Solution to Example 7: How to solve an equation with variables (unknowns) on both sides of the equation

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It is also recommended to familiarize yourself with the basic operations on matrices.

Example No. 1. Exercise. Find the solution to the matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: A·X·B = C.
The determinant of matrix A is equal to detA=-1
Since A is a non-singular matrix, there is an inverse matrix A -1 . Multiply both sides of the equation on the left by A -1: Multiply both sides of this equation on the left by A -1 and on the right by B -1: A -1 ·A·X·B·B -1 = A -1 ·C·B -1 . Since A A -1 = B B -1 = E and E X = X E = X, then X = A -1 C B -1

Inverse matrix A -1:
Let's find the inverse matrix B -1.
Transposed matrix B T:
Inverse matrix B -1:
We look for matrix X using the formula: X = A -1 ·C·B -1

Answer:

Example No. 2. Exercise. Solve matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: A·X = B.
The determinant of matrix A is detA=0
Since A is a singular matrix (the determinant is 0), therefore the equation has no solution.

Example No. 3. Exercise. Find the solution to the matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: X A = B.
The determinant of matrix A is detA=-60
Since A is a non-singular matrix, there is an inverse matrix A -1 . Let's multiply both sides of the equation on the right by A -1: X A A -1 = B A -1, from where we find that X = B A -1
Let's find the inverse matrix A -1 .
Transposed matrix A T:
Inverse matrix A -1:
We look for matrix X using the formula: X = B A -1


Answer: >

In simple algebraic equations, a variable is on only one side of the equation, but in more complex equations, variables can be on both sides of the equation. When solving such equations, always remember that any operation that is performed on one side of the equation must also be performed on the other side. Using this rule, variables can be moved from one side of an equation to the other to isolate them and calculate their values.

Steps

Solving equations with one variable on both sides of the equation

1. Apply the distributive law (if necessary). This law states that a (b + c) = a b + a c (\displaystyle a(b+c)=ab+ac). The distributive law allows you to open the brackets by multiplying the term outside the brackets by each term in the brackets.

   • For example, if given an equation, use the distributive law to multiply the term outside the parentheses by each term in the parentheses:
     2 (10 − 2 x) = 4 (2 x + 2) (\displaystyle 2(10-2x)=4(2x+2))
     

2. Get rid of a variable on one side of the equation. To do this, subtract or add the same term with the variable. For example, if a variable term is subtracted, add the same term to get rid of it; if a term with a variable is added, subtract the same term to get rid of it. It is usually easier to get rid of the variable with the smaller coefficient.

   • For example, in Eq. 20 − 4 x = 8 x + 8 (\displaystyle 20-4x=8x+8) get rid of your penis − 4 x (\displaystyle -4x); for this add 4 x (\displaystyle 4x):
     20 − 4 x + 4 x = 8 x + 8 (\displaystyle 20-4x+4x=8x+8).

3. Make sure that equality is not violated. Any mathematical operation performed on one side of the equation must also be performed on the other side. So if you add or subtract a term to get rid of a variable on one side of the equation, add or subtract the same term on the other side of the equation.

   • For example, if you add to one side of the equation 4 x (\displaystyle 4x) to get rid of a variable, you need to add 4 x (\displaystyle 4x) and to the other side of the equation:
     

4. Simplify the equation by adding or subtracting like terms. At this point, the variable should be on one side of the equation.

   • For example:
     20 − 4 x + 4 x = 8 x + 8 + 4 x (\displaystyle 20-4x+4x=8x+8+4x)
     

5. Move the free terms to one side of the equation (if necessary). It is necessary to make sure that the term with the variable is on one side, and the free term is on the other. To move the dummy term (and get rid of it on one side of the equation), add or subtract it from both sides of the equation.

   • For example, to get rid of a free member + 8 (\displaystyle +8) on the variable side, subtract 8 from both sides of the equation:
     20 = 12 x + 8 (\displaystyle 20=12x+8)
     20 − 8 = 12 x + 8 − 8 (\displaystyle 20-8=12x+8-8)
     

6. Get rid of the coefficient on the variable. To do this, perform the opposite operation of the operation between the coefficient and the variable. In most cases, simply divide both sides of the equation by the coefficient of the variable. Remember that any mathematical operation performed on one side of the equation must also be performed on the other side.

