In a simpler way. To do this, put z out of brackets. You will get: z(аz + b) = 0. The factors can be written: z=0 and аz + b = 0, since both can result in zero. In the notation az + b = 0, we move the second one to the right with a different sign. From here we get z1 = 0 and z2 = -b/a. These are the roots of the original.
If there is an incomplete equation of the form az² + c = 0, in this case they are found by simply moving the free term to the right side of the equation. Also change its sign. The result will be az² = -с. Express z² = -c/a. Take the root and write down two solutions - a positive and a negative square root.
note
If there are fractional coefficients in the equation, multiply the entire equation by the appropriate factor so as to get rid of the fractions.
Knowledge of how to solve quadratic equations is necessary for both schoolchildren and students; sometimes this can also help an adult in everyday life. There are several specific solution methods.
Solving Quadratic Equations
Quadratic equation of the form a*x^2+b*x+c=0. Coefficient x is the desired variable, a, b, c are numerical coefficients. Remember that the “+” sign can change to a “-” sign.In order to solve this equation, it is necessary to use Vieta’s theorem or find the discriminant. The most common method is to find the discriminant, since for some values of a, b, c it is not possible to use Vieta’s theorem.
To find the discriminant (D), you need to write the formula D=b^2 - 4*a*c. The D value can be greater than, less than, or equal to zero. If D is greater or less than zero, then there will be two roots; if D = 0, then only one root remains; more precisely, we can say that D in this case has two equivalent roots. Substitute the known coefficients a, b, c into the formula and calculate the value.
After you have found the discriminant, use the formulas to find x: x(1) = (- b+sqrt(D))/2*a; x(2) = (- b-sqrt(D))/2*a, where sqrt is a function that means taking the square root of a given number. After calculating these expressions, you will find two roots of your equation, after which the equation is considered solved.
If D is less than zero, then it still has roots. This section is practically not studied at school. University students should be aware that a negative number appears under the root. They get rid of it by highlighting the imaginary part, that is, -1 under the root is always equal to the imaginary element “i”, which is multiplied by the root with the same positive number. For example, if D=sqrt(-20), after the transformation we get D=sqrt(20)*i. After this transformation, solving the equation is reduced to the same finding of roots as described above.
Vieta's theorem consists of selecting the values of x(1) and x(2). Two identical equations are used: x(1) + x(2)= -b; x(1)*x(2)=с. Moreover, a very important point is the sign in front of the coefficient b; remember that this sign is opposite to the one in the equation. At first glance, it seems that calculating x(1) and x(2) is very simple, but when solving, you will be faced with the fact that you will have to select the numbers.
Elements of solving quadratic equations
According to the rules of mathematics, some can be factorized: (a+x(1))*(b-x(2))=0, if you managed to transform this quadratic equation in a similar way using mathematical formulas, then feel free to write down the answer. x(1) and x(2) will be equal to the adjacent coefficients in brackets, but with the opposite sign.Also, do not forget about incomplete quadratic equations. You may be missing some of the terms; if so, then all its coefficients are simply equal to zero. If there is nothing in front of x^2 or x, then the coefficients a and b are equal to 1.
", that is, equations of the first degree. In this lesson we will look at what is called a quadratic equation and how to solve it.
What is a quadratic equation?
Important!
The degree of an equation is determined by the highest degree to which the unknown stands.
If the maximum power in which the unknown is “2”, then you have a quadratic equation.
Examples of quadratic equations
- 5x 2 − 14x + 17 = 0
- −x 2 + x +
= 01 3 - x 2 + 0.25x = 0
- x 2 − 8 = 0
Important! The general form of a quadratic equation looks like this:
A x 2 + b x + c = 0
“a”, “b” and “c” are given numbers.- “a” is the first or highest coefficient;
- “b” is the second coefficient;
- “c” is a free term.
To find “a”, “b” and “c” you need to compare your equation with the general form of the quadratic equation “ax 2 + bx + c = 0”.
Let's practice determining the coefficients "a", "b" and "c" in quadratic equations.
| The equation | Odds | |||
|---|---|---|---|---|
|
||||
|
||||
| 1 |
| 3 |
- a = −1
- b = 1
- c =
1 3
- a = 1
- b = 0.25
- c = 0
- a = 1
- b = 0
- c = −8
How to Solve Quadratic Equations
Unlike linear equations, a special method is used to solve quadratic equations. formula for finding roots.
