Solve an equation with square roots. How to solve equations with roots: solving equations with roots
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Quite often the root sign appears in equations and many people mistakenly believe that such equations are difficult to solve. For such equations in mathematics there is special term, which is what equations with a root are called - irrational equations.
The main difference in solving equations with roots from other equations, for example, quadratic, logarithmic, linear, is that they do not have a standard solution algorithm. Therefore, in order to solve an irrational equation, it is necessary to analyze the initial data and choose a more suitable solution.
In most cases, to solve this type of equation, they use the method of raising both sides of the equation to the same power
Let's say the following equation is given:
\[\sqrt((5x-16))=x-2\]
We square both sides of the equation:
\[\sqrt((5x-16)))^2 =(x-2)^2\], from which we consistently obtain:
Having received quadratic equation, we find its roots:
Answer: \
If we substitute these values into the equation, we will obtain the correct equality, which indicates the correctness of the data obtained.
Where can I solve an equation with roots using an online solver?
You can solve the equation on our website https://site. A free online solver will allow you to solve the equation online any complexity in seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.
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Every new action in mathematics instantly generates its opposite. Once upon a time, the ancient Greeks discovered that a square piece of land 2 meters long and 2 meters wide would have an area of 2*2 = 4 square meters(hereinafter will be denoted by m^2). Now, on the contrary, if a Greek knew that his plot of land was square and had an area of 4 m^2, how would he know what the length and width of his plot was? An operation was introduced that was the inverse of the squaring operation and became known as square root extraction. People began to understand that 2 squared (2^2) equals 4. Conversely, the square root of 4 (hereinafter referred to as √(4)) will be equal to two. Models became more complex, and records describing processes with roots also became more complex. The question has arisen many times: how to solve an equation with a root.
Let a certain value x, when multiplied by itself once, give 9. This can be written as x*x=9. Or through a degree: x^2=9. To find x, you need to take the root of 9, which to some extent is already an equation with a radical: x=√(9) . The root can be extracted orally or using a calculator. Next we should consider the inverse problem. A certain quantity, when the square root is taken from it, gives the value 7. If we write this in the form of an irrational equation, we get: √(x) = 7. To solve this problem, it is necessary to square both sides of the expression. Considering that √(x) *√(x) =x, it turns out x = 49. The root is immediately ready in its pure form. Next we need to look at more complex examples equations with roots.
Let us subtract 5 from a certain quantity, then raise the expression to the power of 1/2. As a result, the number 3 was obtained. Now this condition must be written as an equation: √(x-5) =3. Next, you should multiply each part of the equation by itself: x-5 = 3. After raising to the second power, the expression was freed from radicals. Now it’s time to solve the simplest linear equation, moving the five to right side and changing its sign. x = 5+3. x = 8. Unfortunately, not all life processes can be described as follows simple equations. Very often you can find expressions with several radicals; sometimes the degree of the root can be higher than the second. There is no single solution algorithm for such identities. It is worth looking for a special approach to each equation. An example is given in which the equation with the root has the third degree.
The cube root will be denoted by 3√. Find the volume of a container shaped like a cube with a side of 5 meters. Let the volume be x m^3. Then the cube root of the volume will be equal to side cube and equal to five meters. The resulting equation is: 3√(x) =5. To solve it, you need to raise both parts to the third power, x = 125. Answer: 125 cubic meters. Below is an example of an equation with a sum of roots. √(x) +√(x-1) =5. First you need to square both parts. To do this, it is worth remembering the abbreviated multiplication formula for the square of the sum: (a+b) ^2=a^2+2*ab+b^2. Applying to the equation, we get: x + 2*√(x) *√(x-1) + x-1 = 25. Next, the roots are left on the left side, and everything else is transferred to the right: 2*√(x) *√ (x-1) = 26 - 2x. It is convenient to divide both sides of the expression by 2: √((x) (x-1)) = 13 - x. A simpler irrational equation is obtained.
