We continue to study the topic " solving equations" We have already become acquainted with linear equations and are moving on to getting acquainted with quadratic equations.
First, we will look at what a quadratic equation is, how it is written in general form, and give related definitions. After this, we will use examples to examine in detail how incomplete quadratic equations are solved. Next, we will move on to solving complete equations, obtain the root formula, get acquainted with the discriminant of a quadratic equation, and consider solutions to typical examples. Finally, let's trace the connections between the roots and coefficients.
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What is a quadratic equation? Their types
First you need to clearly understand what a quadratic equation is. Therefore, it is logical to start a conversation about quadratic equations with the definition of a quadratic equation, as well as related definitions. After this, you can consider the main types of quadratic equations: reduced and unreduced, as well as complete and incomplete equations.
Definition and examples of quadratic equations
Definition.
Quadratic equation is an equation of the form a x 2 +b x+c=0, where x is a variable, a, b and c are some numbers, and a is non-zero.
Let's say right away that quadratic equations are often called equations of the second degree. This is due to the fact that the quadratic equation is algebraic equation second degree.
The stated definition allows us to give examples of quadratic equations. So 2 x 2 +6 x+1=0, 0.2 x 2 +2.5 x+0.03=0, etc. These are quadratic equations.
Definition.
Numbers a, b and c are called coefficients of the quadratic equation a·x 2 +b·x+c=0, and coefficient a is called the first, or the highest, or the coefficient of x 2, b is the second coefficient, or the coefficient of x, and c is the free term.
For example, let's take a quadratic equation of the form 5 x 2 −2 x −3=0, here the leading coefficient is 5, the second coefficient is equal to −2, and the free term is equal to −3. Please note that when the coefficients b and/or c are negative, as in the example just given, the short form of the quadratic equation is 5 x 2 −2 x−3=0 , rather than 5 x 2 +(−2 )·x+(−3)=0 .
It is worth noting that when the coefficients a and/or b are equal to 1 or −1, then they are usually not explicitly present in the quadratic equation, which is due to the peculiarities of writing such. For example, in the quadratic equation y 2 −y+3=0 the leading coefficient is one, and the coefficient of y is equal to −1.
Reduced and unreduced quadratic equations
Depending on the value of the leading coefficient, reduced and unreduced quadratic equations are distinguished. Let us give the corresponding definitions.
Definition.
A quadratic equation in which the leading coefficient is 1 is called given quadratic equation. Otherwise the quadratic equation is untouched.
According to this definition, quadratic equations x 2 −3·x+1=0, x 2 −x−2/3=0, etc. – given, in each of them the first coefficient is equal to one. A 5 x 2 −x−1=0, etc. - unreduced quadratic equations, their leading coefficients are different from 1.
From any unreduced quadratic equation, by dividing both sides by the leading coefficient, you can go to the reduced one. This action is an equivalent transformation, that is, the reduced quadratic equation obtained in this way has the same roots as the original unreduced quadratic equation, or, like it, has no roots.
Let us look at an example of how the transition from an unreduced quadratic equation to a reduced one is performed.
Example.
From the equation 3 x 2 +12 x−7=0, go to the corresponding reduced quadratic equation.
Solution.
We just need to divide both sides of the original equation by the leading coefficient 3, it is non-zero, so we can perform this action. We have (3 x 2 +12 x−7):3=0:3, which is the same, (3 x 2):3+(12 x):3−7:3=0, and then (3:3) x 2 +(12:3) x−7:3=0, from where . This is how we obtained the reduced quadratic equation, which is equivalent to the original one.
Answer:
Complete and incomplete quadratic equations
The definition of a quadratic equation contains the condition a≠0. This condition is necessary so that the equation a x 2 + b x + c = 0 is quadratic, since when a = 0 it actually becomes a linear equation of the form b x + c = 0.
As for the coefficients b and c, they can be equal to zero, both individually and together. In these cases, the quadratic equation is called incomplete.
Definition.