Remember!
To solve a quadratic equation you need:
- bring the quadratic equation to the general form “ax 2 + bx + c = 0”. That is, only “0” should remain on the right side;
- use formula for roots:
Let's look at an example of how to use the formula to find the roots of a quadratic equation. Let's solve a quadratic equation.
X 2 − 3x − 4 = 0
The equation “x 2 − 3x − 4 = 0” has already been reduced to the general form “ax 2 + bx + c = 0” and does not require additional simplifications. To solve it, we just need to apply formula for finding the roots of a quadratic equation.
Let us determine the coefficients “a”, “b” and “c” for this equation.
x 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =
It can be used to solve any quadratic equation.
In the formula “x 1;2 = ” the radical expression is often replaced
“b 2 − 4ac” for the letter “D” and is called discriminant. The concept of a discriminant is discussed in more detail in the lesson “What is a discriminant”.
Let's look at another example of a quadratic equation.
x 2 + 9 + x = 7x
In this form, it is quite difficult to determine the coefficients “a”, “b” and “c”. Let's first reduce the equation to the general form “ax 2 + bx + c = 0”.
X 2 + 9 + x = 7x
x 2 + 9 + x − 7x = 0
x 2 + 9 − 6x = 0
x 2 − 6x + 9 = 0
Now you can use the formula for the roots.
X 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =
x =
| 6 |
| 2 |
x = 3
Answer: x = 3
There are times when quadratic equations have no roots. This situation occurs when the formula contains a negative number under the root.
Quadratic equation - easy to solve! *Hereinafter referred to as “KU”. Friends, it would seem that there could be nothing simpler in mathematics than solving such an equation. But something told me that many people have problems with him. I decided to see how many on-demand impressions Yandex gives out per month. Here's what happened, look:
What does it mean? This means that about 70,000 people a month are looking for this information, and this is summer, and what will happen during the school year - there will be twice as many requests. This is not surprising, because those guys and girls who graduated from school a long time ago and are preparing for the Unified State Exam are looking for this information, and schoolchildren also strive to refresh their memory.
Despite the fact that there are a lot of sites that tell you how to solve this equation, I decided to also contribute and publish the material. Firstly, I want visitors to come to my site based on this request; secondly, in other articles, when the topic of “KU” comes up, I will provide a link to this article; thirdly, I’ll tell you a little more about his solution than is usually stated on other sites. Let's get started! The content of the article:
A quadratic equation is an equation of the form:
where coefficients a,band c are arbitrary numbers, with a≠0.
In the school course, the material is given in the following form - the equations are divided into three classes:
1. They have two roots.
2. *Have only one root.
3. They have no roots. It is worth especially noting here that they do not have real roots
How are roots calculated? Just!
We calculate the discriminant. Underneath this “terrible” word lies a very simple formula:
The root formulas are as follows:
*You need to know these formulas by heart.
You can immediately write down and solve:
Example:
1. If D > 0, then the equation has two roots.
2. If D = 0, then the equation has one root.
3. If D< 0, то уравнение не имеет действительных корней.
Let's look at the equation:
In this regard, when the discriminant is equal to zero, the school course says that one root is obtained, here it is equal to nine. Everything is correct, it is so, but...
This idea is somewhat incorrect. In fact, there are two roots. Yes, yes, don’t be surprised, you get two equal roots, and to be mathematically precise, then the answer should write two roots:
x 1 = 3 x 2 = 3
But this is so - a small digression. At school you can write it down and say that there is one root.
Now the next example:
As we know, the root of a negative number cannot be taken, so there is no solution in this case.
That's the whole decision process.
Quadratic function.
This shows what the solution looks like geometrically. This is extremely important to understand (in the future, in one of the articles we will analyze in detail the solution to the quadratic inequality).
This is a function of the form:
where x and y are variables
a, b, c – given numbers, with a ≠ 0
The graph is a parabola:
That is, it turns out that by solving a quadratic equation with “y” equal to zero, we find the points of intersection of the parabola with the x axis. There can be two of these points (the discriminant is positive), one (the discriminant is zero) and none (the discriminant is negative). Details about the quadratic function You can view article by Inna Feldman.
