 1 ∙x 2 =c:a=-21:3=-7.
I. ax 2 +bx+c=0– general quadratic equation
Discriminant D=b 2 - 4ac.
If D>0, then we have two real roots:
If D=0, then we have a single root (or two equal roots) x=-b/(2a).
If D<0, то действительных корней нет.
Example 1) 2x 2 +5x-3=0.
Solution. a=2; b=5; c=-3.
D=b 2 - 4ac=5 2 -4∙2∙(-3)=25+24=49=7 2 >0; 2 real roots.
4x 2 +21x+5=0.
Solution. a=4; b=21; c=5.
D=b 2 - 4ac=21 2 - 4∙4∙5=441-80=361=19 2 >0; 2 real roots.
II. ax 2 +bx+c=0 – quadratic equation of particular form with even second
coefficient b
Example 3) 3x 2 -10x+3=0.
Solution. a=3; b=-10 (even number); c=3.
Example 4) 5x 2 -14x-3=0.
Solution. a=5; b= -14 (even number); c=-3.
Example 5) 71x 2 +144x+4=0.
Solution. a=71; b=144 (even number); c=4.
Example 6) 9x 2 -30x+25=0.
Solution. a=9; b=-30 (even number); c=25.
III. ax 2 +bx+c=0 – quadratic equation private type provided: a-b+c=0.
The first root is always equal to minus one, and the second root is always equal to minus With, divided by A:
x 1 =-1, x 2 =-c/a.
Example 7) 2x 2 +9x+7=0.
Solution. a=2; b=9; c=7. Let's check the equality: a-b+c=0. We get: 2-9+7=0 .
Then x 1 =-1, x 2 =-c/a=-7/2=-3.5. Answer: -1; -3,5.
IV. ax 2 +bx+c=0 – quadratic equation of a particular form subject to : a+b+c=0.
The first root is always equal to one, and the second root is equal to With, divided by A:
x 1 =1, x 2 =c/a.
Example 8) 2x 2 -9x+7=0.
Solution. a=2; b=-9; c=7. Let's check the equality: a+b+c=0. We get: 2-9+7=0 .
Then x 1 =1, x 2 =c/a=7/2=3.5. Answer: 1; 3,5.
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Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.
Before studying specific solution methods, note that all quadratic equations can be divided into three classes:
- Have no roots;
- Have exactly one root;
- They have two different roots.
This is an important difference between quadratic equations and linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.
Discriminant
Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.
You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:
- If D< 0, корней нет;
- If D = 0, there is exactly one root;
- If D > 0, there will be two roots.
Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:
Task. How many roots do quadratic equations have:
- x 2 − 8x + 12 = 0;
- 5x 2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16
So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.
The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.
The discriminant is zero - the root will be one.
Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.
By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.
Roots of a quadratic equation
Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:
Basic formula for the roots of a quadratic equation
When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.
- x 2 − 2x − 3 = 0;
- 15 − 2x − x 2 = 0;
- x 2 + 12x + 36 = 0.
First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.
D > 0 ⇒ the equation has two roots. Let's find them:
Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.
D > 0 ⇒ the equation again has two roots. Let's find them
\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]
Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.
D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:
As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of mistakes.
Incomplete quadratic equations
It happens that a quadratic equation is slightly different from what is given in the definition. For example:
- x 2 + 9x = 0;
- x 2 − 16 = 0.
It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:
The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.
Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.
Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:
Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:
- If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
- If (−c /a)< 0, корней нет.
As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.
Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:
Taking the common factor out of bracketsThe product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:
Task. Solve quadratic equations:
- x 2 − 7x = 0;
- 5x 2 + 30 = 0;
- 4x 2 − 9 = 0.
x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.
5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.
4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.
In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems in which certain conditions are introduced on the coefficients of the equation that limit them fall out of sight. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are found more and more often in the Unified State Examination materials and in entrance exams.
Which equation will be called an equation with two variables?
So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are equations in two variables.
Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values is a solution to the equation in question.
Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values of the variables that turn this equation into a true numerical equality.
