Quadratic equation or an equation of the second degree with one unknown is an equation that, after transformations, can be reduced to the following form:

ax 2 + bx + c = 0 - quadratic equation

where x is the unknown, and a, b and c- coefficients of the equation. In quadratic equations a is called the first coefficient ( a ≠ 0), b is called the second coefficient, and c is called a known or free member.

The equation:

ax 2 + bx + c = 0

called complete quadratic equation. If one of the coefficients b or c is zero, or both of these coefficients are equal to zero, then the equation is presented as an incomplete quadratic equation.

Reduced quadratic equation

The complete quadratic equation can be reduced to a more convenient form by dividing all its terms by a, that is, for the first coefficient:

The equation x 2 + px + q= 0 is called a reduced quadratic equation. Therefore, any quadratic equation in which the first coefficient is equal to 1 can be called reduced.

For example, the equation:

x 2 + 10x - 5 = 0

is reduced, and the equation:

3x 2 + 9x - 12 = 0

can be replaced by the above equation by dividing all its terms by -3:

x 2 - 3x + 4 = 0

Solving quadratic equations

To solve a quadratic equation, you need to bring it to one of the following forms:

ax 2 + bx + c = 0

ax 2 + 2kx + c = 0

x 2 + px + q = 0

Each type of equation has its own formula for finding the roots:

Pay attention to the equation:

ax 2 + 2kx + c = 0

this is the converted equation ax 2 + bx + c= 0, in which the coefficient b- even, which allows it to be replaced by type 2 k. Therefore, the formula for finding the roots for this equation can be simplified by substituting 2 k instead of b:

Example 1 Solve the equation:

3x 2 + 7x + 2 = 0

Since in the equation the second coefficient is not an even number, and the first coefficient is not equal to one, then we will look for the roots according to the very first formula, called general formula finding the roots of a quadratic equation. At first

a = 3, b = 7, c = 2

Now, to find the roots of the equation, we simply substitute the values ​​of the coefficients into the formula:

x 1 =	-2	= -	1	, x 2 =	-12	= -2
	6		3		6

Answer: -	1	, -2.
	3

Example 2:

x 2 - 4x - 60 = 0

Let's determine what the coefficients are equal to:

a = 1, b = -4, c = -60

Since the second coefficient in the equation is even number, then we will use the formula for quadratic equations with an even second coefficient:

x 1 = 2 + 8 = 10, x 2 = 2 - 8 = -6

Answer: 10, -6.

Example 3

y 2 + 11y = y - 25

Let's bring the equation to general view:

y 2 + 11y = y - 25

y 2 + 11y - y + 25 = 0

y 2 + 10y + 25 = 0

Let's determine what the coefficients are equal to:

a = 1, p = 10, q = 25

Since the first coefficient is equal to 1, we will look for the roots using the formula for the above equations with an even second coefficient:

Answer: -5.

Example 4

x 2 - 7x + 6 = 0

Let's determine what the coefficients are equal to:

a = 1, p = -7, q = 6

Since the first coefficient is equal to 1, we will look for the roots using the formula for the given equations with an odd second coefficient:

x 1 = (7 + 5) : 2 = 6, x 2 = (7 - 5) : 2 = 1

There are several classes of equations that are solved by reducing them to quadratic equations. One of such equations are biquadratic equations.

Biquadratic Equations

Biquadratic equations are equations of the form a*x^4 + b*x^2 + c = 0, where a is not equal to 0.

Biquadratic equations are solved using the substitution x^2 =t. After such a substitution, we obtain a quadratic equation for t. a*t^2+b*t+c=0. We solve the resulting equation, we have in general case t1 and t2. If at this stage a negative root is obtained, it can be excluded from the solution, since we took t \u003d x ^ 2, and the square of any number is a positive number.

Returning to the original variables, we have x^2 =t1, x^2=t2.

x1,2 = ±√(t1), x3,4=±√(t2).

Let's take a small example:

9*x^4+5*x^2 - 4 = 0.

We introduce the replacement t=x^2. Then the original equation will take the following form:

9*t^2+5*t-4=0.

We solve this quadratic equation by any of the known methods, we find:

t1=4/9, t2=-1.