Let's look at examples:
Example 1: Solve 2x 2 +8 x–192=0
a=2 b=8 c= –192
D=b 2 –4ac = 8 2 –4∙2∙(–192) = 64+1536 = 1600
Answer: x 1 = 8 x 2 = –12
*It was possible to immediately divide the left and right sides of the equation by 2, that is, simplify it. The calculations will be easier.
Example 2: Decide x 2–22 x+121 = 0
a=1 b=–22 c=121
D = b 2 –4ac =(–22) 2 –4∙1∙121 = 484–484 = 0
We found that x 1 = 11 and x 2 = 11
It is permissible to write x = 11 in the answer.
Answer: x = 11
Example 3: Decide x 2 –8x+72 = 0
a=1 b= –8 c=72
D = b 2 –4ac =(–8) 2 –4∙1∙72 = 64–288 = –224
The discriminant is negative, there is no solution in real numbers.
Answer: no solution
The discriminant is negative. There is a solution!
Here we will talk about solving the equation in the case when a negative discriminant is obtained. Do you know anything about complex numbers? I will not go into detail here about why and where they arose and what their specific role and necessity in mathematics is; this is a topic for a large separate article.
The concept of a complex number.
A little theory.
A complex number z is a number of the form
z = a + bi
where a and b are real numbers, i is the so-called imaginary unit.
a+bi – this is a SINGLE NUMBER, not an addition.
The imaginary unit is equal to the root of minus one:
Now consider the equation:
We get two conjugate roots.
Incomplete quadratic equation.
Let's consider special cases, this is when the coefficient “b” or “c” is equal to zero (or both are equal to zero). They can be solved easily without any discriminants.
Case 1. Coefficient b = 0.
The equation becomes:
Let's transform:
Example:
4x 2 –16 = 0 => 4x 2 =16 => x 2 = 4 => x 1 = 2 x 2 = –2
Case 2. Coefficient c = 0.
The equation becomes:
Let's transform and factorize:
*The product is equal to zero when at least one of the factors is equal to zero.
Example:
9x 2 –45x = 0 => 9x (x–5) =0 => x = 0 or x–5 =0
x 1 = 0 x 2 = 5
Case 3. Coefficients b = 0 and c = 0.
Here it is clear that the solution to the equation will always be x = 0.
Useful properties and patterns of coefficients.
There are properties that allow you to solve equations with large coefficients.
Ax 2 + bx+ c=0 equality holds
a + b+ c = 0, That
- if for the coefficients of the equation Ax 2 + bx+ c=0 equality holds
a+ c =b, That
These properties help solve a certain type of equation.
Example 1: 5001 x 2 –4995 x – 6=0
The sum of the odds is 5001+( – 4995)+(– 6) = 0, which means
Example 2: 2501 x 2 +2507 x+6=0
Equality holds a+ c =b, Means
Regularities of coefficients.
1. If in the equation ax 2 + bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 + (a 2 +1)∙x+ a= 0 = > x 1 = –a x 2 = –1/a.
Example. Consider the equation 6x 2 + 37x + 6 = 0.
x 1 = –6 x 2 = –1/6.
2. If in the equation ax 2 – bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 – (a 2 +1)∙x+ a= 0 = > x 1 = a x 2 = 1/a.
Example. Consider the equation 15x 2 –226x +15 = 0.
x 1 = 15 x 2 = 1/15.
3. If in Eq. ax 2 + bx – c = 0 coefficient “b” is equal to (a 2 – 1), and coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 + (a 2 –1)∙x – a= 0 = > x 1 = – a x 2 = 1/a.
Example. Consider the equation 17x 2 +288x – 17 = 0.
x 1 = – 17 x 2 = 1/17.
4. If in the equation ax 2 – bx – c = 0 the coefficient “b” is equal to (a 2 – 1), and the coefficient c is numerically equal to the coefficient “a”, then its roots are equal
ax 2 – (a 2 –1)∙x – a= 0 = > x 1 = a x 2 = – 1/a.
Example. Consider the equation 10x 2 – 99x –10 = 0.
x 1 = 10 x 2 = – 1/10
Vieta's theorem.
Vieta's theorem is named after the famous French mathematician Francois Vieta. Using Vieta's theorem, we can express the sum and product of the roots of an arbitrary KU in terms of its coefficients.