An equation with two unknowns can:
A) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);
b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);
V) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;
G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is equal to 3. The set of solutions to this equation can be written in the form (k; 3 – k), where k is any real number.
The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.
Factorization
Example 1.
Solve the equation: xy – 2 = 2x – y.
Solution.
We group the terms for the purpose of factorization:
(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:
y(x + 1) – 2(x + 1) = 0;
(x + 1)(y – 2) = 0. We have:
y = 2, x – any real number or x = -1, y – any real number.
Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.
Equality of non-negative numbers to zero
Example 2.
Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).
Solution.
Grouping:
(9x 2 – 12x + 4) + (4y 2 – 12y + 9) = 0. Now each bracket can be folded using the squared difference formula.
(3x – 2) 2 + (2y – 3) 2 = 0.
The sum of two non-negative expressions is zero only if 3x – 2 = 0 and 2y – 3 = 0.
This means x = 2/3 and y = 3/2.
Answer: (2/3; 3/2).
Estimation method
Example 3.
Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.
Solution.
In each bracket we select a complete square:
((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate 
the meaning of the expressions in parentheses.
(x + 1) 2 + 1 ≥ 1 and (y – 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:
(x + 1) 2 + 1 = 1 and (y – 2) 2 + 2 = 2, which means x = -1, y = 2.
Answer: (-1; 2).
Let's get acquainted with another method for solving equations with two variables of the second degree. This method consists of treating the equation as square with respect to some variable.
Example 4.
Solve the equation: x 2 – 6x + y – 4√y + 13 = 0.
Solution.
Let's solve the equation as a quadratic equation for x. Let's find the discriminant:
D = 36 – 4(y – 4√y + 13) = -4y + 16√y – 16 = -4(√y – 2) 2 . The equation will have a solution only when D = 0, that is, if y = 4. We substitute the value of y into the original equation and find that x = 3.
Answer: (3; 4).
Often in equations with two unknowns they indicate restrictions on variables.
Example 5.
Solve the equation in whole numbers: x 2 + 5y 2 = 20x + 2.
Solution.
Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation when divided by 5 gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number not divisible by 5 gives a remainder of 1 or 4. Thus, equality is impossible and there are no solutions.
Answer: no roots.
Example 6.
Solve the equation: (x 2 – 4|x| + 5)(y 2 + 6y + 12) = 3.
Solution.
Let's highlight the complete squares in each bracket:
((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.
Answer: (2; -3) and (-2; -3).
Example 7.
For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). In your answer, indicate the smallest amount.
Solution.
Let's select complete squares:
(x 2 – 2xy + y 2) + (y 2 + 4y + 4) = 37;
(x – y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. We get the sum of the squares of two integers equal to 37 if we add 1 + 36. Therefore:
(x – y) 2 = 36 and (y + 2) 2 = 1 
(x – y) 2 = 1 and (y + 2) 2 = 36.
Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).
Answer: -17.
Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle any equation.
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Online calculator for finding the roots of a cubic equation. You enter the coefficients of a cubic equation and get its solution.
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Cubic Equation Roots Calculator
Description of the online calculator
The calculator calculates the roots of the cubic equation:
(1)
.
To find the roots of this equation, enter the values of the coefficients A, B, C, D in the form fields and click the “Calculate roots” button. After this, the calculation results will appear below. If the coefficients are entered incorrectly, the input field is highlighted in red and the roots are not calculated. Correct the highlighted value and click the “Calculate Roots” button again.
Rules for entering numbers
To enter a number, enter the following in the input field:
-6.626e-34
That is The separator between the integer and fractional parts of a number is a dot.
The order of the number is entered after the Latin letter e.
Calculation method
Let us have a cubic equation:
.
Let's divide it into:
(1)
,
Where , , . Let's make a substitution:
.
We obtain an incomplete equation:
(4)
,
Where
(5)
;
.
We calculate the determinant:
.
If , then we calculate the roots using the Cardano formula:
(6)
,
,
Where
(7)
;
.
When the roots are real. We calculate them using Vieta's formula:
(9)
;
(10)
;
(11)
,
Where
(12)
;
.
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