The quadratic equation a x 2 +b x+c=0 is called incomplete, if at least one of the coefficients b, c is equal to zero.
In turn
Definition.
Complete quadratic equation is an equation in which all coefficients are different from zero.
Such names were not given by chance. This will become clear from the following discussions.
If the coefficient b is zero, then the quadratic equation takes the form a·x 2 +0·x+c=0, and it is equivalent to the equation a·x 2 +c=0. If c=0, that is, the quadratic equation has the form a·x 2 +b·x+0=0, then it can be rewritten as a·x 2 +b·x=0. And with b=0 and c=0 we get the quadratic equation a·x 2 =0. The resulting equations differ from the complete quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Hence their name - incomplete quadratic equations.
So the equations x 2 +x+1=0 and −2 x 2 −5 x+0.2=0 are examples of complete quadratic equations, and x 2 =0, −2 x 2 =0, 5 x 2 +3=0 , −x 2 −5 x=0 are incomplete quadratic equations.
Solving incomplete quadratic equations
From the information in the previous paragraph it follows that there is three types of incomplete quadratic equations:
- a·x 2 =0, the coefficients b=0 and c=0 correspond to it;
- a x 2 +c=0 when b=0 ;
- and a·x 2 +b·x=0 when c=0.
Let us examine in order how incomplete quadratic equations of each of these types are solved.
a x 2 =0
Let's start with solving incomplete quadratic equations in which the coefficients b and c are equal to zero, that is, with equations of the form a x 2 =0. The equation a·x 2 =0 is equivalent to the equation x 2 =0, which is obtained from the original by dividing both parts by a non-zero number a. Obviously, the root of the equation x 2 =0 is zero, since 0 2 =0. This equation has no other roots, which is explained by the fact that for any non-zero number p the inequality p 2 >0 holds, which means that for p≠0 the equality p 2 =0 is never achieved.
So, the incomplete quadratic equation a·x 2 =0 has a single root x=0.
As an example, we give the solution to the incomplete quadratic equation −4 x 2 =0. It is equivalent to the equation x 2 =0, its only root is x=0, therefore, the original equation has a single root zero.
A short solution in this case can be written as follows:
−4 x 2 =0 ,
x 2 =0,
x=0 .
a x 2 +c=0
Now let's look at how incomplete quadratic equations are solved in which the coefficient b is zero and c≠0, that is, equations of the form a x 2 +c=0. We know that moving a term from one side of the equation to the other with the opposite sign, as well as dividing both sides of the equation by a non-zero number, gives an equivalent equation. Therefore, we can carry out the following equivalent transformations of the incomplete quadratic equation a x 2 +c=0:
- move c to the right side, which gives the equation a x 2 =−c,
- and divide both sides by a, we get .
The resulting equation allows us to draw conclusions about its roots. Depending on the values of a and c, the value of the expression can be negative (for example, if a=1 and c=2, then ) or positive (for example, if a=−2 and c=6, then ), it is not equal to zero , since by condition c≠0. We will separately analyze the cases and.
If , then the equation has no roots. This statement follows from the fact that the square of any number is a non-negative number. It follows from this that when , then for any number p the equality cannot be true.
If , then the situation with the roots of the equation is different. In this case, if we remember about , then the root of the equation immediately becomes obvious; it is the number, since . It’s easy to guess that the number is also the root of the equation, indeed, . This equation has no other roots, which can be shown, for example, by contradiction. Let's do this.
Let us denote the roots of the equation just announced as x 1 and −x 1 . Suppose that the equation has one more root x 2, different from the indicated roots x 1 and −x 1. It is known that substituting its roots into an equation instead of x turns the equation into a correct numerical equality. For x 1 and −x 1 we have , and for x 2 we have . The properties of numerical equalities allow us to perform term-by-term subtraction of correct numerical equalities, so subtracting the corresponding parts of the equalities gives x 1 2 −x 2 2 =0. The properties of operations with numbers allow us to rewrite the resulting equality as (x 1 −x 2)·(x 1 +x 2)=0. We know that the product of two numbers is equal to zero if and only if at least one of them is equal to zero. Therefore, from the resulting equality it follows that x 1 −x 2 =0 and/or x 1 +x 2 =0, which is the same, x 2 =x 1 and/or x 2 =−x 1. So we came to a contradiction, since at the beginning we said that the root of the equation x 2 is different from x 1 and −x 1. This proves that the equation has no roots other than and .