Next, both sides should be squared again: x*(x-1) = 169 - 26x + x^2. It is necessary to open the brackets and bring similar terms: x^2 - x = 169 - 26x + x^2. The second degree disappears, hence 25x = 169. x = 169/25 = 6.6. After checking, substituting the resulting root into the original equation: √(6.6) +√(6.6-1) = 2.6 + √(5.6) = 2.6 + 2.4 = 5, you can get a satisfactory answer. It is also very important to understand that an expression with a root of even degree cannot be negative. Indeed, by multiplying any number by itself an even number of times, it is impossible to obtain a value less than zero. Therefore, equations such as √(x^2+7x-11) = -3 can be safely not solved, but written that the equation has no roots. As mentioned above, solving equations with radicals can take a variety of forms.
A simple example of an equation where it is necessary to change variables. √(y) - 5*4√(y) +6 = 0, where 4√(y) is the fourth root of y. The proposed replacement looks like this: x = 4√(y) . After doing this, we get: x^2 - 5x + 6 = 0. The resulting quadratic equation is obtained. Its discriminant: 25 - 4*6 = 25 - 24 = 1. The first root x1 will be equal to (5 + √1) /2 = 6/2 = 3. The second root x2 = (5 - √1) /2 = 4/ 2 = 2. You can also find the roots using a corollary of Vieta’s theorem. Roots have been found, should be carried out reverse replacement. 4√(y) = 3, hence y1 = 1.6. Also 4√(y) = 2, taking the 4th root turns out that y2 = 1.9. Values calculated using a calculator. But you don’t have to do them, leaving the answer in the form of radicals.
Solution irrational equations.
In this article we will talk about solutions simplest irrational equations.
Irrational equation is an equation that contains an unknown under the root sign.
Let's look at two types irrational equations, which are very similar at first glance, but in essence are very different from each other.
(1)
(2)
In the first equation we see that the unknown is under the sign of the root of the third degree. We can take the odd root of negative number, therefore, in this equation there are no restrictions either on the expression under the root sign or on the expression on the right side of the equation. We can raise both sides of the equation to the third power to get rid of the root. We get an equivalent equation:
When raising the right and left sides of the equation to an odd power, we can not be afraid of getting extraneous roots.
Example 1. Let's solve the equation 
Let's raise both sides of the equation to the third power. We get an equivalent equation:
Let's move all the terms to one side and put x out of brackets:
Equating each factor to zero, we get:
Answer: (0;1;2)
Let's look closely at the second equation: . On the left side of the equation is the square root, which takes only non-negative values. Therefore, for the equation to have solutions, the right-hand side must also be non-negative. Therefore, the condition is imposed on the right side of the equation:
Title="g(x)>=0"> - это !} condition for the existence of roots.
To solve an equation of this type, you need to square both sides of the equation:
(3)
Squaring can lead to the appearance of extraneous roots, so we need the equations:
Title="f(x)>=0"> (4)!}
However, inequality (4) follows from condition (3): if on the right side of the equality there is a square of some expression, and the square of any expression can only take non-negative values, therefore left side must also be non-negative. Therefore, condition (4) automatically follows from condition (3) and our equation is equivalent to the system:
Title="delim(lbrace)(matrix(2)(1)((f(x)=g^2((x))) (g(x)>=0) ))( )">!}
Example 2. Let's solve the equation:
.
Let's move on to an equivalent system:
Title="delim(lbrace)(matrix(2)(1)((2x^2-7x+5=((1-x))^2) (1-x>=0) ))( )">!}
Let's solve the first equation of the system and check which roots satisfy the inequality.
Inequality title="1-x>=0">удовлетворяет только корень !}
Answer: x=1
Attention! If in the process of solving we square both sides of the equation, then we must remember that extraneous roots may appear. Therefore, you either need to move on to an equivalent system, or at the end of the solution, DO A CHECK: find the roots and substitute them into the original equation.
Example 3. Let's solve the equation:
To solve this equation, we also need to square both sides. Let's not bother with the ODZ and the condition for the existence of roots in this equation, but simply do a check at the end of the solution.
Let's square both sides of the equation:
Let's move the term containing the root to the left, and all other terms to the right:
Let's square both sides of the equation again:
On Vieta's theme:
Let's do a check. To do this, we substitute the found roots into the original equation. Obviously, at , the right-hand side of the original equation is negative, and the left-hand side is positive.
At we obtain the correct equality.
Equations in which a variable is contained under the root sign are called irrational.