45 = 1∙45 45 = 3∙15 45 = 5∙9.
In total, the number 14 gives only 5 and 9. These are roots. With a certain skill, using the presented theorem, you can solve many quadratic equations orally immediately.
Vieta's theorem, in addition. It is convenient in that after solving a quadratic equation in the usual way (through a discriminant), the resulting roots can be checked. I recommend doing this always.
TRANSPORTATION METHOD
With this method, the coefficient “a” is multiplied by the free term, as if “thrown” to it, which is why it is called "transfer" method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.
If A± b+c≠ 0, then the transfer technique is used, for example:
2X 2 – 11x+ 5 = 0 (1) => X 2 – 11x+ 10 = 0 (2)
Using Vieta's theorem in equation (2), it is easy to determine that x 1 = 10 x 2 = 1
The resulting roots of the equation must be divided by 2 (since the two were “thrown” from x 2), we get
x 1 = 5 x 2 = 0.5.
What is the rationale? Look what's happening.
The discriminants of equations (1) and (2) are equal:
If you look at the roots of the equations, you only get different denominators, and the result depends precisely on the coefficient of x 2:
The second (modified) one has roots that are 2 times larger.
Therefore, we divide the result by 2.
*If we reroll the three, we will divide the result by 3, etc.
Answer: x 1 = 5 x 2 = 0.5
Sq. ur-ie and Unified State Examination.
I’ll tell you briefly about its importance - YOU MUST BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of roots and discriminants by heart. Many of the problems included in the Unified State Examination tasks boil down to solving a quadratic equation (geometric ones included).
Something worth noting!
1. The form of writing an equation can be “implicit”. For example, the following entry is possible:
15+ 9x 2 - 45x = 0 or 15x+42+9x 2 - 45x=0 or 15 -5x+10x 2 = 0.
You need to bring it to a standard form (so as not to get confused when solving).
2. Remember that x is an unknown quantity and it can be denoted by any other letter - t, q, p, h and others.
An incomplete quadratic equation differs from classical (complete) equations in that its factors or free term are equal to zero. The graphs of such functions are parabolas. Depending on their general appearance, they are divided into 3 groups. The principles of solution for all types of equations are the same.
There is nothing complicated in determining the type of an incomplete polynomial. It is best to consider the main differences using visual examples:
- If b = 0, then the equation is ax 2 + c = 0.
- If c = 0, then the expression ax 2 + bx = 0 should be solved.
- If b = 0 and c = 0, then the polynomial turns into an equality like ax 2 = 0.
The latter case is more of a theoretical possibility and never occurs in knowledge testing tasks, since the only correct value of the variable x in the expression is zero. In the future, methods and examples of solving incomplete quadratic equations of 1) and 2) types will be considered.
General algorithm for searching variables and examples with solutions
Regardless of the type of equation, the solution algorithm is reduced to the following steps:
- Reduce the expression to a form convenient for finding roots.
- Perform calculations.
- Write down the answer.
The easiest way to solve incomplete equations is to factor the left side and leave a zero on the right. Thus, the formula for an incomplete quadratic equation for finding roots is reduced to calculating the value of x for each of the factors.
You can only learn how to solve it through practice, so let’s consider a specific example of finding the roots of an incomplete equation:
As you can see, in this case b = 0. Let’s factorize the left side and get the expression:
4(x – 0.5) ⋅ (x + 0.5) = 0.
Obviously, the product is equal to zero when at least one of the factors is equal to zero. The values of the variable x1 = 0.5 and (or) x2 = -0.5 meet similar requirements.
In order to easily and quickly cope with the problem of factoring a quadratic trinomial, you should remember the following formula:
If there is no free term in the expression, the problem is greatly simplified. It will be enough just to find and bracket the common denominator. For clarity, consider an example of how to solve incomplete quadratic equations of the form ax2 + bx = 0.
Let's take the variable x out of brackets and get the following expression:
x ⋅ (x + 3) = 0.
Guided by logic, we come to the conclusion that x1 = 0, and x2 = -3.
Traditional solution method and incomplete quadratic equations
What happens if you apply the discriminant formula and try to find the roots of a polynomial with coefficients equal to zero? Let's take an example from a collection of standard tasks for the Unified State Exam in Mathematics 2017, solve it using standard formulas and the factorization method.
7x 2 – 3x = 0.