Let us summarize the information in this paragraph. The incomplete quadratic equation a x 2 +c=0 is equivalent to the equation that
- has no roots if ,
- has two roots and , if .
Let's consider examples of solving incomplete quadratic equations of the form a·x 2 +c=0.
Let's start with the quadratic equation 9 x 2 +7=0. After moving the free term to the right side of the equation, it will take the form 9 x 2 =−7. Dividing both sides of the resulting equation by 9, we arrive at . Since the right side has a negative number, this equation has no roots, therefore, the original incomplete quadratic equation 9 x 2 +7 = 0 has no roots.
Let's solve another incomplete quadratic equation −x 2 +9=0. We move the nine to the right side: −x 2 =−9. Now we divide both sides by −1, we get x 2 =9. On the right side there is a positive number, from which we conclude that or . Then we write down the final answer: the incomplete quadratic equation −x 2 +9=0 has two roots x=3 or x=−3.
a x 2 +b x=0
It remains to deal with the solution of the last type of incomplete quadratic equations for c=0. Incomplete quadratic equations of the form a x 2 + b x = 0 allows you to solve factorization method. Obviously, we can, located on the left side of the equation, for which it is enough to take the common factor x out of brackets. This allows us to move from the original incomplete quadratic equation to an equivalent equation of the form x·(a·x+b)=0. And this equation is equivalent to a set of two equations x=0 and a·x+b=0, the latter of which is linear and has a root x=−b/a.
So, the incomplete quadratic equation a·x 2 +b·x=0 has two roots x=0 and x=−b/a.
To consolidate the material, we will analyze the solution to a specific example.
Example.
Solve the equation.
Solution.
Taking x out of brackets gives the equation . It is equivalent to two equations x=0 and . We solve the resulting linear equation: , and by dividing the mixed number by an ordinary fraction, we find . Therefore, the roots of the original equation are x=0 and .
After gaining the necessary practice, solutions to such equations can be written briefly:
Answer:
x=0 , .
Discriminant, formula for the roots of a quadratic equation
To solve quadratic equations, there is a root formula. Let's write it down formula for the roots of a quadratic equation: , Where D=b 2 −4 a c- so-called discriminant of a quadratic equation. The entry essentially means that .
It is useful to know how the root formula was derived and how it is used in finding the roots of quadratic equations. Let's figure this out.
Derivation of the formula for the roots of a quadratic equation
Let us need to solve the quadratic equation a·x 2 +b·x+c=0. Let's perform some equivalent transformations:
- We can divide both sides of this equation by a non-zero number a, resulting in the following quadratic equation.
- Now select a complete square on its left side: . After this, the equation will take the form .
- At this stage, it is possible to transfer the last two terms to the right side with the opposite sign, we have .
- And let’s also transform the expression on the right side: .
As a result, we arrive at an equation that is equivalent to the original quadratic equation a·x 2 +b·x+c=0.
We have already solved equations similar in form in the previous paragraphs, when we examined. This allows us to draw the following conclusions regarding the roots of the equation:
- if , then the equation has no real solutions;
- if , then the equation has the form , therefore, , from which its only root is visible;
- if , then or , which is the same as or , that is, the equation has two roots.
Thus, the presence or absence of roots of the equation, and therefore the original quadratic equation, depends on the sign of the expression on the right side. In turn, the sign of this expression is determined by the sign of the numerator, since the denominator 4·a 2 is always positive, that is, by the sign of the expression b 2 −4·a·c. This expression b 2 −4 a c was called discriminant of a quadratic equation and designated by the letter D. From here the essence of the discriminant is clear - based on its value and sign, they conclude whether the quadratic equation has real roots, and if so, what is their number - one or two.
