Quadratic equations often appear in a number of problems in mathematics and physics, so every student must be able to solve them. This article discusses in detail the basic methods for solving quadratic equations, as well as examples of their use.
Which equation is called quadratic?
First of all, we will answer the question of this paragraph in order to better understand what will be discussed in the article. So, the quadratic equation has the following general form: c + b * x + a * x 2 = 0, where a, b, c are some numbers called coefficients. Here a ≠ 0 is an indispensable condition; otherwise, the indicated equation degenerates into a linear one. The remaining coefficients (b, c) can take absolutely any value, including zero. So, expressions of the type a * x 2 = 0, where b = 0 and c = 0 or c + a * x 2 = 0, where b = 0, or b * x + a * x 2 = 0, where c = 0 - these are also quadratic equations, which are called incomplete, because in them either the linear coefficient b is equal to zero, the free term c is zero, or both of them vanish.
The equation in which a = 1 is called reduced, that is, it has the form: x 2 + c / a + (b / a) * x = 0.
The solution of the quadratic equation is to find such values of x that satisfy its equality. These values are called roots. Since the equation in question is an expression of the second degree, this means that the maximum number of its roots cannot exceed two.
What methods of solving quadratic equations exist
In general, there are 4 solution methods. Their names are listed below:
- Factorization.
- Addition to the square.
- Using a well-known formula (through discriminant).
- The solution method is geometric.
As you can see from the list, the first three methods are algebraic, so they are used more often than the last, which involves plotting a function.
There is another way to solve the quadratic equations by the Vieta theorem. It could be included 5th in the list above, however, this has not been done, since the Vieta theorem is a simple consequence of the 3rd method.
Further in the article, we will consider the above-mentioned methods of solution in more detail, as well as give examples of their use to find the roots of specific equations.
Method number 1. Factorization
There is a beautiful name for this method in the mathematics of quadratic equations: factorization. The essence of this method is as follows: it is necessary to present the quadratic equation in the form of the product of two terms (expressions), which should be zero. After such a representation, we can use the property of the product, which will be equal to zero only when one or more (all) of its members are zero.
Now consider the sequence of specific actions that must be performed to find the roots of the equation:
- Transfer all terms to one part of the expression (for example, to the left) so that only 0 remains in the other part of the expression (right).
- Present the sum of the members in one part of the equality in the form of the product of two linear equations.
- Equate each of the linear expressions to zero and solve them.
As you can see, the factorization algorithm is quite simple, however, most students have difficulties during the implementation of the second paragraph, so we explain it in more detail.
To guess which 2 linear expressions, when multiplied by each other, will give the desired quadratic equation, you need to remember two simple rules:
- The linear coefficients of two linear expressions when multiplied by each other should give the first coefficient of the quadratic equation, that is, the number a.
- The free terms of linear expressions in their product should give the number c of the desired equation.
After all the numbers of the factors are selected, they should be multiplied, and if they give the desired equation, then go to step 3 in the above algorithm, otherwise you should change the factors, but you need to do this so that the rules given are always satisfied.
Factorization Solution Example
Let us show clearly how to compose an algorithm for solving the quadratic equation and find unknown roots. Let an arbitrary expression be given, for example, 2 * x-5 + 5 * x 2 -2 * x 2 = x 2 + 2 + x 2 +1. Let us proceed to its solution, observing the sequence of items from 1 to 3, which are set out in the previous paragraph of the article.
Point 1. Move all the terms to the left side and arrange them in the classical sequence for the quadratic equation. We have the following equality: 2 * x + (- 8) + x 2 = 0.
Point 2. We break it into the product of linear equations. Since a = 1, and c = -8, we choose, for example, such a product (x-2) * (x + 4). It satisfies the rules for finding prospective factors set out in the paragraph above. If we open the brackets, we get: -8 + 2 * x + x 2 , that is, we get exactly the same expression as on the left side of the equation. This means that we correctly guessed the factors, and we can proceed to the third point of the algorithm.
Point 3. Equating each factor to zero, we obtain: x = -4 and x = 2.
If there are any doubts about the result, it is recommended to perform a check by substituting the found roots in the original equation. In this case, we have: 2 * 2 + 2 2 -8 = 0 and 2 * (- 4) + (- 4) 2 -8 = 0. The roots are found correctly.
Thus, using the factorization method, we found that the given equation has two different roots: 2 and -4.
Method number 2. Supplement to full square
In the algebra of equations of quadratic equations, the multiplier method cannot always be used, since in the case of fractional values of the coefficients of the quadratic equation, difficulties arise in implementing paragraph 2 of the algorithm.
The full square method, in turn, is universal and can be applied to quadratic equations of any type. Its essence is to perform the following operations:
- The terms of the equation containing the coefficients a and b must be transferred to one part of the equality, and the free term c to the other.
- Next, it is necessary to divide the parts of the equality (left and right) by the coefficient a, that is, present the equation in the reduced form (a = 1).
- The sum of the terms with coefficients a and b is represented as the square of the linear equation. Since a = 1, then the linear coefficient will be equal to 1, as for the free term of the linear equation, then it should be equal to half the linear coefficient of the reduced quadratic equation. After the square of the linear expression is compiled, it is necessary to add to the right side of the equality, where the free term is, the corresponding number, which is obtained when the square is opened.
- Take the square root with the signs "+" and "-" and solve the equation already obtained is linear.
At first glance, the described algorithm can be perceived as quite complex, however, in practice it is easier to implement than the factorization method.