   • For example, to get rid of the factor 12, divide both sides of the equation by 12:
     12 = 12 x (\displaystyle 12=12x)
     12 12 = 12 x 12 (\displaystyle (\frac (12)(12))=(\frac (12x)(12)))
     1 = x (\displaystyle 1=x)

7. Check the answer. To do this, substitute the found value into the original equation. If equality is satisfied, the answer is correct.

   • For example, if 1 = x (\displaystyle 1=x), substitute 1 (instead of the variable) into the original equation:
     2 (10 − 2 x) = 4 (2 x + 2) (\displaystyle 2(10-2x)=4(2x+2))
     2 (10 − 2 (1)) = 4 (2 (1) + 2) (\displaystyle 2(10-2(1))=4(2(1)+2))
     2 (10 − 2) = 4 (2 + 2) (\displaystyle 2(10-2)=4(2+2))
     20 − 4 = 8 + 8 (\displaystyle 20-4=8+8)
     16 = 16 (\displaystyle 16=16)

   Solving a system of equations with two variables

   1. Isolate the variable in one equation. Perhaps in one of the equations the variable will already be isolated; otherwise, use mathematical operations to isolate the variable on one side of the equation. Remember that any mathematical operation performed on one side of the equation must also be performed on the other side.

      • For example, the equation is given. To isolate a variable y (\displaystyle y), subtract 1 from both sides of the equation:
        y + 1 = x − 1 (\displaystyle y+1=x-1)
        y + 1 − 1 = x − 1 − 1 (\displaystyle y+1-1=x-1-1)
        

   2. Substitute the value (as an expression) of the isolated variable into the other equation. Be sure to substitute the entire expression. The result is an equation with one variable that is easy to solve.

      • For example, the first equation has the form , and the second equation is reduced to the form y = x − 2 (\displaystyle y=x-2). In this case, in the first equation instead y (\displaystyle y) substitute x − 2 (\displaystyle x-2):
        2 x = 20 − 2 y (\displaystyle 2x=20-2y)
        

   3. Find the value of the variable. To do this, move the variable to one side of the equation. Then move the free terms to the other side of the equation. Then isolate the variable using a multiplication or division operation.

      • For example:
        2 x = 20 − 2 (x − 2) (\displaystyle 2x=20-2(x-2))
        2 x = 20 − 2 x + 4 (\displaystyle 2x=20-2x+4)
        2 x = 24 − 2 x (\displaystyle 2x=24-2x)
        2 x + 2 x = 24 − 2 x + 2 x (\displaystyle 2x+2x=24-2x+2x)
        4 x = 24 (\displaystyle 4x=24)
        4 x 4 = 24 4 (\displaystyle (\frac (4x)(4))=(\frac (24)(4)))
        x = 6 (\displaystyle x=6)

   4. Find the value of another variable. To do this, substitute the found value of the variable into one of the equations. The result is an equation with one variable that is easy to solve. Keep in mind that the found value of a variable can be substituted into any equation.

      • For example, if x = 6 (\displaystyle x=6), substitute 6 (instead of x (\displaystyle x)) into the second equation:
        y = x − 2 (\displaystyle y=x-2)
        y = (6) − 2 (\displaystyle y=(6)-2)
        y = 4 (\displaystyle y=4)

   5. Check the answer. To do this, substitute the values ​​of both variables into one of the equations. If equality is satisfied, the answer is correct.

      • For example, if you find that x = 6 (\displaystyle x=6) And y = 4 (\displaystyle y=4), substitute these values ​​into one of the original equations:
        2 x = 20 − 2 y (\displaystyle 2x=20-2y)
        2 (6) = 20 − 2 (4) (\displaystyle 2(6)=20-2(4))
        12 = 20 − 8 (\displaystyle 12=20-8)
        12 = 12 (\displaystyle 12=12)

   Solving equations

   1. Solve the following one-variable equation using the distributive law: .

      • 
        5 (x + 4) = 6 x − 5 (\displaystyle 5(x+4)=6x-5)
        
      • Get rid of 5 x (\displaystyle 5x) on the left side of the equation; to do this, subtract 5 x (\displaystyle 5x) from both sides of the equation:
        5 x + 20 = 6 x − 5 (\displaystyle 5x+20=6x-5)
        5 x + 20 − 5 x = 6 x − 5 − 5 x (\displaystyle 5x+20-5x=6x-5-5x)
        