Let's return to the equation and rewrite it using the discriminant notation: . And we draw conclusions:
- if D<0 , то это уравнение не имеет действительных корней;
- if D=0, then this equation has a single root;
- finally, if D>0, then the equation has two roots or, which can be rewritten in the form or, and after expanding and bringing the fractions to a common denominator we obtain.
So we derived the formulas for the roots of the quadratic equation, they have the form , where the discriminant D is calculated by the formula D=b 2 −4·a·c.
With their help, with a positive discriminant, you can calculate both real roots of a quadratic equation. When the discriminant is zero, both formulas give the same value of the root, corresponding to a unique solution to the quadratic equation. And with a negative discriminant, when we try to use the formula for the roots of a quadratic equation, we are faced with extracting the square root of a negative number, which takes us beyond the scope of the school curriculum. With a negative discriminant, the quadratic equation has no real roots, but has a pair complex conjugate roots, which can be found using the same root formulas we obtained.
Algorithm for solving quadratic equations using root formulas
In practice, when solving quadratic equations, you can immediately use the root formula to calculate their values. But this is more related to finding complex roots.
However, in a school algebra course we usually talk not about complex, but about real roots of a quadratic equation. In this case, it is advisable, before using the formulas for the roots of a quadratic equation, to first find the discriminant, make sure that it is non-negative (otherwise, we can conclude that the equation does not have real roots), and only then calculate the values of the roots.
The above reasoning allows us to write algorithm for solving a quadratic equation. To solve the quadratic equation a x 2 +b x+c=0, you need to:
- using the discriminant formula D=b 2 −4·a·c, calculate its value;
- conclude that a quadratic equation has no real roots if the discriminant is negative;
- calculate the only root of the equation using the formula if D=0;
- find two real roots of a quadratic equation using the root formula if the discriminant is positive.
Here we just note that if the discriminant is equal to zero, you can also use the formula; it will give the same value as .
You can move on to examples of using the algorithm for solving quadratic equations.
Examples of solving quadratic equations
Let's consider solutions to three quadratic equations with a positive, negative and zero discriminant. Having dealt with their solution, by analogy it will be possible to solve any other quadratic equation. Let's begin.
Example.
Find the roots of the equation x 2 +2·x−6=0.
Solution.
In this case, we have the following coefficients of the quadratic equation: a=1, b=2 and c=−6. According to the algorithm, you first need to calculate the discriminant; to do this, we substitute the indicated a, b and c into the discriminant formula, we have D=b 2 −4·a·c=2 2 −4·1·(−6)=4+24=28. Since 28>0, that is, the discriminant is greater than zero, the quadratic equation has two real roots. Let's find them using the root formula, we get, here you can simplify the resulting expressions by doing moving the multiplier beyond the root sign followed by reduction of the fraction:
Answer:
Let's move on to the next typical example.
Example.
Solve the quadratic equation −4 x 2 +28 x−49=0 .
Solution.
We start by finding the discriminant: D=28 2 −4·(−4)·(−49)=784−784=0. Therefore, this quadratic equation has a single root, which we find as , that is,
Answer:
x=3.5.
It remains to consider solving quadratic equations with a negative discriminant.
Example.
Solve the equation 5·y 2 +6·y+2=0.
Solution.
Here are the coefficients of the quadratic equation: a=5, b=6 and c=2. We substitute these values into the discriminant formula, we have D=b 2 −4·a·c=6 2 −4·5·2=36−40=−4. The discriminant is negative, therefore, this quadratic equation has no real roots.
If you need to indicate complex roots, then we apply the well-known formula for the roots of a quadratic equation, and perform operations with complex numbers:
Answer:
there are no real roots, complex roots are: .
