We will analyze two types of solving systems of equations:
1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.
In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. From any equation, we express one variable.
2. Substitute. We substitute in another equation instead of the expressed variable, the resulting value.
3. We solve the resulting equation with one variable. We find a solution to the system.
To solve system by term-by-term addition (subtraction) need:
1. Select a variable for which we will make the same coefficients.
2. We add or subtract the equations, as a result we get an equation with one variable.
3. We solve the resulting linear equation. We find a solution to the system.
The solution of the system is the intersection points of the graphs of the function.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by the substitution method
Solving the system of equations by the substitution method2x+5y=1 (1 equation)
x-10y=3 (2nd equation)
1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y
2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1
3. We solve the resulting equation with one variable.
2(3+10y)+5y=1 (open brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution of the system of equations is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first paragraph where we expressed we substitute y there.
x=3+10y
x=3+10*(-0.2)=1
It is customary to write points in the first place, we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)
Example #2:
Let's solve by term-by-term addition (subtraction).
Solving a system of equations by the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. Select a variable, let's say we select x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.
3x-2y=1 |*2
6x-4y=2
2x-3y=-10 |*3
6x-9y=-30
2. From the first equation, subtract the second to get rid of the variable x. Solve the linear equation.
__6x-4y=2
5y=32 | :5
y=6.4
3. Find x. We substitute the found y in any of the equations, let's say in the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6
The point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)
Do you want to prepare for exams for free? Tutor online for free. No kidding.
Goals:
- To systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
- To deepen knowledge by completing a series of tasks, some of which are not familiar either in their type or in the method of solving.
- Formation of interest in mathematics through the study of new chapters of mathematics, education of graphic culture through the construction of graphs of equations.
Lesson type: combined.
Equipment: graph projector.
Visibility: table "Vieta's theorem".
During the classes
1. Mental account
a) What is the remainder of the division of the polynomial p n (x) \u003d a n x n + a n-1 x n-1 + ... + a 1 x 1 + a 0 by the binomial x-a?
b) How many roots can a cubic equation have?
c) With what help do we solve the equation of the third and fourth degree?
d) If b is an even number in the quadratic equation, then what is D and x 1; x 2
2. Independent work (in groups)
Make an equation if the roots are known (answers to tasks are coded) Use the "Vieta Theorem"
1 group
Roots: x 1 = 1; x 2 \u003d -2; x 3 \u003d -3; x 4 = 6
Write an equation:
B=1 -2-3+6=2; b=-2
c=-2-3+6+6-12-18=-23; c= -23
d=6-12+36-18=12; d=-12
e=1(-2)(-3)6=36
x 4 -2 x 3 - 23 x 2 - 12 x + 36 = 0(this equation is then solved by group 2 on the board)
Decision . We are looking for integer roots among the divisors of the number 36.
p = ±1; ±2; ±3; ±4; ±6…
p 4 (1)=1-2-23-12+36=0 The number 1 satisfies the equation, therefore =1 is the root of the equation. Horner's scheme
p 3 (x) = x 3 -x 2 -24x -36
p 3 (-2) \u003d -8 -4 +48 -36 \u003d 0, x 2 \u003d -2
p 2 (x) \u003d x 2 -3x -18 \u003d 0
x 3 \u003d -3, x 4 \u003d 6
Answer: 1; -2; -3; 6 the sum of the roots 2 (P)
2 group
Roots: x 1 \u003d -1; x 2 = x 3 =2; x 4 \u003d 5
Write an equation:
B=-1+2+2+5-8; b=-8
c=2(-1)+4+10-2-5+10=15; c=15
D=-4-10+20-10=-4; d=4
e=2(-1)2*5=-20;e=-20
8 + 15 + 4x-20 \u003d 0 (group 3 solves this equation on the board)
p = ±1; ±2; ±4; ±5; ±10; ±20.
p 4 (1)=1-8+15+4-20=-8
p 4 (-1)=1+8+15-4-20=0
p 3 (x) \u003d x 3 -9x 2 + 24x -20
p 3 (2) \u003d 8 -36 + 48 -20 \u003d 0
p 2 (x) \u003d x 2 -7x + 10 \u003d 0 x 1 \u003d 2; x 2 \u003d 5
Answer: -1;2;2;5 sum of roots 8(P)
3 group
Roots: x 1 \u003d -1; x 2 =1; x 3 \u003d -2; x 4 \u003d 3
Write an equation:
B=-1+1-2+3=1;b=-1
s=-1+2-3-2+3-6=-7; s=-7
D=2+6-3-6=-1; d=1
e=-1*1*(-2)*3=6
x 4 - x 3- 7x 2 + x + 6 = 0(this equation is solved later on the board by group 4)
Decision. We are looking for integer roots among the divisors of the number 6.