Let's calculate the discriminant value: D = (-3)2 – 4 ⋅ (-7) ⋅ 0 = 9. It turns out that the polynomial has two roots:
Now, let's solve the equation by factoring and compare the results.
X ⋅ (7x + 3) = 0,
2) 7x + 3 = 0,
7x = -3,
x = -.
As you can see, both methods give the same result, but solving the equation using the second method was much easier and faster.
Vieta's theorem
But what to do with Vieta’s favorite theorem? Can this method be used when the trinomial is incomplete? Let's try to understand the aspects of bringing incomplete equations to the classical form ax2 + bx + c = 0.
In fact, it is possible to apply Vieta's theorem in this case. It is only necessary to bring the expression to its general form, replacing the missing terms with zero.
For example, with b = 0 and a = 1, in order to eliminate the possibility of confusion, the task should be written in the form: ax2 + 0 + c = 0. Then the ratio of the sum and product of the roots and factors of the polynomial can be expressed as follows:
Theoretical calculations help to get acquainted with the essence of the issue, and always require the development of skills when solving specific problems. Let's turn again to the reference book of standard tasks for the Unified State Exam and find a suitable example:
Let us write the expression in a form convenient for applying Vieta’s theorem:
x 2 + 0 – 16 = 0.
The next step is to create a system of conditions:
Obviously, the roots of the quadratic polynomial will be x 1 = 4 and x 2 = -4.
Now, let's practice bringing the equation to its general form. Let's take the following example: 1/4× x 2 – 1 = 0
In order to apply Vieta's theorem to an expression, it is necessary to get rid of the fraction. Let’s multiply the left and right sides by 4, and look at the result: x2– 4 = 0. The resulting equality is ready to be solved by Vieta’s theorem, but it is much easier and faster to get the answer by simply moving c = 4 to the right side of the equation: x2 = 4.
To summarize, the best way to solve incomplete equations is factorization, which is the simplest and fastest method. If difficulties arise in the process of searching for roots, you can turn to the traditional method of finding roots through a discriminant.
This topic may seem complicated at first due to the many not-so-simple formulas. Not only do the quadratic equations themselves have long notations, but the roots are also found through the discriminant. In total, three new formulas are obtained. Not very easy to remember. This is possible only after solving such equations frequently. Then all the formulas will be remembered by themselves.
General view of a quadratic equation
Here we propose their explicit recording, when the largest degree is written first, and then in descending order. There are often situations when the terms are inconsistent. Then it is better to rewrite the equation in descending order of the degree of the variable.
Let us introduce some notation. They are presented in the table below.
If we accept these notations, all quadratic equations are reduced to the following notation.
Moreover, the coefficient a ≠ 0. Let this formula be designated number one.
When an equation is given, it is not clear how many roots there will be in the answer. Because one of three options is always possible:
- the solution will have two roots;
- the answer will be one number;
- the equation will have no roots at all.
And until the decision is finalized, it is difficult to understand which option will appear in a particular case.
Types of recordings of quadratic equations
There may be different entries in tasks. They will not always look like the general quadratic equation formula. Sometimes it will be missing some terms. What was written above is the complete equation. If you remove the second or third term in it, you get something else. These records are also called quadratic equations, only incomplete.
Moreover, only terms with coefficients “b” and “c” can disappear. The number "a" cannot be equal to zero under any circumstances. Because in this case the formula turns into a linear equation. The formulas for the incomplete form of equations will be as follows:
So, there are only two types; in addition to complete ones, there are also incomplete quadratic equations. Let the first formula be number two, and the second - three.
Discriminant and dependence of the number of roots on its value
You need to know this number in order to calculate the roots of the equation. It can always be calculated, no matter what the formula of the quadratic equation is. In order to calculate the discriminant, you need to use the equality written below, which will have number four.
After substituting the coefficient values into this formula, you can get numbers with different signs. If the answer is yes, then the answer to the equation will be two different roots. If the number is negative, there will be no roots of the quadratic equation. If it is equal to zero, there will be only one answer.
How to solve a complete quadratic equation?
In fact, consideration of this issue has already begun. Because first you need to find a discriminant. After it is determined that there are roots of the quadratic equation, and their number is known, you need to use formulas for the variables. If there are two roots, then you need to apply the following formula.
