Radicals are one of the key elements of algebra, which frequently appears on the YÖS exam. It is necessary to understand not only the definition of a root, but also all of its properties, to be able to simplify radical expressions and to rationalize the denominator.
---
Definition: The arithmetic square root of a non-negative number $a$ is the non-negative number whose square equals $a$.
$$\sqrt{a} = b, \text{ если } b^2 = a \text{ и } b \geq 0$$
Examples:- $\sqrt{9} = 3$ (because $3^2 = 9$ and $3 \geq 0$)- $\sqrt{16} = 4$ (because $4^2 = 16$ and $4 \geq 0$)- $\sqrt{0} = 0$- $\sqrt{0.25} = 0.5$
Important remarks:- The radicand (the expression under the root) must be non-negative: $a \geq 0$- $\sqrt{a} \geq 0$ always (the arithmetic root cannot be negative)- $\sqrt{(-3)^2} = \sqrt{9} = 3 \neq -3$
For non-negative $a$:$$(\sqrt{a})^2 = a$$
Examples:- $(\sqrt{5})^2 = 5$- $(\sqrt{0.7})^2 = 0.7$
For any real $a$:$$\sqrt{a^2} = |a|$$
Examples:- $\sqrt{5^2} = |5| = 5$- $\sqrt{(-5)^2} = |-5| = 5$- $\sqrt{x^2} = |x|$
Definition: The $n$-th root of a number $a$ is a number $b$ such that $b^n = a$.
$$\sqrt[n]{a} = b, \text{ если } b^n = a$$
Notation: $\sqrt[n]{a}$ or $a^{1/n}$
Examples:- $\sqrt[3]{8} = 2$ (because $2^3 = 8$)- $\sqrt[3]{-27} = -3$ (because $(-3)^3 = -27$)- $\sqrt[4]{16} = 2$ (because $2^4 = 16$)
For even $n$, the arithmetic root $\sqrt[n]{a}$ is defined only for $a \geq 0$ and $\sqrt[n]{a} \geq 0$.
For odd $n$, the root $\sqrt[n]{a}$ is defined for any real $a$.
Examples:- $\sqrt[2]{9} = 3$ (only the positive value)- $\sqrt[3]{-8} = -2$ (odd root, may be negative)- $\sqrt[4]{16} = 2$ (only the positive value)
Property: The root of a product equals the product of the roots (provided that they are defined).
$$\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$$
Examples:$$\sqrt{4 \cdot 9} = \sqrt{4} \cdot \sqrt{9} = 2 \cdot 3 = 6$$$$\sqrt{9 \cdot 16} = \sqrt{9} \cdot \sqrt{16} = 3 \cdot 4 = 12$$$$\sqrt[3]{8 \cdot 27} = \sqrt[3]{8} \cdot \sqrt[3]{27} = 2 \cdot 3 = 6$$
Practical application: Simplifying the radicand$$\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$$$$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$$
Property: The root of a quotient equals the quotient of the roots.
$$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \quad (b \neq 0)$$
Examples:$$\sqrt{\frac{9}{4}} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2}$$$$\sqrt[3]{\frac{8}{27}} = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}$$$$\sqrt{\frac{x^2}{y^2}} = \frac{\sqrt{x^2}}{\sqrt{y^2}} = \frac{|x|}{|y|}$$
Property:$$\sqrt[n]{a^m} = a^{m/n}$$
Examples:$$\sqrt{5^2} = 5$$$$\sqrt[3]{2^6} = 2^{6/3} = 2^2 = 4$$$$\sqrt[4]{x^8} = x^{8/4} = x^2$$
Property:$$(\sqrt[n]{a})^m = a^{m/n} = \sqrt[n]{a^m}$$
Examples:$$(\sqrt{3})^2 = 3$$$$(\sqrt[3]{2})^3 = 2$$$$(\sqrt{5})^4 = (5^{1/2})^4 = 5^2 = 25$$
Property:$$\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} = a^{1/(mn)}$$
Examples:$$\sqrt{\sqrt{16}} = \sqrt[4]{16} = 2$$$$\sqrt[3]{\sqrt{64}} = \sqrt[6]{64} = \sqrt[6]{2^6} = 2$$
Algorithm:1. Factor the radicand so that at least one factor is a perfect power of the required degree2. Apply the product-of-radicals property3. Extract the root of the perfect power
Examples:
$$\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$$
$$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$$
$$\sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = \sqrt[3]{8} \cdot \sqrt[3]{2} = 2\sqrt[3]{2}$$
$$\sqrt{45x^2} = \sqrt{9x^2 \cdot 5} = 3|x|\sqrt{5}$$
(if $x \geq 0$, then $= 3x\sqrt{5}$)
Example: Simplify $\sqrt{200}$
$$200 = 8 \cdot 25 = 2^3 \cdot 5^2$$
We need perfect squares:$$200 = 4 \cdot 50 = 4 \cdot 25 \cdot 2 = 100 \cdot 2$$
$$\sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}$$
Example: Simplify $\sqrt[3]{135}$
$$135 = 27 \cdot 5 = 3^3 \cdot 5$$
$$\sqrt[3]{135} = \sqrt[3]{27 \cdot 5} = 3\sqrt[3]{5}$$
Algorithm:1. Raise the factor to the power equal to the degree of the root2. Place the result under the radical sign3. Multiply the radicands
$$3\sqrt{2} = \sqrt{3^2 \cdot 2} = \sqrt{9 \cdot 2} = \sqrt{18}$$
$$2\sqrt[3]{3} = \sqrt[3]{2^3 \cdot 3} = \sqrt[3]{8 \cdot 3} = \sqrt[3]{24}$$
$$x\sqrt{5} = \sqrt{x^2 \cdot 5} = \sqrt{5x^2}$$ (for $x \geq 0$)
This is often used to compare radical expressions:
$$3\sqrt{2} \text{ и } 2\sqrt{3}$$
Bring in:$$3\sqrt{2} = \sqrt{18}, \quad 2\sqrt{3} = \sqrt{12}$$
Since $18 > 12$, we have $3\sqrt{2} > 2\sqrt{3}$
Method: Multiply the numerator and denominator by that root.