The root -1 is not suitable, since the equation x^2 = -1 does not make sense.

There remains the second root 4/9. Passing to the original variables, we have the following equation:

x^2 = 4/9.

x1=-2/3, x2=2/3.

This will be the solution to the equation.

Answer: x1=-2/3, x2=2/3.

Another type of equations that can be reduced to quadratic equations are fractional rational equations. Rational equations are equations whose left and right sides are rational expressions. If in a rational equation the left or right sides are fractional expressions, then this rational equation called fractional.

Scheme for solving a fractional rational equation

General scheme for solving a fractional rational equation.

1. Find common denominator all the fractions that are in the equation.

2. Multiply both sides of the equation by a common denominator.

3. Solve the resulting whole equation.

4. Check the roots, and exclude those that turn the common denominator to zero.

Consider an example:

Solve a fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).

We will adhere to the general scheme. Let us first find the common denominator of all fractions.

We get x*(x-5).

Multiply each fraction by a common denominator and write the resulting whole equation.

x*(x+3) + (x-5) = (x+5);

Let's simplify the resulting equation. We get

x^2+3*x + x-5 - x - 5 =0;

x^2+3*x-10=0;

Got simple reduced quadratic equation. We solve it by any of the known methods, we get the roots x=-2 and x=5. Now we check the obtained solutions. We substitute the numbers -2 and 5 in the common denominator.

At x=-2, the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. So the number -2 will be the root of the original fractional rational equation.

At x=5, the common denominator x*(x-5) becomes zero. Therefore, this number is not the root of the original fractional rational equation, since there will be division by zero.

Answer: x=-2.

General theory of problem solving using equations

Before moving on to specific types of problems, we first give general theory to solve various problems using equations. First of all, problems in such disciplines as economics, geometry, physics and many others are reduced to equations. The general procedure for solving problems using equations is as follows:

• All the quantities we are looking for from the condition of the problem, as well as any auxiliary ones, are denoted by variables convenient for us. Most often, these variables are the last letters of the Latin alphabet.
• Using data in tasks numerical values, as well as verbal relationships, one or more equations are compiled (depending on the condition of the problem).
• They solve the resulting equation or their system and throw out “illogical” solutions. For example, if you need to find the area, then a negative number, obviously, will be an extraneous root.
• We get the final answer.

An example of a problem in algebra

Here we give an example of a problem that reduces to a quadratic equation without relying on any particular area.

Example 1

Find two such irrational numbers, when added together, the squares of which will be five, and when they are usually added to each other, three.

Let's denote these numbers by the letters $x$ and $y$. According to the condition of the problem, it is quite easy to compose two equations $x^2+y^2=5$ and $x+y=3$. We see that one of them is square. To find a solution, you need to solve the system:

$\cases(x^2+y^2=5,\\x+y=3.)$

First, we express from the second $x$

Substituting into the first and performing elementary transformations

$(3-y)^2 +y^2=5$

$9-6y+y^2+y^2=5$

We have moved on to solving a quadratic equation. Let's do it with formulas. Let's find the discriminant:

First root

$y=\frac(3+\sqrt(17))(2)$

Second root

$y=\frac(3-\sqrt(17))(2)$

Let's find the second variable.

For the first root:

$x=3-\frac(3+\sqrt(17))(2)=\frac(3-\sqrt(17))(2)$

For the second root:

$x=3-\frac(3-\sqrt(17))(2)=\frac(3+\sqrt(17))(2)$

Since the sequence of numbers is not important to us, we get one pair of numbers.

Answer: $\frac(3-\sqrt(17))(2)$ and $\frac(3+\sqrt(17))(2)$.

An example of a problem in physics

Consider an example of a problem that leads to the solution of a quadratic equation in physics.

Example 2

A helicopter flying uniformly in calm weather has a speed of $250$ km/h. He needs to fly from his base to the fire site, which is $70$ km away from it, and return back. At this time, the wind was blowing towards the base, slowing down the movement of the helicopter towards the forest. Because of what he got back to the base 1 hour earlier. Find the wind speed.

