- How to solve equations with roots
- How to solve irrational equations
- How to solve an irrational equation

Unlike other types of equations, for example, quadratic or linear systems of equations, there is no standard algorithm for solving equations with roots, or more precisely, irrational equations. In each particular case, it is necessary to choose the most suitable solution method based on the “appearance” and features of the equation.

The raising of parts of the equation to the same degree.

Most often, to solve equations with roots (irrational equations), the raising of both sides of the equation to the same degree is used. As a rule, to a degree equal to the degree of the root (squared for square root, cube for cubic root). It should be borne in mind that when raising the left and right sides of the equation to an even degree, he may have “extra” roots. Therefore, in this case, one should check the obtained roots by substituting them in the equation. Particular attention in solving equations with square (even) roots should be given to the region of admissible values of the variable (ODZ). Sometimes, the estimation of ODZ alone is enough to solve or significantly simplify the equation.

Example. Solve the equation:

We square both sides of the equation:

(√ (5x-16)) ² = (x-2) ², whence we successively get:

Solving the obtained quadratic equation, we find its roots:

Substituting both found roots into the original equation, we obtain the correct equality. Therefore, both numbers are solutions of the equation.

Method for introducing a new variable.

Sometimes it’s more convenient to find the roots of the “equation with roots” (an irrational equation) by introducing new variables. In fact, the essence of this method is reduced to a more compact record of the solution, i.e. instead of writing a bulky expression each time, it is replaced by a legend.

Example. Solve the equation: 2x + √x-3 = 0

You can solve this equation by squaring both sides. However, the calculations themselves will look rather cumbersome. With the introduction of a new variable, the decision process will turn out to be much more elegant:

We introduce a new variable: y = √ x

Then we get the ordinary quadratic equation:

2y² + y-3 = 0, with variable y.

Solving the resulting equation, we find two roots:

substituting the found roots in the expression for the new variable (y), we obtain:

Since the square root value cannot be a negative number (if you do not touch the area of complex numbers), we get the only solution:

## Solution of the irrational equation

We’ll deal with our irrational equation given at the very beginning of the lesson. Here the root is already solitary: to the left of the equal sign there is nothing but the root. We square both sides:

2 x 2 - 14 x + 13 = (5 - x) 2

2 x 2 - 14 x + 13 = 25 - 10 x + x 2

x 2 - 4 x - 12 = 0

We solve the obtained quadratic equation through the discriminant:

D = b 2 - 4 ac = (−4) 2 - 4 · 1 · (−12) = 16 + 48 = 64

x _{1} = 6, x _{2} = −2

It remains only to substitute these numbers in the original equation, i.e. run check. But here you can do the right thing to simplify the final decision.

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## How to simplify the solution

Let's think: why do we do a check at the end of solving an irrational equation? We want to make sure that when substituting our roots, a non-negative number will be to the right of the equal sign. After all, we already know for sure that precisely a non-negative number is on the left, because the arithmetic square root (because of which our equation is called irrational) by definition cannot be less than zero.

Therefore, all we need to check is that the function g (x) = 5 - x, which is to the right of the equal sign, is non-negative:

We substitute our roots in this function and get:

g (x _{1}) = g (6) = 5 - 6 = −1 g (x _{2}) = g (−2) = 5 - (−2) = 5 + 2 = 7> 0

From the obtained values it follows that the root x _{1} = 6 does not suit us, since when substituting into the right side of the original equation, we get a negative number. And here is the root x _{2} = −2 is quite suitable for us, because:

- This root is a solution to the quadratic equation obtained by erecting both sides
*irrational equation*squared. - The right side of the original irrational equation when substituting the root x
_{2}= −2 turns into a positive number, i.e. the value range of the arithmetic root is not violated.

That's the whole algorithm! As you can see, solving equations with radicals is not so difficult. The main thing is not to forget to check the received roots, otherwise it is very likely to get extra answers.