Let us note once again that if the discriminant of a quadratic equation is negative, then in school they usually immediately write down an answer in which they indicate that there are no real roots, and complex roots are not found.
Root formula for even second coefficients
The formula for the roots of a quadratic equation, where D=b 2 −4·a·c allows you to obtain a formula of a more compact form, allowing you to solve quadratic equations with an even coefficient for x (or simply with a coefficient of the form 2·n, for example, or 14· ln5=2·7·ln5 ). Let's get her out.
Let's say we need to solve a quadratic equation of the form a x 2 +2 n x+c=0. Let's find its roots using the formula we know. To do this, we calculate the discriminant D=(2 n) 2 −4 a c=4 n 2 −4 a c=4 (n 2 −a c), and then we use the root formula:
Let us denote the expression n 2 −a c as D 1 (sometimes it is denoted D "). Then the formula for the roots of the quadratic equation under consideration with the second coefficient 2 n will take the form 
, where D 1 =n 2 −a·c.
It is easy to see that D=4·D 1, or D 1 =D/4. In other words, D 1 is the fourth part of the discriminant. It is clear that the sign of D 1 is the same as the sign of D . That is, the sign D 1 is also an indicator of the presence or absence of roots of the quadratic equation.
So, to solve a quadratic equation with a second coefficient 2·n, you need
- Calculate D 1 =n 2 −a·c ;
- If D 1<0 , то сделать вывод, что действительных корней нет;
- If D 1 =0, then calculate the only root of the equation using the formula;
- If D 1 >0, then find two real roots using the formula.
Let's consider solving the example using the root formula obtained in this paragraph.
Example.
Solve the quadratic equation 5 x 2 −6 x −32=0 .
Solution.
The second coefficient of this equation can be represented as 2·(−3) . That is, you can rewrite the original quadratic equation in the form 5 x 2 +2 (−3) x−32=0, here a=5, n=−3 and c=−32, and calculate the fourth part of the discriminant: D 1 =n 2 −a·c=(−3) 2 −5·(−32)=9+160=169. Since its value is positive, the equation has two real roots. Let's find them using the appropriate root formula:
Note that it was possible to use the usual formula for the roots of a quadratic equation, but in this case more computational work would have to be performed.
Answer:
Simplifying the form of quadratic equations
Sometimes, before starting to calculate the roots of a quadratic equation using formulas, it doesn’t hurt to ask the question: “Is it possible to simplify the form of this equation?” Agree that in terms of calculations it will be easier to solve the quadratic equation 11 x 2 −4 x−6=0 than 1100 x 2 −400 x−600=0.
Typically, simplifying the form of a quadratic equation is achieved by multiplying or dividing both sides by a certain number. For example, in the previous paragraph it was possible to simplify the equation 1100 x 2 −400 x −600=0 by dividing both sides by 100.
A similar transformation is carried out with quadratic equations, the coefficients of which are not . In this case, both sides of the equation are usually divided by the absolute values of its coefficients. For example, let's take the quadratic equation 12 x 2 −42 x+48=0. absolute values of its coefficients: GCD(12, 42, 48)= GCD(GCD(12, 42), 48)= GCD(6, 48)=6. Dividing both sides of the original quadratic equation by 6, we arrive at the equivalent quadratic equation 2 x 2 −7 x+8=0.
And multiplying both sides of a quadratic equation is usually done to get rid of fractional coefficients. In this case, multiplication is carried out by the denominators of its coefficients. For example, if both sides of the quadratic equation are multiplied by LCM(6, 3, 1)=6, then it will take the simpler form x 2 +4·x−18=0.
In conclusion of this point, we note that they almost always get rid of the minus at the highest coefficient of a quadratic equation by changing the signs of all terms, which corresponds to multiplying (or dividing) both sides by −1. For example, usually one moves from the quadratic equation −2 x 2 −3 x+7=0 to the solution 2 x 2 +3 x−7=0 .