Let's denote the wind speed as $v$. Then we get that the helicopter will fly towards the forest with a real speed equal to $250-v$, and back its real speed will be $250+v$. Let's calculate the time for the way there and the way back.

$t_1=\frac(70)(250-v)$

$t_2=\frac(70)(250+v)$

Since the helicopter got back to the base $1$ an hour earlier, we will have

$\frac(70)(250-v)-\frac(70)(250+v)=1$

We reduce the left side to a common denominator, apply the proportion rule and perform elementary transformations:

$\frac(17500+70v-17500+70v)((250-v)(250+v))=1$

$140v=62500-v^2$

$v^2+140v-62500=0$

Received a quadratic equation to solve this problem. Let's solve it.

We will solve it using the discriminant:

$D=19600+250000=269600≈519^2$

The equation has two roots:

$v=\frac(-140-519)(2)=-329.5$ and $v=\frac(-140+519)(2)=189.5$

Since we were looking for speed (which cannot be negative), it is obvious that the first root is superfluous.

Answer: $189.5$

An example of a problem in geometry

Consider an example of a problem that leads to the solution of a quadratic equation in geometry.

Example 3

Find the area right triangle, which satisfies the following conditions: its hypotenuse is $25$, and the length of the legs is $4$ to $3$.

In order to find the desired area, we need to find the legs. We mark one part of the leg through $x$. Then expressing the legs in terms of this variable, we get that their lengths are equal to $4x$ and $3x$. Thus, from the Pythagorean theorem, we can compose the following quadratic equation:

$(4x)^2+(3x)^2=625$

(the root $x=-5$ can be ignored, since the leg cannot be negative)

We got that the legs are equal to $20$ and $15$ respectively, so the area is

$S=\frac(1)(2)\cdot 20\cdot 15=150$

There are several classes of equations that are solved by reducing them to quadratic equations. One of such equations are biquadratic equations.

Biquadratic Equations

Biquadratic equations are equations of the form a*x^4 + b*x^2 + c = 0, where a is not equal to 0.

Biquadratic equations are solved using the substitution x^2 =t. After such a substitution, we obtain a quadratic equation for t. a*t^2+b*t+c=0. We solve the resulting equation, in the general case we have t1 and t2. If at this stage a negative root is obtained, it can be excluded from the solution, since we took t \u003d x ^ 2, and the square of any number is a positive number.

Returning to the original variables, we have x^2 =t1, x^2=t2.

x1,2 = ±√(t1), x3,4=±√(t2).

Let's take a small example:

9*x^4+5*x^2 - 4 = 0.

We introduce the replacement t=x^2. Then the original equation will take the following form:

We solve this quadratic equation by any of the known methods, we find:

The root -1 is not suitable, since the equation x^2 = -1 does not make sense.

There remains the second root 4/9. Passing to the original variables, we have the following equation:

x1=-2/3, x2=2/3.

This will be the solution to the equation.

Answer: x1=-2/3, x2=2/3.

Another type of equations that can be reduced to quadratic equations are fractional rational equations. Rational equations are equations in which the left and right sides are rational expressions. If in a rational equation the left or right parts are fractional expressions, then such a rational equation is called fractional.

Scheme for solving a fractional rational equation

1. Find the common denominator of all fractions that are included in the equation.

2. Multiply both sides of the equation by a common denominator.

3. Solve the resulting whole equation.

4. Check the roots, and exclude those that turn the common denominator to zero.

Consider an example:

Solve a fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).

We will adhere to the general scheme. Let us first find the common denominator of all fractions.

We get x*(x-5).

Multiply each fraction by a common denominator and write the resulting whole equation.

x*(x+3) + (x-5) = (x+5);

Let's simplify the resulting equation. We get

x^2+3*x + x-5 - x - 5 =0;

Got simple reduced quadratic equation. We solve it by any of the known methods, we get the roots x=-2 and x=5. Now we check the obtained solutions. We substitute the numbers -2 and 5 in the common denominator.

At x=-2, the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. So the number -2 will be the root of the original fractional rational equation.

