Square trinomial ax 2 +bx+c can be factorized into linear factors using the formula:
ax 2 +bx+c=a (x-x 1)(x-x 2), Where x 1, x 2- roots of quadratic equation ax 2 +bx+c=0.
Expand quadratic trinomial to linear factors:
Example 1). 2x 2 -7x-15.
Solution. 2x 2 -7x-15=0.
a=2; b=-7; c=-15. This general case for a complete quadratic equation. Finding the discriminant D.
D=b 2 -4ac=(-7) 2 -4∙2∙(-15)=49+120=169=13 2 >0; 2 real roots.
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
2x 2 -7x-15=2 (x+1.5)(x-5)=(2x+3)(x-5). We introduced this trinomial 2x 2 -7x-15 2x+3 And x-5.
Answer: 2x 2 -7x-15= (2x+3)(x-5).
Example 2). 3x 2 +2x-8.
Solution. Let's find the roots of the quadratic equation:
a=3; b=2;c=-8. This special case for a complete quadratic equation with an even second coefficient ( b=2). Finding the discriminant D 1.
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
We introduced the trinomial 3x 2 +2x-8 as a product of binomials x+2 And 3x-4.
Answer: 3x 2 +2x-8 =(x+2)(3x-4).
Example 3). 5x 2 -3x-2.
Solution. Let's find the roots of the quadratic equation:
a=5; b=-3; c=-2. This is a special case for a complete quadratic equation with the following condition: a+b+c=0(5-3-2=0). In such cases first root Always equal to one, A second root equal to the quotient of the free term divided by the first coefficient:
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
5x 2 -3x-2=5 (x-1)(x+0.4)=(x-1)(5x+2). We introduced the trinomial 5x 2 -3x-2 as a product of binomials x-1 And 5x+2.
Answer: 5x 2 -3x-2= (x-1)(5x+2).
Example 4). 6x 2 +x-5.
Solution. Let's find the roots of the quadratic equation:
a=6; b=1; c=-5. This is a special case for a complete quadratic equation with the following condition: a-b+c=0(6-1-5=0). In such cases first root is always equal to minus one, and second root is equal to the minus quotient of dividing the free term by the first coefficient:
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
We introduced the trinomial 6x 2 +x-5 as a product of binomials x+1 And 6x-5.
Answer: 6x 2 +x-5= (x+1)(6x-5).
Example 5). x 2 -13x+12.
Solution. Let's find the roots of the given quadratic equation:
x 2 -13x+12=0. Let's check if it can be applied. To do this, let’s find a discriminant and make sure that it is perfect square whole number.
a=1; b=-13; c=12. Finding the discriminant D.
D=b 2 -4ac=13 2 -4∙1∙12=169-48=121=11 2 .
Let's apply Vieta's theorem: the sum of the roots must be equal to the second coefficient taken from opposite sign, and the product of the roots must be equal to the free term:
x 1 + x 2 =13; x 1 ∙x 2 =12. It is obvious that x 1 =1; x 2 =12.
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2).
x 2 -13x+12=(x-1)(x-12).
Answer: x 2 -13x+12= (x-1)(x-12).
Example 6). x 2 -4x-6.
Solution. Let's find the roots of the given quadratic equation:
a=1; b=-4; c=-6. Second coefficient - even number. Find the discriminant D 1.
The discriminant is not a perfect square of an integer, therefore, Vieta’s theorem will not help us, and we will find the roots using the formulas for the even second coefficient:
Let's apply the formula: ax 2 +bx+c=a (x-x 1)(x-x 2) and write down the answer.
Goals:
- Systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
- Deepen your knowledge by completing a number of tasks, some of which are unfamiliar either in type or method of solution.
- Forming an interest in mathematics through the study of new chapters of mathematics, nurturing a graphic culture through the construction of graphs of equations.
Lesson type: combined.
Equipment: graphic projector.
Visibility: table "Viete's Theorem".
During the classes
1. Oral counting
a) What is the remainder when dividing the polynomial p n (x) = 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) How do we solve equations of the third and fourth degrees?
d) If b is an even number in a quadratic equation, then what is the value of D and x 1; x 2
2. Independent work(in groups)
Write an equation if the roots are known (answers to tasks are coded) “Vieta’s Theorem” is used
1 group
Roots: x 1 = 1; x 2 = -2; x 3 = -3; x 4 = 6
Make up 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 - 23x 2 - 12 x + 36 = 0(this equation is then solved by group 2 on the board)
Solution . We look for whole roots among the divisors of the number 36.