p = ±1; ±2; ±3; ±6
p 4 (1)=1-1-7+1+6=0
p 3 (x) = x 3 - 7x -6
p 3 (-1) \u003d -1 + 7-6 \u003d 0
p 2 (x) = x 2 -x -6=0; x 1 \u003d -2; x 2 \u003d 3
Answer: -1; 1; -2; 3 The sum of the roots 1 (O)
4 group
Roots: x 1 = -2; x 2 \u003d -2; x 3 \u003d -3; x 4 = -3
Write an equation:
B=-2-2-3+3=-4; b=4
c=4+6-6+6-6-9=-5; c=-5
D=-12+12+18+18=36; d=-36
e=-2*(-2)*(-3)*3=-36; e=-36
x 4+4x 3 - 5x 2 - 36x -36 = 0(this equation is then solved by group 5 on the board)
Decision. We are looking for integer roots among the divisors of the number -36
p = ±1; ±2; ±3…
p(1)= 1 + 4-5-36-36 = -72
p 4 (-2) \u003d 16 -32 -20 + 72 -36 \u003d 0
p 3 (x) \u003d x 3 + 2x 2 -9x-18 \u003d 0
p 3 (-2) \u003d -8 + 8 + 18-18 \u003d 0
p 2 (x) = x 2 -9 = 0; x=±3
Answer: -2; -2; -3; 3 Sum of roots-4 (F)
5 group
Roots: x 1 \u003d -1; x 2 \u003d -2; x 3 \u003d -3; x 4 = -4
Write an equation
x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by the 6th group on the board)
Decision . We are looking for integer roots among the divisors of the number 24.
p = ±1; ±2; ±3
p 4 (-1) = 1 -10 + 35 -50 + 24 = 0
p 3 (x) \u003d x- 3 + 9x 2 + 26x + 24 \u003d 0
p 3 (-2) \u003d -8 + 36-52 + 24 \u003d O
p 2 (x) \u003d x 2 + 7x + 12 \u003d 0
Answer: -1; -2; -3; -4 sum-10 (I)
6 group
Roots: x 1 = 1; x 2 = 1; x 3 \u003d -3; x 4 = 8
Write an equation
B=1+1-3+8=7;b=-7
c=1 -3+8-3+8-24= -13
D=-3-24+8-24=-43; d=43
x 4 - 7x 3- 13x 2 + 43x - 24 = 0 (this equation is then solved by 1 group on the board)
Decision . We are looking for integer roots among the divisors of the number -24.
p 4 (1)=1-7-13+43-24=0
p 3 (1)=1-6-19+24=0
p 2 (x) \u003d x 2 -5x - 24 \u003d 0
x 3 \u003d -3, x 4 \u003d 8
Answer: 1; 1; -3; 8 sum 7 (L)
3. Solution of equations with a parameter
1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is (-1)
Answer in ascending order
R=P 3 (-1)=-1+3-m-15=0
x 3 + 3x 2 -13x - 15 = 0; -1+3+13-15=0
By condition x 1 = - 1; D=1+15=16
P 2 (x) \u003d x 2 + 2x-15 \u003d 0
x 2 \u003d -1-4 \u003d -5;
x 3 \u003d -1 + 4 \u003d 3;
Answer: - 1; -5; 3
In ascending order: -5;-1;3. (b n s)
2. Find all the roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders of its division into binomials x-1 and x + 2 are equal.
Solution: R \u003d R 3 (1) \u003d R 3 (-2)
P 3 (1) \u003d 1-3 + a- 2a + 6 \u003d 4-a
P 3 (-2) \u003d -8-12-2a-2a + 6 \u003d -14-4a
x 3 -3x 2 -6x + 12 + 6 \u003d x 3 -3x 2 -6x + 18
x 2 (x-3)-6(x-3) = 0
(x-3)(x 2 -6) = 0
The product of two factors is equal to zero if and only if at least one of these factors is equal to zero, while the other makes sense.
2 group. Roots: -3; -2; one; 2;3 group. Roots: -1; 2; 6; ten;
4 group. Roots: -3; 2; 2; 5;
5 group. Roots: -5; -2; 2; 4;
6 group. Roots: -8; -2; 6; 7.
Equations
How to solve equations?
In this section, we will recall (or study - as anyone likes) the most elementary equations. So what is an equation? Speaking in human language, this is some kind of mathematical expression, where there is an equals sign and an unknown. Which is usually denoted by the letter "X". solve the equation is to find such x-values that, when substituting into initial expression, will give us the correct identity. Let me remind you that identity is an expression that does not raise doubts even for a person who is absolutely not burdened with mathematical knowledge. Like 2=2, 0=0, ab=ab etc. So how do you solve equations? Let's figure it out.
There are all sorts of equations (I was surprised, right?). But all their infinite variety can be divided into only four types.
4. Other.)
All the rest, of course, most of all, yes ...) This includes cubic, and exponential, and logarithmic, and trigonometric, and all sorts of others. We will work closely with them in the relevant sections.