Relationship between roots and coefficients of a quadratic equation
The formula for the roots of a quadratic equation expresses the roots of the equation through its coefficients. Based on the root formula, you can obtain other relationships between roots and coefficients.
The most well-known and applicable formulas from Vieta’s theorem are of the form and . In particular, for the given quadratic equation, the sum of the roots is equal to the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by the form of the quadratic equation 3 x 2 −7 x + 22 = 0 we can immediately say that the sum of its roots is equal to 7/3, and the product of the roots is equal to 22/3.
Using the already written formulas, you can obtain a number of other connections between the roots and coefficients of the quadratic equation. For example, you can express the sum of the squares of the roots of a quadratic equation through its coefficients: .
References.
- Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
- Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
At first glance, it may seem that the procedure for factoring a square root is complex and inaccessible. But that's not true. In this article, we'll show you how to approach square roots and factors, and solve square roots with ease using two proven methods.
Yandex.RTB R-A-339285-1
Factoring a root
First, let's define the purpose of the square root factorization procedure. Target- simplify the square root and write it in a form convenient for calculations.
Definition 1
Factoring a square root is finding two or more numbers that, when multiplied by each other, will give a number equal to the original. For example: 4x4 = 16.
If you can find the factors, you can easily simplify the square root expression or eliminate it altogether:
Example 1
Divide the radical number by 2 if it is even.
The radical number should always be divided by prime numbers, since any prime number value can be factorized into prime factors. If you have an odd number, try dividing it by 3. Not divisible by 3? Continue dividing by 5, 7, 9, etc.
Write the expression as the root of the product of two numbers.
For example, you can simplify 98 in this way: = 98 ÷ 2 = 49. It follows that 2 × 49 = 98, so we can rewrite the problem as follows: 98 = (2 × 49).
Continue decomposing the numbers until the product of two identical numbers and other numbers remains under the root.
Let's take our example (2 × 49):
Since 2 is already maximally simplified, it is necessary to simplify 49. We are looking for a prime number that can be divided by 49. Obviously, neither 3 nor 5 are suitable. That leaves 7: 49 ÷ 7 = 7, so 7 × 7 = 49.
We write the example in the following form: (2 × 49) = (2 × 7 × 7) .
Simplify the square root expression.
Since in brackets we have the product of 2 and two identical numbers (7), we can take the number 7 out of the root sign.
Example 2
(2 × 7 × 7) = (2) × (7 × 7) = (2) × 7 = 7 (2) .
At the moment when there are two identical numbers under the root, stop factoring the numbers. Of course, if you have used all the possibilities to the maximum.
Remember: there are roots that can be simplified many times.
In this case, the numbers that we take out from under the root and the numbers that stand in front of it are multiplied.
Example 3
180 = (2 × 90) 180 = (2 × 2 × 45) 180 = 2 45
but 45 can be factorized and the root simplified again.
180 = 2 (3 × 15) 180 = 2 (3 × 3 × 5) 180 = 2 × 3 5 180 = 6 5
When it is impossible to obtain two identical numbers under the root sign, this means that such a root cannot be simplified.
If, after decomposing the radical expression into a product of prime numbers, you were unable to obtain two identical numbers, then such a root cannot be simplified.
Example 4
70 = 35 × 2, so 70 = (35 × 2)
35 = 7 × 5, so (35 × 2) = (7 × 5 × 2)
As you can see, all three factors are prime numbers that cannot be factorized. There are no identical numbers among them, so it is not possible to remove an integer from under the root. Simplify 70 it is forbidden.
Full square
Memorize some squares of prime numbers.
The square of a number is obtained by multiplying it by itself, i.e. when squaring. If you remember ten squares of prime numbers, it will greatly simplify your life in further simplifying the roots.
Example 5
1 2 = 1 2 2 = 4 3 2 = 9 4 2 = 16 5 2 = 25 6 2 = 36 7 2 = 49 8 2 = 64 9 2 = 81 10 2 = 100
If there is a complete square under the square root sign, then you should remove the root sign and write down the square root of this complete square.