р = ±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. According to Horner's scheme
p 3 (x) = x 3 - x 2 -24x -36
p 3 (-2) = -8 -4 +48 -36 = 0, x 2 = -2
p 2 (x) = x 2 -3x -18=0
x 3 =-3, x 4 =6
Answer: 1;-2;-3;6 sum of roots 2 (P)
2nd group
Roots: x 1 = -1; x 2 = x 3 =2; x 4 =5
Make up 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=0 (group 3 solves this equation on the board)
р = ±1;±2;±4;±5;±10;±20.
p 4 (1)=1-8+15+4-20=-8
р 4 (-1)=1+8+15-4-20=0
p 3 (x) = x 3 -9x 2 +24x -20
p 3 (2) = 8 -36+48 -20=0
p 2 (x) = x 2 -7x +10 = 0 x 1 = 2; x 2 =5
Answer: -1;2;2;5 sum of roots 8(P)
3 group
Roots: x 1 = -1; x 2 =1; x 3 = -2; x 4 =3
Make up an equation:
В=-1+1-2+3=1;В=-1
с=-1+2-3-2+3-6=-7;с=-7
D=2+6-3-6=-1; d=1
e=-1*1*(-2)*3=6
x 4 - x 3- 7x 2 + x + 6 = 0(group 4 solves this equation later on the board)
Solution. We look for whole roots among the divisors of the number 6.
р = ±1;±2;±3;±6
p 4 (1)=1-1-7+1+6=0
p 3 (x) = x 3 - 7x -6
р 3 (-1) = -1+7-6=0
p 2 (x) = x 2 - x -6 = 0; x 1 = -2; x 2 =3
Answer: -1;1;-2;3 Sum of roots 1(O)
4 group
Roots: x 1 = -2; x 2 = -2; x 3 = -3; x 4 = -3
Make up an equation:
B=-2-2-3+3=-4; b=4
c=4+6-6+6-6-9=-5; с=-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)
Solution. We look for whole roots among the divisors of the number -36
р = ±1;±2;±3…
p(1)= 1 + 4-5-36-36 = -72
p 4 (-2) = 16 -32 -20 + 72 -36 = 0
p 3 (x) = x 3 +2x 2 -9x-18 = 0
p 3 (-2) = -8 + 8 + 18-18 = 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 = -1; x 2 = -2; x 3 = -3; x 4 = -4
Write an equation
x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by group 6 on the board)
Solution . We look for whole roots among the divisors of the number 24.
р = ±1;±2;±3
p 4 (-1) = 1 -10 + 35 -50 + 24 = 0
p 3 (x) = x- 3 + 9x 2 + 26x+ 24 = 0
p 3 (-2) = -8 + 36-52 + 24 = O
p 2 (x) = x 2 + 7x+ 12 = 0
Answer: -1;-2;-3;-4 sum-10 (I)
6 group
Roots: x 1 = 1; x 2 = 1; x 3 = -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 group 1 on the board)
Solution . We look for whole 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)= x 2 -5x - 24 = 0
x 3 =-3, x 4 =8
Answer: 1;1;-3;8 sum 7 (L)
3. Solving equations with a parameter
1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is equal to (-1)
Write the 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) = x 2 +2x-15 = 0
x 2 = -1-4 = -5;
x 3 = -1 + 4 = 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 from its division into binomials x-1 and x +2 are equal.
Solution: R=P 3 (1) = P 3 (-2)
P 3 (1) = 1-3 + a- 2a + 6 = 4-a
P 3 (-2) = -8-12-2a-2a + 6 = -14-4a
x 3 -Zx 2 -6x + 12 + 6 = x 3 -Zx 2 -6x + 18
x 2 (x-3)-6(x-3) = 0
(x-3)(x 2 -6) = 0
3) a=0, x 2 -0*x 2 +0 = 0; x 2 =0; x 4 =0
a=0; x=0; x=1
a>0; x=1; x=a ± √a
2. Write an equation
1 group. Roots: -4; -2; 1; 7;
2nd group. Roots: -3; -2; 1; 2;
3 group. Roots: -1; 2; 6; 10;
4 group. Roots: -3; 2; 2; 5;
5 group. Roots: -5; -2; 2; 4;
6 group. Roots: -8; -2; 6; 7.