      • Isolate the variable; To do this, add 5 to both sides of the equation:
        20 = x − 5 (\displaystyle 20=x-5)
        20 + 5 = x − 5 + 5 (\displaystyle 20+5=x-5+5)
        25 = x (\displaystyle 25=x)

   2. Solve the following fraction equation: .

      • Get rid of the fraction. To do this, multiply both sides of the equation by the expression (or number) in the denominator of the fraction:
        − 7 + 3 x = 7 − x 2 (\displaystyle -7+3x=(\frac (7-x)(2)))
        2 (− 7 + 3 x) = 2 (7 − x 2) (\displaystyle 2(-7+3x)=2((\frac (7-x)(2))))
        
      • Get rid of − x (\displaystyle -x) on the right side of the equation; for this add x (\displaystyle x) to both sides of the equation:
        − 14 + 6 x = 7 − x (\displaystyle -14+6x=7-x)
        − 14 + 6 x + x = 7 − x + x (\displaystyle -14+6x+x=7-x+x)
        
      • Move the free terms to one side of the equation; To do this, add 14 to both sides of the equation:
        − 14 + 7 x = 7 (\displaystyle -14+7x=7)
        − 14 + 7 x + 14 = 7 + 14 (\displaystyle -14+7x+14=7+14)
        
      • Get rid of the coefficient on the variable; To do this, divide both sides of the equation by 7:
        7 x = 21 (\displaystyle 7x=21)
        7 x 7 = 21 7 (\displaystyle (\frac (7x)(7))=(\frac (21)(7)))
        x = 3 (\displaystyle x=3)

   3. Solve the following system of equations: 9 x + 15 = 12 y ; 9 y = 9 x + 27 (\displaystyle 9x+15=12y;9y=9x+27)

      • Isolate a variable y (\displaystyle y) in the second equation:
        
        9 y = 9 (x + 3) (\displaystyle 9y=9(x+3))
        9 y 9 = 9 (x + 3) 9 (\displaystyle (\frac (9y)(9))=(\frac (9(x+3))(9)))
        y = x + 3 (\displaystyle y=x+3)
      • In the first equation instead y (\displaystyle y) substitute x + 3 (\displaystyle x+3):
        9 x + 15 = 12 y (\displaystyle 9x+15=12y)
        9 x + 15 = 12 (x + 3) (\displaystyle 9x+15=12(x+3))
      • Use the distributive law to open the brackets:
        
      • Get rid of the variable on the left side of the equation; to do this, subtract 9 x (\displaystyle 9x) from both sides of the equation:
        9 x + 15 = 12 x + 36 (\displaystyle 9x+15=12x+36)
        9 x + 15 − 9 x = 12 x + 36 − 9 x (\displaystyle 9x+15-9x=12x+36-9x)
        
      • Move the free terms to one side of the equation; To do this, subtract 36 from both sides of the equation:
        15 = 3 x + 36 (\displaystyle 15=3x+36)
        15 − 36 = 3 x + 36 − 36 (\displaystyle 15-36=3x+36-36)
        
      • Get rid of the coefficient on the variable; To do this, divide both sides of the equation by 3:
        − 21 = 3 x (\displaystyle -21=3x)
        − 21 3 = 3 x 3 (\displaystyle (\frac (-21)(3))=(\frac (3x)(3)))
        − 7 = x (\displaystyle -7=x)
      • Find the value y (\displaystyle y); To do this, substitute the found value x (\displaystyle x) into one of the equations:
        9 y = 9 x + 27 (\displaystyle 9y=9x+27)
        9 y = 9 (− 7) + 27 (\displaystyle 9y=9(-7)+27)
        9 y = − 63 + 27 (\displaystyle 9y=-63+27)
        9 y = − 36 (\displaystyle 9y=-36)
        9 y 9 = − 36 9 (\displaystyle (\frac (9y)(9))=(\frac (-36)(9)))
        y = − 4 (\displaystyle y=-4)

I. ax 2 =0 – incomplete quadratic equation (b=0, c=0 ). Solution: x=0. Answer: 0.

Solve equations.

2x·(x+3)=6x-x 2 .

Solution. Let's open the brackets by multiplying 2x for each term in brackets:

2x 2 +6x=6x-x 2 ; We move the terms from the right side to the left:

2x 2 +6x-6x+x 2 =0; Here are similar terms:

3x 2 =0, hence x=0.