Difficult? No:
Example 6
1 = 1 4 = 2 9 = 3 16 = 4 25 = 5 36 = 6 49 = 7 64 = 8 81 = 9 100 = 10
Try to decompose the number under the root sign into the product of a perfect square and another number.
If you see that the radical expression is decomposed into the product of a perfect square and some number, then by remembering a few examples, you will significantly save time and nerves:
Example 7
50 = (25 × 2) = 5 2. If the radical number ends in 25, 50 or 75, you can always factor it into the product of 25 and some number.
1700 = (100 × 17) = 10 17. If the radical number ends in 00, you can always factor it into the product of 100 and some number.
72 = (9 × 8) = 3 8. If the sum of the digits of a radical number is 9, you can always factor it into the product of 9 and some number.
Try to decompose the radical number into the product of several complete squares: take them out from under the root sign and multiply.
Example 8
72 = (9 × 8) 72 = (9 × 4 × 2) 72 = 9 × 4 × 2 72 = 3 × 2 × 2 72 = 6 2
If you notice an error in the text, please highlight it and press Ctrl+Enter
In 8th grade, schoolchildren in mathematics lessons are introduced to the concept of “radical” or, simply put, “root”. It was then that they first encountered the problem of simplifying complex radicals. Complex radicals are expressions in which one root is under another. Therefore, they are sometimes called nested radicals. In this article, the mathematics and physics tutor talks in detail about how to simplify a complex radical.
Methods for simplifying complex radicals
To simplify a complex radical means to get rid of the outer root. It is best to start studying this topic by simplifying double radicals. After all, if we learn to simplify double radicals, then we will also be able to simplify more complex ones.
How do we get rid of the outer root? It is clear that for this you need to transform the radical expression, presenting it in the form of a complete square. To do this, we will use the well-known formula “Square of the difference”:
Here, as you can see, the negative term has a factor on the right. Therefore, let's get this factor under the root. To do this, we present it as a product of:
Then and. It remains only to pay attention to the fact that 
. Now we can see that under the root we have a squared difference:
Now let's remember that. Exactly the module. This is very important here because the square root is a positive number. Then we get:
Well, since title="Rendered by QuickLaTeX.com" height="21" width="61" style="vertical-align: -3px;">, модуль раскрывается со знаком минус. В результате в ответе получаем:!}
This is how we managed to simplify this radical. But there are also more complex cases when it is not immediately possible to guess how to represent a radical expression in the form of a complete square. For example, in the following example.
In order not to rack your brains for a long time, you can use the following method.
Let me remind you that our goal is to represent the expression under the root as a perfect square. Specifically in this example, in the form of a square of the sum:
Well, the square of the sum is revealed according to the well-known formula, which we already wrote today:
So, the idea, in fact, is to take the irrational part of the radical expression for, and the rational part for. Then we get the following system of equations:
It is clear that . Otherwise, the second equation of the system is not satisfied. Then we express the coefficient from the second equation:
The denominator of this fraction is not equal to zero, which means its numerator is equal to zero. We obtain a biquadratic equation, which can be solved in the standard way (for more details, see the attached video). Solving it, we get as many as 4 roots. You can take any one. I like it better. Then . So, we finally get:
Here is a way to simplify a complex radical. There is one more. For those who like to memorize complex formulas, which I am not. But for the sake of completeness, I’ll tell you about him too.
Formula of complex radicals
This is what the formula looks like:
Quite scary, isn't it? But don't be afraid, it can actually be used successfully in some cases. Let's look at an example:
We substitute the corresponding values into the formula:
This is the answer.
So, today in class I talked about how to simplify a complex radical. If you did not previously know the methods discussed today, then most likely you still have a lot to learn in order to feel confident in the Unified State Exam or the entrance exam in mathematics. But don't worry, I can teach you all this. All the necessary information about my classes is on. Good luck to you!
Material prepared by Sergey Valerievich