Instructions
Substitution MethodExpress one variable and substitute it into another equation. You can express any variable at your discretion. For example, express y from the second equation:
x-y=2 => y=x-2Then substitute everything into the first equation:
2x+(x-2)=10 Move everything without “x” to the right side and calculate:
2x+x=10+2
3x=12 Next, to get x, divide both sides of the equation by 3:
x=4. So, you found “x. Find "y. To do this, substitute “x” into the equation from which you expressed “y”:
y=x-2=4-2=2
y=2.
Do a check. To do this, substitute the resulting values into the equations:
2*4+2=10
4-2=2
The unknowns have been found correctly!
A way to add or subtract equations Get rid of any variable right away. In our case, this is easier to do with “y.
Since in “y” there is a “+” sign, and in the second one “-”, then you can perform the addition operation, i.e. left side add it to the left one and add the right one to the right one:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4Substitute “x” into any equation and find “y”:
2*4+y=10
8+y=10
y=10-8
y=2By the 1st method you can see that they were found correctly.
If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have “2x”, and in the second we simply have “x”. In order for x to be reduced during addition, multiply the second equation by 2:
x-y=2
2x-2y=4Then subtract the second from the first equation:
2x+y-(2x-2y)=10-4 Note that if there is a minus before the bracket, then after opening, change it to the opposite:
2x+y-2x+2y=6
3у=6
find y=2x by expressing from any equation, i.e.
x=4
Video on the topic
Tip 2: How to solve a linear equation in two variables
The equation, written in general form ax+bу+c=0, is called a linear equation with two variables. This equation itself contains infinite set solutions, therefore in problems it is always supplemented with something - another equation or limiting conditions. Depending on the conditions provided by the problem, solve a linear equation with two variables should different ways.
You will need
- - linear equation with two variables;
- - second equation or additional conditions.
Instructions
If given a system of two linear equations, solve it in the following way. Choose one of the equations in which the coefficients are variables smaller and express one of the variables, for example, x. Then substitute this value containing y into the second equation. In the resulting equation there will be only one variable y, move all parts with y to the left side, and free ones to the right. Find y and substitute into any of the original equations to find x.
There is another way to solve a system of two equations. Multiply one of the equations by a number so that the coefficient of one of the variables, such as x, is the same in both equations. Then subtract one of the equations from the other (if the right-hand side is not equal to 0, remember to subtract the right-hand sides in the same way). You will see that the x variable has disappeared and only one y variable remains. Solve the resulting equation, and substitute the found value of y into any of the original equalities. Find x.
The third way to solve a system of two linear equations is graphical. Draw a coordinate system and graph two straight lines whose equations are given in your system. To do this, substitute any two x values into the equation and find the corresponding y - these will be the coordinates of the points belonging to the line. The most convenient way to find the intersection with the coordinate axes is to simply substitute the values x=0 and y=0. The coordinates of the point of intersection of these two lines will be the tasks.
If there is only one linear equation in the problem conditions, then you have been given additional conditions through which you can find a solution. Read the problem carefully to find these conditions. If variables x and y indicate distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y hides the number of apples, etc. – then the values can only be . If x is the son's age, it is clear that he cannot be older than father, so indicate this in the task conditions.
Sources:
- how to solve an equation with one variable
By itself the equation with three unknown has many solutions, so most often it is supplemented by two more equations or conditions. Depending on what the initial data are, the course of the decision will largely depend.
You will need
- - a system of three equations with three unknowns.
Instructions
If two of the three systems have only two of the three unknowns, try to express some variables in terms of the others and substitute them into the equation with three unknown. Your goal in this case is to turn it into normal the equation with an unknown person. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.
Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of or a variable so that two unknowns are canceled at once. If there is such an opportunity, take advantage of it; most likely, the subsequent solution will not be difficult. Remember that when multiplying by a number, you must multiply both the left side and the right side. Likewise, when subtracting equations, you must remember that the right-hand side must also be subtracted.