Answer: 0.

II. ax 2 +bx=0 –incomplete quadratic equation (c=0 ). Solution: x (ax+b)=0 → x 1 =0 or ax+b=0 → x 2 =-b/a. Answer: 0; -b/a.

5x 2 -26x=0.

Solution. Let's take out the common factor X outside the brackets:

x(5x-26)=0; each factor can be equal to zero:

x=0 or 5x-26=0→ 5x=26, divide both sides of the equality by 5 and we get: x=5.2.

Answer: 0; 5,2.

Example 3. 64x+4x 2 =0.

Solution. Let's take out the common factor 4x outside the brackets:

4x(16+x)=0. We have three factors, 4≠0, therefore, or x=0 or 16+x=0. From the last equality we get x=-16.

Answer: -16; 0.

Example 4.(x-3) 2 +5x=9.

Solution. Applying the formula for the square of the difference of two expressions, we will open the brackets:

x 2 -6x+9+5x=9; transform to the form: x 2 -6x+9+5x-9=0; Let us present similar terms:

x 2 -x=0; we'll take it out X outside the brackets, we get: x (x-1)=0. From here or x=0 or x-1=0→ x=1.

Answer: 0; 1.

III. ax 2 +c=0 –incomplete quadratic equation (b=0 ); Solution: ax 2 =-c → x 2 =-c/a.

If (-c/a)<0 , then there are no real roots. If (-с/а)>0

Example 5. x 2 -49=0.

Solution.

x 2 =49, from here x=±7. Answer:-7; 7.

Example 6. 9x 2 -4=0.

Solution.

Often you need to find the sum of squares (x 1 2 +x 2 2) or the sum of cubes (x 1 3 +x 2 3) of the roots of a quadratic equation, less often - the sum of the reciprocal values ​​of the squares of the roots or the sum of arithmetic square roots of the roots of a quadratic equation:

Vieta's theorem can help with this:

x 2 +px+q=0

x 1 + x 2 = -p; x 1 ∙x 2 =q.

Let's express through p And q:

1) sum of squares of the roots of the equation x 2 +px+q=0;

2) sum of cubes of the roots of the equation x 2 +px+q=0.

Solution.

1) Expression x 1 2 +x 2 2 obtained by squaring both sides of the equation x 1 + x 2 = -p;

(x 1 +x 2) 2 =(-p) 2 ; open the brackets: x 1 2 +2x 1 x 2 + x 2 2 =p 2 ; we express the required amount: x 1 2 +x 2 2 =p 2 -2x 1 x 2 =p 2 -2q. We got a useful equality: x 1 2 +x 2 2 =p 2 -2q.

2) Expression x 1 3 +x 2 3 Let us represent the sum of cubes using the formula:

(x 1 3 +x 2 3)=(x 1 +x 2)(x 1 2 -x 1 x 2 +x 2 2)=-p·(p 2 -2q-q)=-p·(p 2 -3q).

Another useful equation: x 1 3 +x 2 3 = -p·(p 2 -3q).

Examples.

3) x 2 -3x-4=0. Without solving the equation, calculate the value of the expression x 1 2 +x 2 2.

Solution.

x 1 +x 2 =-p=3, and the work x 1 ∙x 2 =q=in example 1) equality:

x 1 2 +x 2 2 =p 2 -2q. We have -p=x 1 +x 2 = 3 → p 2 =3 2 =9; q= x 1 x 2 = -4. Then x 1 2 +x 2 2 =9-2·(-4)=9+8=17.

Answer: x 1 2 +x 2 2 =17.

4) x 2 -2x-4=0. Calculate: x 1 3 +x 2 3 .

Solution.

By Vieta's theorem, the sum of the roots of this reduced quadratic equation is x 1 +x 2 =-p=2, and the work x 1 ∙x 2 =q=-4. Let's apply what we have received ( in example 2) equality: x 1 3 +x 2 3 =-p·(p 2 -3q)= 2·(2 2 -3·(-4))=2·(4+12)=2·16=32.

Answer: x 1 3 +x 2 3 =32.

Question: what if we are given an unreduced quadratic equation? Answer: it can always be “reduced” by dividing term by term by the first coefficient.

5) 2x 2 -5x-7=0. Without deciding, calculate: x 1 2 +x 2 2.

