There are several ways to solve quadratic equations, however, the use of a formula that relates the coefficients of equalities of the named type is universal. This method is often called the "through discriminant" method. Examples of solving quadratic equations using it are given in this article. Every high school student should know about them.
Quadratic equations
The discriminant examples relate to solving quadratic equations. Such equations have the form shown in the photo below.
Here a, b and c are some coefficients (numbers), which are called the quadratic, linear and free terms, respectively. If the X values are known for which equality in the photo is true, then they are said to be the roots of this equation.
As you can see, this equation is called quadratic because "2" is the maximum degree to which x is raised. If a = 0, then the equation turns into a linear one.
Since the maximum degree of the equation is two, only 0, 1, or 2 of its roots can exist, which will take real numerical values.
To solve the above equation, you can use several methods. However, the simplest and most reliable of them is the use of the discriminant formula.
Which formula should I use?
The formula for the method of solving the quadratic equations through discriminant is written as shown in the figure below.
You can see that for its use it is necessary to know all three coefficients of the equation, and the “±” sign in front of the root indicates that the formula allows you to find two different roots at the same time.
The root expression is called the discriminant. It is usually denoted by the Latin letter D or the Greek Δ. Why is this part isolated in the presented formula? The fact is that the sign of D determines how many roots the corresponding equation will have and what they will be.
So, if D is positive, then the expression leads to two different solutions of the quadratic equation, if D is negative, then there are no real numbers that would satisfy the original equality. In this case, we speak of imaginary roots, expressed as complex numbers. Finally, if D = 0, then the formula leads to the existence of a single root.
Important root properties in the discriminant method
Before proceeding to consider specific examples of equations with a discriminant, it is necessary to give two important properties of the roots obtained by the solution method using the formula in question.
The first property is that their sum (x 1 + x 2 ) is equal to the ratio of the linear coefficient (b) to the first or quadratic coefficient (a), taken with the opposite sign, that is, -b / a.
The second property is that the product x 1 * x 2 is always equal to the ratio of the free term (c) to the first coefficient (a), that is, c / a.
The above equalities, which connect the roots of the equation with its coefficients, are the essence of the so-called Vieta theorem.
Note that these formulas are valid for any quadratic equation (including the incomplete one, that is, for which b or / and c is equal to zero).
Further in the article, we consider the use of the formula with the discriminant of the quadratic equation in the examples, which will be formulated in the form of problems of practical importance.
Task number 1. The product and the sum of numbers
The first example of an equation with a discriminant is the following: it is necessary to name two numbers, the sum of which is 34, and the product is 273.
According to the condition of the problem, we compose a system of equations, designating two unknowns as unknowns, as x 1 and x 2 . We get:
x 1 + x 2 = 34
x 1 * x 2 = 273.
Expressing x 2 through x 1 in the first equation, and substituting it in the second, we have: (34 -x 1 ) * x 1 = 273. Opening the brackets, we get: (x 1 ) 2 - 34 * x 1 + 273 = 0. That is, the condition of the problem was reduced to solving the quadratic equation.
We solve this example with the discriminant formula: D = (-34) 2 - 4 * 1 * 273 = 64. It turned out to be convenient for calculating the square root. The solutions of this equation will have the form: x 1 = (34 ± √64) / 2 = (21; 13). We substitute each of the obtained numbers x 1 into the first equation of the above system, we get: x 2 = (34 - 21 = 13; 34 - 13 = 21).
Thus, only one pair of numbers (13 and 21) satisfies the condition of the problem. Since we have already checked the amount, we will now check the product: 13 * 21 = 273.
Task number 2. Preparation and solution of the equation for a given condition
In the following example, a discriminant formula will also be required to solve it. So, the condition is formulated as follows: find a number whose double square exceeds it by 45. We write this condition in the language of mathematics: 2 * x 2 - x = 45. That is, again the problem reduces to finding the unknown x in the quadratic equation.
We transfer all terms to the left side of the equality and calculate the discriminant: D = 1 - 4 * 2 * (-45) = 361. The root of this number is 19. Therefore, the solutions of the equation are numbers: x = (1 ± 19) / (2 * 2 ) = (5; -4.5).