Since it contains a “±” sign, there will be two values. The expression under the square root sign is the discriminant. Therefore, the formula can be rewritten differently.
Formula number five. From the same record it is clear that if the discriminant is equal to zero, then both roots will take the same values.
If solving quadratic equations has not yet been worked out, then it is better to write down the values of all coefficients before applying the discriminant and variable formulas. Later this moment will not cause difficulties. But at the very beginning there is confusion.
How to solve an incomplete quadratic equation?
Everything is much simpler here. There is not even a need for additional formulas. And those that have already been written down for the discriminant and the unknown will not be needed.
First, let's look at incomplete equation number two. In this equality, it is necessary to take the unknown quantity out of brackets and solve the linear equation, which will remain in brackets. The answer will have two roots. The first one is necessarily equal to zero, because there is a multiplier consisting of the variable itself. The second one will be obtained by solving a linear equation.
Incomplete equation number three is solved by moving the number from the left side of the equality to the right. Then you need to divide by the coefficient facing the unknown. All that remains is to extract the square root and remember to write it down twice with opposite signs.
Below are some steps that will help you learn how to solve all kinds of equalities that turn into quadratic equations. They will help the student to avoid mistakes due to inattention. These shortcomings can cause poor grades when studying the extensive topic “Quadratic Equations (8th Grade).” Subsequently, these actions will not need to be performed constantly. Because a stable skill will appear.
- First you need to write the equation in standard form. That is, first the term with the largest degree of the variable, and then - without a degree, and last - just a number.
- If a minus appears before the coefficient “a”, it can complicate the work for a beginner studying quadratic equations. It's better to get rid of it. For this purpose, all equality must be multiplied by “-1”. This means that all terms will change sign to the opposite.
- It is recommended to get rid of fractions in the same way. Simply multiply the equation by the appropriate factor so that the denominators cancel out.
Examples
It is required to solve the following quadratic equations:
x 2 − 7x = 0;
15 − 2x − x 2 = 0;
x 2 + 8 + 3x = 0;
12x + x 2 + 36 = 0;
(x+1) 2 + x + 1 = (x+1)(x+2).
The first equation: x 2 − 7x = 0. It is incomplete, therefore it is solved as described for formula number two.
After taking it out of brackets, it turns out: x (x - 7) = 0.
The first root takes the value: x 1 = 0. The second will be found from the linear equation: x - 7 = 0. It is easy to see that x 2 = 7.
Second equation: 5x 2 + 30 = 0. Again incomplete. Only it is solved as described for the third formula.
After moving 30 to the right side of the equation: 5x 2 = 30. Now you need to divide by 5. It turns out: x 2 = 6. The answers will be the numbers: x 1 = √6, x 2 = - √6.
The third equation: 15 − 2x − x 2 = 0. Hereinafter, solving quadratic equations will begin by rewriting them in standard form: − x 2 − 2x + 15 = 0. Now it’s time to use the second useful tip and multiply everything by minus one . It turns out x 2 + 2x - 15 = 0. Using the fourth formula, you need to calculate the discriminant: D = 2 2 - 4 * (- 15) = 4 + 60 = 64. It is a positive number. From what is said above, it turns out that the equation has two roots. They need to be calculated using the fifth formula. It turns out that x = (-2 ± √64) / 2 = (-2 ± 8) / 2. Then x 1 = 3, x 2 = - 5.
The fourth equation x 2 + 8 + 3x = 0 is transformed into this: x 2 + 3x + 8 = 0. Its discriminant is equal to this value: -23. Since this number is negative, the answer to this task will be the following entry: “There are no roots.”
The fifth equation 12x + x 2 + 36 = 0 should be rewritten as follows: x 2 + 12x + 36 = 0. After applying the formula for the discriminant, the number zero is obtained. This means that it will have one root, namely: x = -12/ (2 * 1) = -6.
The sixth equation (x+1) 2 + x + 1 = (x+1)(x+2) requires transformations, which consist in the fact that you need to bring similar terms, first opening the brackets. In place of the first there will be the following expression: x 2 + 2x + 1. After the equality, this entry will appear: x 2 + 3x + 2. After similar terms are counted, the equation will take the form: x 2 - x = 0. It has become incomplete . Something similar to this has already been discussed a little higher. The roots of this will be the numbers 0 and 1.