Solution. We are given a complete quadratic equation. Divide both sides of the equality by 2 (the first coefficient) and obtain the following quadratic equation: x 2 -2.5x-3.5=0.

According to Vieta's theorem, the sum of the roots is equal to 2,5 ; the product of the roots is equal -3,5 .

We solve it in the same way as the example 3) using the equality: x 1 2 +x 2 2 =p 2 -2q.

x 1 2 +x 2 2 =p 2 -2q= 2,5 2 -2∙(-3,5)=6,25+7=13,25.

Answer: x 1 2 + x 2 2 = 13,25.

6) x 2 -5x-2=0. Find:

Let us transform this equality and, using Vieta’s theorem, replace the sum of roots through -p, and the product of the roots through q, we get another useful formula. When deriving the formula, we used equality 1): x 1 2 +x 2 2 =p 2 -2q.

In our example x 1 +x 2 =-p=5; x 1 ∙x 2 =q=-2. We substitute these values ​​into the resulting formula:

7) x 2 -13x+36=0. Find:

Let's transform this sum and get a formula that can be used to find the sum of arithmetic square roots from the roots of a quadratic equation.

We have x 1 +x 2 =-p=13; x 1 ∙x 2 =q=36. We substitute these values ​​into the resulting formula:

Advice : always check the possibility of finding the roots of a quadratic equation using a suitable method, because 4 reviewed useful formulas allow you to quickly complete a task, especially in cases where the discriminant is an “inconvenient” number. In all simple cases, find the roots and operate on them. For example, in the last example we select the roots using Vieta’s theorem: the sum of the roots should be equal to 13 , and the product of the roots 36 . What are these numbers? Certainly, 4 and 9. Now calculate the sum of the square roots of these numbers: 2+3=5. That's it!

I. Vieta's theorem for the reduced quadratic equation.

Sum of roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term:

x 1 + x 2 = -p; x 1 ∙x 2 =q.

Find the roots of the given quadratic equation using Vieta's theorem.

Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), second coefficient p=-1, and the free member q=-30. First, let's make sure that this equation has roots, and that the roots (if any) will be expressed in integers. To do this, it is enough that the discriminant be a perfect square of an integer.

Finding the discriminant D=b 2 — 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .

Now, according to Vieta’s theorem, the sum of the roots must be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:

x 1 +x 2 =1; x 1 ∙x 2 =-30. We need to choose two numbers such that their product is equal to -30 , and the amount is unit. These are numbers -5 And 6 . Answer: -5; 6.

Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Let's make sure that there are integer roots. Let's find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us select the roots using Vieta’s theorem: the sum of the roots is equal to –р=-6, and the product of the roots is equal to q=8. These are numbers -4 And -2 .

In fact: -4-2=-6=-р; -4∙(-2)=8=q. Answer: -4; -2.

Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free member q=-4. Let's find the discriminant D 1, since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, so we do conclusion: The roots of this equation are not integers and cannot be found using Vieta’s theorem. This means that we solve this equation, as usual, using the formulas (in this case, using the formulas). We get:

Example 4). Write a quadratic equation using its roots if x 1 =-7, x 2 =4.

Solution. The required equation will be written in the form: x 2 +px+q=0, and, based on Vieta’s theorem –p=x 1 +x 2=-7+4=-3 → p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x 2 +3x-28=0.

Example 5). Write a quadratic equation using its roots if:

II. Vieta's theorem for a complete quadratic equation ax 2 +bx+c=0.

The sum of the roots is minus b, divided by A, the product of the roots is equal to With, divided by A:

x 1 + x 2 = -b/a; x 1 ∙x 2 =c/a.

Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11=0.

Solution.

We make sure that this equation will have roots. To do this, it is enough to create an expression for the discriminant, and, without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 . Now let's use theorem Vieta for complete quadratic equations.

x 1 +x 2 =-b:a=- (-7):2=3,5.

Example 7). Find the product of the roots of a quadratic equation 3x 2 +8x-21=0.

Solution.

Let's find the discriminant D 1, since the second coefficient ( 8 ) is an even number. D 1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to Vieta’s theorem, the product of roots x 1 ∙x 2 =c:a=-21:3=-7.

I. ax 2 +bx+c=0– general quadratic equation

Discriminant D=b 2 - 4ac.