If the previous methods did not help, use in a general way solutions to any equations with three unknown. To do this, rewrite the equations in the form a11x1+a12x2+a13x3=b1, a21x1+a22x2+a23x3=b2, a31x1+a32x2+a33x3=b3. Now create a matrix of coefficients for x (A), a matrix of unknowns (X) and a matrix of free variables (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix of free terms, that is, A*X=B.
Find matrix A to the power (-1) by first finding , note that it should not be equal to zero. After this, multiply the resulting matrix by matrix B, as a result you will receive the desired matrix X, indicating all the values.
You can also find a solution to a system of three equations using Cramer's method. To do this, find the third-order determinant ∆ corresponding to the system matrix. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values of the free terms instead of the values of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.
Sources:
- solutions to equations with three unknowns
Solving a system of equations is challenging and exciting. How more complex system, the more interesting it is to solve it. Most often in mathematics high school there are systems of equations with two unknowns, but in higher mathematics there may be more variables. Systems can be solved using several methods.
Instructions
The most common method for solving a system of equations is substitution. To do this, you need to express one variable in terms of another and substitute it into the second the equation systems, thus leading the equation to one variable. For example, given the following equations: 2x-3y-1=0;x+y-3=0.
From the second expression it is convenient to express one of the variables, moving everything else to the right side of the expression, not forgetting to change the sign of the coefficient: x = 3-y.
Open the brackets: 6-2y-3y-1=0;-5y+5=0;y=1. We substitute the resulting value y into the expression: x=3-y;x=3-1;x=2.
In the first expression all terms are 2, you can put 2 out of brackets distributive property multiplication: 2*(2x-y-3)=0. Now both parts of the expression can be reduced by this number, and then expressed as y, since the modulus coefficient for it is equal to one: -y = 3-2x or y = 2x-3.
Just as in the first case, we substitute this expression in the second the equation and we get: 3x+2*(2x-3)-8=0;3x+4x-6-8=0;7x-14=0;7x=14;x=2. Substitute the resulting value into the expression: y=2x -3;y=4-3=1.
We see that the coefficient for y is the same in value, but different in sign, therefore, if we add these equations, we will completely get rid of y: 4x+3x-2y+2y-6-8=0; 7x-14=0; x=2. Substitute the value of x into any of the two equations of the system and get y=1.
Video on the topic
Biquadratic the equation represents the equation fourth degree, general form which is represented by the expression ax^4 + bx^2 + c = 0. Its solution is based on the use of the method of substitution of unknowns. IN in this case x^2 is replaced by another variable. Thus, the result is an ordinary square the equation, which needs to be solved.
Instructions
Solve the quadratic the equation, resulting from the replacement. To do this, first calculate the value in accordance with the formula: D = b^2? 4ac. In this case, the variables a, b, c are the coefficients of our equation.
Find the roots biquadratic equation. To do this, take the square root of the solutions obtained. If there was one solution, then there will be two - a positive and negative value of the square root. If there were two solutions, the biquadratic equation will have four roots.
Video on the topic
One of the classical methods for solving systems of linear equations is the Gauss method. It lies in consistent exclusion variables when a system of equations using simple transformations is translated into a stepwise system, from which all variables are sequentially found, starting with the last.
Instructions
First, bring the system of equations into such a form when all the unknowns are in strict order. in a certain order. For example, all unknown X's will appear first on each line, all Y's will come after X's, all Z's will come after Y's, and so on. There should be no unknowns on the right side of each equation. Mentally determine the coefficients in front of each unknown, as well as the coefficients on the right side of each equation.
In order to learn how to solve equations with a modulus, you need to remember and learn the definition of a modulus.
From the definition it is clear that the modulus of any number is non-negative. In addition, the definition shows how it is possible get rid of the modulus sign in Eq.
In practice this is done like this:
1) Find the values of the variable at which the expressions under the modulus sign turn to zero.
2) Mark all zeros on the number line. They will divide this line into rays and intervals on which all submodular expressions have a constant sign.
3) We determine the signs of submodular expressions at each interval and expand all modules (replacing them with submodular expressions with a plus sign or a minus sign, depending on the sign of the submodular expression).