I must say right away that sometimes the equations of the first three types are so wound up that you don’t recognize them ... Nothing. We will learn how to unwind them.
And why do we need these four types? And then what linear equations solved in one way square others fractional rational - the third, a rest not solved at all! Well, it’s not that they don’t decide at all, I offended mathematics in vain.) It’s just that they have their own special techniques and methods.
But for any (I repeat - for any!) equations is a reliable and trouble-free basis for solving. Works everywhere and always. This base - Sounds scary, but the thing is very simple. And very (very!) important.
Actually, the solution of the equation consists of these same transformations. At 99%. Answer to the question: " How to solve equations?" lies, just in these transformations. Is the hint clear?)
Identity transformations of equations.
AT any equations to find the unknown, it is necessary to transform and simplify the original example. Moreover, so that when changing the appearance the essence of the equation has not changed. Such transformations are called identical or equivalent.
Note that these transformations are just for the equations. In mathematics, there are still identical transformations expressions. This is another topic.
Now we will repeat all-all-all basic identical transformations of equations.
Basic because they can be applied to any equations - linear, quadratic, fractional, trigonometric, exponential, logarithmic, etc. etc.
First identical transformation: both sides of any equation can be added (subtracted) any(but the same!) a number or an expression (including an expression with an unknown!). The essence of the equation does not change.
By the way, you constantly used this transformation, you only thought that you were transferring some terms from one part of the equation to another with a sign change. Type:
The matter is familiar, we move the deuce to the right, and we get:
Actually you taken away from both sides of the equation deuce. The result is the same:
x+2 - 2 = 3 - 2
The transfer of terms to the left-right with a change of sign is simply an abbreviated version of the first identical transformation. And why do we need such deep knowledge? - you ask. Nothing in the equations. Move it, for God's sake. Just don't forget to change the sign. But in inequalities, the habit of transference can lead to a dead end ....
Second identity transformation: both sides of the equation can be multiplied (divided) by the same non-zero number or expression. An understandable limitation already appears here: it is stupid to multiply by zero, but it is impossible to divide at all. This is the transformation you use when you decide something cool like
Understandably, X= 2. But how did you find it? Selection? Or just lit up? In order not to pick up and wait for insight, you need to understand that you are just divide both sides of the equation by 5. When dividing the left side (5x), the five was reduced, leaving a pure X. Which is what we needed. And when dividing the right side of (10) by five, it turned out, of course, a deuce.
That's all.
It's funny, but these two (only two!) identical transformations underlie the solution all equations of mathematics. How! It makes sense to look at examples of what and how, right?)
Examples of identical transformations of equations. Main problems.
Let's start with first identical transformation. Move left-right.
An example for the little ones.)
Let's say we need to solve the following equation:
3-2x=5-3x
Let's remember the spell: "with X - to the left, without X - to the right!" This spell is an instruction for applying the first identity transformation.) What expression with x do we have on the right? 3x? The answer is wrong! On our right - 3x! Minus three x! Therefore, when shifting to the left, the sign will change to a plus. Get:
3-2x+3x=5
So, the X's were put together. Let's do the numbers. Three on the left. What sign? The answer "with none" is not accepted!) In front of the triple, indeed, nothing is drawn. And this means that in front of the triple is plus. So the mathematicians agreed. Nothing is written, so plus. Therefore, the triple will be transferred to the right side with a minus. We get:
-2x+3x=5-3
There are empty spaces left. On the left - give similar ones, on the right - count. The answer is immediately:
In this example, one identical transformation was enough. The second was not needed. Well, okay.)
An example for the elders.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.
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, we note that all quadratic equations can be divided into three classes:
- Have no roots;
- They have exactly one root;
- They have two different roots.
This is an important difference between quadratic and linear equations, 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 .
This formula must be known 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 think. 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;
- 5x2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
We write 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 the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.
The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.
The discriminant is equal to zero - the root will be one.
Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.
By the way, if you “fill your hand”, after a while you will no longer need to write out 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 so much.
The roots of a quadratic equation
Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:
The basic formula for the roots of a quadratic equation
When D = 0, you can use any of these formulas - you 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 - x2 = 0;
- x2 + 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 be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.
Incomplete quadratic equations
It happens that the quadratic equation is somewhat different from what is given in the definition. For example:
- x2 + 9x = 0;
- x2 − 16 = 0.
It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate 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 \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.
Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:
Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:
- If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
- If (−c / a )< 0, корней нет.
As you can see, the 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 of 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 negative, there will be no roots at all.
Now let's deal with 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 factorize the polynomial:
Taking the common factor out of the bracketThe product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:
Task. Solve quadratic equations:
- x2 − 7x = 0;
- 5x2 + 30 = 0;
- 4x2 − 9 = 0.
x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.
5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the 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 \u003d -1.5.