MUNICIPAL INSTITUTION OF EDUCATION TUMANOVSKAYA SECONDARY EDUCATIONAL SCHOOL OF MOSKALENSKY MUNICIPAL DISTRICT OF OMSK REGION

Lesson topic: EQUATIONS REDUCED TO SQUARE

Developed by the teacher of mathematics, physics Tumanovskaya secondary school TATYANA VIKTOROVNA

2008

The purpose of the lesson: 1) consider ways to solve equations that are reduced to quadratic ones; learn how to solve these equations. 2) to develop the speech and thinking of students, attentiveness, logical thinking. 3) instill an interest in mathematics,

Lesson type: Lesson learning new material

Lesson plan: 1. organizational stage
2. oral work
3. practical work
4. Summing up the lesson

DURING THE CLASSES
Today in the lesson we will get acquainted with the topic "Equations reducible to square". Each student should be able to correctly and rationally solve equations, learn to apply various ways when solving the given quadratic equations.
1. Oral work 1. Which of the numbers: -3, -2, -1, 0, 1, 2, 3 are the roots of the equation: a) x 3 - x \u003d 0; b) y 3 - 9y = 0; c) y 3 + 4y = 0? How many solutions can an equation of the third degree have? What method did you use to solve these equations?2. Check the equation solution: x 3 - 3x 2 + 4x - 12 = 0 x 2 (x - 3) + 4 (x - 3) = 0(x - 3) (x 2 + 4) = 0 (x - 3) (x - 2) (x + 2) = 0 Answer: x = 3, x = -2, x = 2 Students explain their mistake. I summarize the oral work. So, you were able to solve the three proposed equations orally, find the mistake made in solving the fourth equation. When solving equations orally, the following two methods were used: taking the common factor out of the bracket sign and factoring. Now let's try to apply these methods when doing written work.
2. Practical work 1. One student solves the equation on the board 25x 3 - 50x 2 - x + 2 = 0 When solving, he pays special attention to the change of signs in the second bracket. Speaks the whole solution and finds the roots of the equation.2. The equation x 3 - x 2 - 4 (x - 1) 2 \u003d 0 is proposed to be solved by stronger students. When checking the solution, I pay special attention to the most important points for students.3. Board work. solve the equation (x 2 + 2x) 2 - 2 (x 2 + 2x) - 3 \u003d 0 When solving this equation, students find out that it is necessary to use a “new” way - the introduction of a new variable.Denote by the variable y \u003d x 2 + 2x and substitute into this equation. y 2 - 2y - 3 = 0. Let's solve the quadratic equation for the variable y. Then we find the value of x.4 . Consider the equation (x 2 - x + 1) (x 2 - x - 7) = 65. Let's answer the questions:- what degree is this equation?- what is the most rational way to solve it?- what new variable should be introduced? (x 2 - x + 1) (x 2 - x - 7) = 65 Denote y \u003d x 2 - x (y + 1) (y - 7) \u003d 65The class then solves the equation on its own. We check the solutions of the equation at the blackboard.5. For strong students, I suggest solving the equation x 6 - 3x 4 - x 2 - 3 = 0 Answer: -1, 1 6. The equation (2x 2 + 7x - 8) (2x 2 + 7x - 3) - 6 = 0 class proposes to solve as follows: the strongest students decide on their own; for the rest, one of the students on the board decides.Solve: 2x 2 + 7x = y(y - 8) (y - 3) - 6 = 0 We find: y1 \u003d 2, y2 \u003d 9 We substitute into our equation and find the values ​​of x, for this we solve the equations:2x 2 + 7x = 2 2x 2 + 7x = 9As a result of solving two equations, we find four values ​​of x, which are the roots of this equation.7. At the end of the lesson, I propose to verbally solve the equation x 6 - 1 = 0. When solving, it is necessary to apply the formula for the difference of squares, it is easy to find the roots.(x 3) 2 - 1 \u003d 0 (x 3 - 1) (x 3 + 1) \u003d 0 Answer: -1, 1.
3. Summing up the lesson Once again, I draw students' attention to the methods that were used in solving equations that are reduced to square ones. The work of students in the lesson is evaluated, I comment on the assessments and point out the mistakes made. We write down our homework. As a rule, the lesson takes place at a fast pace, the performance of students is high. Many thanks to all for the good work.