Example solution using the complement to the full square
We give an example of the quadratic equation for training its solution by the method described in the previous paragraph. Let the quadratic equation -10 - 6 * x + 5 * x 2 = 0 be given. We begin to solve it by following the algorithm described above.
Point 1. We use the transfer method to solve the quadratic equations, we get: - 6 * x + 5 * x 2 = 10.
Paragraph 2. The reduced form of this equation is obtained by dividing by each member its number (if both parts are divided or multiplied by the same number, then the equality will be preserved). As a result of the transformations, we obtain: x 2 - 6/5 * x = 2.
Point 3. Half of the coefficient - 6/5 is -6/10 = -3/5, we use this number to compose a full square, we get: (-3 / 5 + x) 2 . We will reveal it and the obtained free term should be subtracted from the left-hand side of the equality to satisfy the original form of the quadratic equation, which is equivalent to adding it to the right-hand side. As a result, we get: (-3 / 5 + x) 2 = 59/25.
Point 4. Calculate the square root with positive and negative signs and find the roots: x = 3/5 ± √59 / 5 = (3 ± √59) / 5. The two found roots have values: x 1 = (√59 + 3) / 5 and x 1 = (3-√59) / 5.
Since the calculations are related to the roots, it is likely to make a mistake. Therefore, it is recommended to check the correctness of the roots x 2 and x 1 . We get for x 1 : 5 * ((3 + √59) / 5) 2 -6 * (3 + √59) / 5 - 10 = (9 + 59 + 6 * √59) / 5 - 18/5 - 6 * √59 / 5-10 = 68 / 5-68 / 5 = 0. Now we substitute x 2 : 5 * ((3-√59) / 5) 2 -6 * (3-√59) / 5 - 10 = (9 + 59-6 * √59) / 5 - 18/5 + 6 * √59 / 5-10 = 68 / 5-68 / 5 = 0.
Thus, we have shown that the found roots of the equation are true.
Method number 3. The use of the well-known formula
This method of solving quadratic equations is perhaps the simplest, since it consists in substituting the coefficients in a well-known formula. To use it, you don’t need to think about compiling decision algorithms, it is enough to remember only one formula. It is shown in the figure above.
In this formula, the radical expression (b 2 -4 * a * c) is called the discriminant (D). From its value depends on what roots will be obtained. There are 3 possible cases:
- D> 0, then the equation of the root two has real and different.
- D = 0, then we get one root, which can be calculated from the expression x = -b / (a * 2).
- D <0, then we get two different imaginary roots, which are represented as complex numbers. For example, the number 3-5 * i is complex, while the imaginary unit i satisfies the property: i 2 = -1.
Example solution through discriminant calculation
Here is an example of a quadratic equation for training using the above formula. Find the roots for -3 * x 2 -6 + 3 * x + 4 * x = 0. First, calculate the value of the discriminant, we get: D = b 2 -4 * a * c = 7 2 -4 * (- 3) * (-6) = -23.
Since D <0 is obtained, it follows that the roots of the equation in question are complex numbers. We find them by substituting the found value of D in the formula given in the previous paragraph (it is also presented in the photo above). We get: x = 7/6 ± √ (-23) / (- 6) = (7 ± i * √23) / 6.
Method number 4. Using function graph
It is also called the graphical method for solving quadratic equations. It should be said that it is used, as a rule, not for quantitative, but for a qualitative analysis of the equation in question.
The essence of the method is to plot the quadratic function y = f (x), which is a parabola. Then, it is necessary to determine at what points the abscissa axis (X) of the parabola intersects, they will be the roots of the corresponding equation.
To say whether the parabola will cross the X axis, it is enough to know the position of its minimum (maximum) and the direction of its branches (they can either increase or decrease). Two properties of this curve should be remembered:
- If a> 0 - the parabolas of the branch are directed upwards, on the contrary, if a <0, then they go down.
- The coordinate of the minimum (maximum) of the parabola is always equal to x = -b / (2 * a).
For example, it is necessary to determine whether the equation -4 * x + 5 * x 2 +10 = 0 has roots. The corresponding parabola will be directed upward, since a = 5> 0. Its extremum has coordinates: x = 4/10 = 2/5, y = -4 * 2/5 + 5 * (2/5) 2 +10 = 9.2. Since the minimum of the curve lies above the abscissa axis (y = 9.2), it does not intersect the latter at any x. That is, the equation does not have real roots.
Vieta Theorem
As noted above, this theorem is a consequence of method No. 3, which is based on the application of the discriminant formula. The essence of the Vieta theorem is that it allows you to tie into the equality the coefficients of the equation and its roots. We obtain the corresponding equalities.
We use the formula to calculate the roots through the discriminant. Add two roots, we get: x 1 + x 2 = -b / a. Now we multiply the roots by each other: x 1 * x 2 , after a series of simplifications we get the number c / a.
Thus, to solve the quadratic equations by the Vieta theorem, we can use the resulting two equalities. If all three coefficients of the equation are known, then the roots can be found by solving the corresponding system of these two equations.
An example of using the Vieta theorem
It is necessary to draw up a quadratic equation if it is known that it has the form x 2 + c = -b * x and its roots are 3 and -4.
Since in the considered equation a = 1, then the Vieta formulas will have the form: x 2 + x 1 = -b and x 2 * x 1 = s. Substituting the known values of the roots, we get: b = 1 and c = -12. As a result, the restored quadratic equation reduced will look like: x 2 -12 = -1 * x. You can substitute the meaning of the roots in it and make sure that the equality holds.
The reverse application of the Vieta theorem, that is, the calculation of the roots using the well-known form of the equation, makes it possible to quickly (intuitively) find solutions for small integers a, b and c.