Methods for solving irrational equations are usually based on the possibility of replacing (with the help of some transformations) an irrational equation with a rational equation that is either equivalent to the original irrational equation or is a consequence of it. Most often, both sides of the equation are raised to the same power. This produces an equation that is a consequence of the original one.
When solving irrational equations, the following must be taken into account:
1) if the root exponent is an even number, then the radical expression must be non-negative; in this case, the value of the root is also non-negative (definition of a root with an even exponent);
2) if the root exponent is an odd number, then the radical expression can be any real number; in this case, the sign of the root coincides with the sign of the radical expression.
Example 1. Solve the equation
Let's square both sides of the equation.
x 2 - 3 = 1;
Let's move -3 from the left side of the equation to the right and perform a reduction of similar terms.
x 2 = 4;
The resulting incomplete quadratic equation has two roots -2 and 2.
Let's check the obtained roots by substituting the values of the variable x into the original equation.
Examination.
When x 1 = -2 - true:
When x 2 = -2- true.
It follows that the original irrational equation has two roots -2 and 2.
Example 2. Solve the equation 
.
This equation can be solved using the same method as in the first example, but we will do it differently.
Let's find the ODZ of this equation. From the definition of the square root it follows that in this equation two conditions must be simultaneously satisfied:
ODZ of this uranium: x.
Answer: no roots.
Example 3. Solve the equation 
=+ 2.
Finding the ODZ in this equation is a rather difficult task. Let's square both sides of the equation:
x 3 + 4x - 1 - 8= x 3 - 1 + 4+ 4x;
=0;
x 1 =1; x 2 =0.
After checking, we establish that x 2 =0 is an extra root.
Answer: x 1 =1.
Example 4. Solve the equation x =.
In this example, the ODZ is easy to find. ODZ of this equation: x[-1;).
Let's square both sides of this equation, and as a result we get the equation x 2 = x + 1. The roots of this equation are:
It is difficult to verify the roots found. But, despite the fact that both roots belong to the ODZ, it is impossible to assert that both roots are roots of the original equation. This will result in an error. IN in this case An irrational equation is equivalent to a combination of two inequalities and one equation:
x+10 And x0 And x 2 = x + 1, from which it follows that the negative root for the irrational equation is extraneous and must be discarded.
Example 5. Solve equation += 7.
Let's square both sides of the equation and perform the reduction of similar terms, transfer the terms from one side of the equation to the other and multiply both sides by 0.5. As a result, we get the equation
= 12, (*) which is a consequence of the original one. Let's square both sides of the equation again. We obtain the equation (x + 5)(20 - x) = 144, which is a consequence of the original one. The resulting equation is reduced to the form x 2 - 15x + 44 =0.
This equation (also a consequence of the original one) has roots x 1 = 4, x 2 = 11. Both roots, as verification shows, satisfy the original equation.
Rep. x 1 = 4, x 2 = 11.
Comment. When squaring equations, students often multiply radical expressions in equations like (*), i.e., instead of equation = 12, they write the equation 
= 12. This does not lead to errors, since the equations are consequences of the equations. It should, however, be borne in mind that in the general case, such multiplication of radical expressions gives unequal equations.
In the examples discussed above, one could first move one of the radicals to the right side of the equation. Then there will be one radical left on the left side of the equation, and after squaring both sides of the equation, a rational function will be obtained on the left side of the equation. This technique (isolation of the radical) is quite often used when solving irrational equations.
Example 6. Solve equation-= 3.
Isolating the first radical, we obtain the equation
=+ 3, equivalent to the original one.
By squaring both sides of this equation, we get the equation
x 2 + 5x + 2 = x 2 - 3x + 3 + 6, equivalent to the equation
4x - 5 = 3(*). This equation is a consequence of the original equation. By squaring both sides of the equation, we arrive at the equation
16x 2 - 40x + 25 = 9(x 2 - 3x + 3), or
7x 2 - 13x - 2 = 0.
This equation is a consequence of equation (*) (and therefore the original equation) and has roots. The first root x 1 = 2 satisfies the original equation, but the second root x 2 = does not.
Answer: x = 2.
Note that if we immediately, without isolating one of the radicals, squared both sides of the original equation, we would have to perform rather cumbersome transformations.
When solving irrational equations, in addition to the isolation of radicals, other methods are used. Let's consider an example of using the method of replacing the unknown (method of introducing an auxiliary variable).
