Solving formulas online. Examples of identical transformations of equations

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 of 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 of 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 formulas (in this case, using 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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