4) We solve the resulting equations on each interval (as many intervals, as many equations). Please note that we necessarily select only those solutions that are in a given interval (the resulting solutions may not belong to the interval).
Enough theory already, it’s time to look at examples to see how equations with modulus are solved. Let's start with something simpler.
Solving equations with moduli
Example 1. Solve the equation.
Solution. Since, then. If , then , and the equation takes the form .
From here we get .
Example 2. Solve the equation.
Solution. It follows from the equation that .
Therefore , , , and the equation takes the form or .
Since , the original equation has no roots.
Answer: no roots.
Example 3. Solve the equation.
Solution. Let's rewrite the equation in equivalent form.
The resulting equation belongs to equations of type .
It is known that an equation of this type is equivalent to an inequality. Therefore, here we have or .
Answer: .
I think you have already figured out how to solve this type of equation with a modulus. Let's try to deal with more complex equation.
Example 4. Solve the equation: |x 2 + 2x| – |2 – x| = |x 2 – x|
Finding zeros of submodular expressions:
x 2 + 2x = 0, x(x + 2) = 0, x = 0 or x = ‒ 2. In this case, the parabola y = x 2 + 2x is positive on the intervals (–∞; –2) and (0; +∞ ), and on the interval (–2; 0) it is negative (see figure).
x 2 ‒ x = 0, x(x – 1) = 0, x = 0 or x = 1. This parabola y = x 2 ‒ x is positive on the intervals (–∞; 0) and (1; +∞), and on the interval (0; 1) it is negative (see figure).
2 – x = 0, x = 2, the modulus is positive on the interval (–∞; 0) and takes negative values on the interval (2; +∞) (see figure).
Now we solve the equations on intervals:
1) x ≤ ‒2: x = 1/2
2) –2 ≤ x<0: ‒(x 2 + 2x) – (2 – x) = x 2 ‒ x, ‒x 2 ‒ 2x – 2 + x = x 2 ‒ x, ‒2 x 2 = 2, x 2 = ‒1, there are no solutions.
3) 0 ≤ x<1:
x 2 + 2x ‒ (2 – x) = ‒ (x 2 ‒ x), x 2 + 2x ‒ 2 + x = ‒x 2 + x, 2x 2 + 2x – 2 = 0, x 2 + x – 1 = 0, √D = √5,
x 1 = (‒1 ‒ √5)/2 and x 2 = (‒1 + √5)/2.
Since the first root is negative, it does not belong to our interval, and the second root is greater than zero and less than one; this is our solution on this interval.
4) 1 ≤ x<2: x 2 + 2x – (2 – x) = x 2 – x, x 2 + 2x – 2 + x = x 2 – x, 4x = 2, x= 1/2(not included in the period under consideration)
5) x ≥ 2: x 2 + 2x –(‒(2 – x)) = x 2 – x, x 2 + 2x + 2 – x = x 2 – x, 2x = – 2, x = ‒1(not included in the period under consideration).
Answer: (‒1 + √5)/2 .
You noticed that this equation is solved in the same way as the previous ones, the difference is in the number of intervals. Since there are quadratic expressions under the module, there are more roots, and, accordingly, more gaps.
But how to solve an equation in which the module is under the module? Let's look at an example.
Example 5. Solve the equation |3 – |x – 2|| = 1
The submodular expression can take the value either 1 or – 1. We get two equations:
3 ‒ |x ‒ 2|= ‒1 or 3 ‒ |x ‒ 2|= 1
We solve each equation separately.
1)
3 ‒ |x ‒ 2|= ‒1, ‒|x ‒ 2|= ‒1 – 3, ‒|x ‒ 2|= ‒4, |x ‒ 2|= 4,
x ‒ 2= 4 or x ‒ 2= ‒ 4, from where we get x 1 = 6, x 2 = ‒2.
2)
3 ‒ |x ‒ 2|= 1, ‒|x ‒ 2|= 1 ‒ 3, ‒|x – 2|= ‒2, |x – 2|= 2,
x – 2 = 2 or x – 2 = ‒2,
x 3 = 4, x 4 = 0.
I hope that after studying this article you will be able to successfully solve modulo equations. If you have any questions, sign up for lessons with me. Tutor Valentina Galinevskaya.
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