Let's check this result: 2 * 5 2 = 50, which really exceeds the number 5 by 45; 2 * (-4.5) 2 = 40.5, this number also satisfies the condition (40.5 - (-4.5) = 45).
Task number 3. Defining the sides of a right triangle
Another example with the discriminant of the quadratic equation is the following problem: it is known that the difference between the two sides of the rectangle is 70 cm. It is necessary to find its sides if the diagonal of the figure is 130 cm.
The condition of the problem allows you to compose a system of two equations:
x 1 - x 2 = 70
(x 1 ) 2 + (x 2 ) 2 = 130 2 .
Here x 1 and x 2 are the unknown sides of the rectangle. Let us explain where the second equation came from. Since the diagonal of the rectangle forms a triangle with an angle of 90 o with its two sides, its sides, which are equal to x 1 and x 2 , are legs, so you can use their connection with the diagonal-hypotenuse (Pythagorean theorem).
Expressing x 2 from the first equation, substituting its value in the second equation, and opening the brackets in it, we obtain: 2 * (x 1 ) 2 - 140 * x 1 - 12 000 = 0. We solve this classical quadratic equation: D = (140 ) 2 - 4 * 2 * (-12,000) = 115600. Using the calculator allows you to calculate the root of this number, it is 340. The roots of this equation are: x 1 = (140 ± 340) / 4 = (120; -50) . A negative number should be immediately discarded, since the side of the rectangle is a positive value.
Substituting x 1 = 120 cm into the first equation of the system, we obtain x 2 = 50 cm.
Thus, the unknown sides of the rectangle are 120 cm and 50 cm.
Task number 4. Two motorcyclists
The following example of the equation through discriminant is related to the solution of the problem of two motorcyclists. It is known that each of them went to meet the other. The initial distance between them was 130 km, the speed of one was 30 km / h, and the other rode at a speed of 33 km / h more than the number of hours through which they met. You need to find how long motorcyclists will meet.
Denote the unknown time by the letter t. From the conditions of the problem it follows that the speed of the second motorcyclist was 33 + t. Before the meeting, each motorcyclist traveled a distance of 30 * t and (33 + t) * t. Obviously, at the time of the meeting, both vehicles covered the total distance of 130 km (see the problem statement). Then we get the equation: 30 * t + (33 + t) * t = 130. Opening the brackets, we get the following form: t 2 + 63 * t - 130 = 0. In this example, we calculate the discriminant: D = (63) 2 -4 * 1 * (-130) = 4489. The root of it will be 67. The values of t that satisfy the equation will be equal to: t = (-63 ± 67) / 2 = (2; -65). Since time cannot be negative, we get the answer to the problem: motorcyclists will meet in 2 hours.
Task number 5. Boat rental with a group of young people
I would like to complete this article with an example and a solution through the discriminant of one interesting problem: several young people decided to rent a boat for 14,000 rubles. They divided this amount into everyone. However, at the very last moment, three people refused to sail in a boat, so each of the remaining ones was forced to pay another 1,500 rubles. How many people wanted to rent a boat initially?
Suppose initially there were x young people. Then each of them had to pay the amount of 14000 / x rubles. As soon as three people refused to sail, the last amount for each remaining became equal to 14000 / (x-3). Since the last amount has increased compared to the original one per person by 1,500 rubles, then you can draw up the following equation: 14000 / (x-3) - 14000 / x = 1500.
Let us reduce this equation to quadratic. We have: 14000 * x - 14000 * x + 14000 * 3 = 1500 * x * (x-3). Opening the brackets and simplifying the expression, we get: 1500 * x 2 - 4500 * x - 42 000 = 0. Dividing both sides of the equality by 1500, we get the expression: x 2 - 3 * x - 28 = 0. We solve this example with the discriminant: D = 9 - 4 * 1 * (-28) = 121. Then x = (3 ± 11) / 2 = (7; -4).
Thus, initially the group of young people consisted of 7 people.