If D>0, then we have two real roots:

If D=0, then we have a single root (or two equal roots) x=-b/(2a).

If D<0, то действительных корней нет.

Example 1) 2x 2 +5x-3=0.

Solution. a=2; b=5; c=-3.

D=b 2 - 4ac=5 2 -4∙2∙(-3)=25+24=49=7 2 >0; 2 real roots.

4x 2 +21x+5=0.

Solution. a=4; b=21; c=5.

D=b 2 - 4ac=21 2 - 4∙4∙5=441-80=361=19 2 >0; 2 real roots.

II. ax 2 +bx+c=0 – quadratic equation of particular form with even second

coefficient b

Example 3) 3x 2 -10x+3=0.

Solution. a=3; b=-10 (even number); c=3.

Example 4) 5x 2 -14x-3=0.

Solution. a=5; b= -14 (even number); c=-3.

Example 5) 71x 2 +144x+4=0.

Solution. a=71; b=144 (even number); c=4.

Example 6) 9x 2 -30x+25=0.

Solution. a=9; b=-30 (even number); c=25.

III. ax 2 +bx+c=0 – quadratic equation private type provided: a-b+c=0.

The first root is always equal to minus one, and the second root is always equal to minus With, divided by A:

x 1 =-1, x 2 =-c/a.

Example 7) 2x 2 +9x+7=0.

Solution. a=2; b=9; c=7. Let's check the equality: a-b+c=0. We get: 2-9+7=0 .

Then x 1 =-1, x 2 =-c/a=-7/2=-3.5. Answer: -1; -3,5.

IV. ax 2 +bx+c=0 – quadratic equation of a particular form subject to : a+b+c=0.

The first root is always equal to one, and the second root is equal to With, divided by A:

x 1 =1, x 2 =c/a.

Example 8) 2x 2 -9x+7=0.

Solution. a=2; b=-9; c=7. Let's check the equality: a+b+c=0. We get: 2-9+7=0 .

Then x 1 =1, x 2 =c/a=7/2=3.5. Answer: 1; 3,5.

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In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.

First, let's define: what is a linear equation and which one is called the simplest?

A linear equation is one in which there is only one variable, and only to the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest using the algorithm:

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Give similar terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$.

Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to expand the parentheses, if there are any (as in our last example);
2. Then combine similar
3. Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.

Then, as a rule, you need to give similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.

In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.

Scheme for solving simple linear equations

First, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the brackets, if any.
2. We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
3. We present similar terms.
4. We divide everything by the coefficient of “x”.

Of course, this scheme does not always work; there are certain subtleties and tricks in it, and now we will get to know them.

Solving real examples of simple linear equations

Task No. 1

The first step requires us to open the brackets. But they are not in this example, so we skip this step. In the second step we need to isolate the variables. Please note: we are talking only about individual terms. Let's write it down:

We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:

\[\frac(6x)(6)=-\frac(72)(6)\]

So we got the answer.

Task No. 2

We can see the parentheses in this problem, so let's expand them:

Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:

Here are some similar ones:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task No. 3

The third linear equation is more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's do the math:

We carry out the last step - divide everything by the coefficient of “x”:

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, there may be zero among them - there is nothing wrong with that.

Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.

Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it using standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such things is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.

Example No. 1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take a look at privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some similar ones:

Obviously, this equation has no solutions, so we’ll write this in the answer:

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some similar ones:

Obviously, this linear equation has no solution, so we’ll write it this way:

\[\varnothing\],

or there are no roots.

Nuances of the solution

Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.

But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.

Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task No. 1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do some privacy:

Here are some similar ones:

Let's complete the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.

Task No. 2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:

Now let’s carefully perform the multiplication in each term:

Let’s move the terms with “X” to the left, and those without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.

About the algebraic sum

With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.

Solving equations with fractions

To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:

1. Open the brackets.
2. Separate variables.
3. Bring similar ones.
4. Divide by the ratio.

Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:

1. Get rid of fractions.
2. Open the brackets.
3. Separate variables.
4. Bring similar ones.
5. Divide by the ratio.

What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.

Example No. 1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, let's move on to the second equation.

Example No. 2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

The problem is solved.

That, in fact, is all I wanted to tell you today.

Key points

Key findings are:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Don't worry if you have quadratic functions somewhere; most likely, they will be reduced in the process of further transformations.
• There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!