$$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\frac{3}{\sqrt{5}} = \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$$
$$\frac{2}{\sqrt[3]{4}} = \frac{2}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{2\sqrt[3]{2}}{\sqrt[3]{4 \cdot 2}} = \frac{2\sqrt[3]{2}}{\sqrt[3]{8}} = \frac{2\sqrt[3]{2}}{2} = \sqrt[3]{2}$$
$$\frac{1}{2\sqrt{3}} = \frac{1}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{6}$$
$$\frac{5}{3\sqrt{2}} = \frac{5}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{3 \cdot 2} = \frac{5\sqrt{2}}{6}$$
Method: Multiply by the conjugate expression.
The conjugate of $(a + \sqrt{b})$ is $(a - \sqrt{b})$
We use: $(a + b)(a - b) = a^2 - b^2$
$$\frac{1}{1 + \sqrt{2}} = \frac{1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{1 - 2} = \frac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1$$
$$\frac{3}{2 - \sqrt{3}} = \frac{3}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{3(2 + \sqrt{3})}{4 - 3} = 3(2 + \sqrt{3}) = 6 + 3\sqrt{3}$$
$$\frac{2}{\sqrt{5} - 1} = \frac{2}{\sqrt{5} - 1} \cdot \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{2(\sqrt{5} + 1)}{5 - 1} = \frac{2(\sqrt{5} + 1)}{4} = \frac{\sqrt{5} + 1}{2}$$
$$\frac{1}{\sqrt{a} + \sqrt{b}} = \frac{1}{\sqrt{a} + \sqrt{b}} \cdot \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} - \sqrt{b}}{a - b}$$
Example:$$\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}$$
Example 1: Simplify $\sqrt{7 + 4\sqrt{3}}$
Let us try to represent it as $(\sqrt{a} + \sqrt{b})^2$:$$(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$$
We need: $a + b = 7$ and $2\sqrt{ab} = 4\sqrt{3}$
From the second equation: $\sqrt{ab} = 2\sqrt{3}$, so $ab = 12$
We solve the system: $a + b = 7, ab = 12$$$t^2 - 7t + 12 = 0$$$$(t - 3)(t - 4) = 0$$
So $a = 3, b = 4$ (or vice versa)
$$\sqrt{7 + 4\sqrt{3}} = \sqrt{(\sqrt{3} + \sqrt{4})^2} = \sqrt{(\sqrt{3} + 2)^2} = \sqrt{3} + 2$$
(we choose the positive value)
Example 2: Simplify $\sqrt{8 - 2\sqrt{15}}$
$$8 - 2\sqrt{15} = (\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab}$$
$a + b = 8, ab = 15$
$$t^2 - 8t + 15 = 0$$$$(t - 3)(t - 5) = 0$$
So $a = 5, b = 3$
$$\sqrt{8 - 2\sqrt{15}} = \sqrt{(\sqrt{5} - \sqrt{3})^2} = |\sqrt{5} - \sqrt{3}| = \sqrt{5} - \sqrt{3}$$
(since $\sqrt{5} > \sqrt{3}$)
Fundamental relation:$$\sqrt[n]{a} = a^{1/n}$$
This allows all properties of powers to be applied to radicals.
$$\sqrt{x^3} = x^{3/2}$$
$$\sqrt[3]{x^4} = x^{4/3}$$
$$\frac{1}{\sqrt{x}} = x^{-1/2}$$
$$\sqrt[4]{x^2} = x^{1/2} = \sqrt{x}$$ (for $x \geq 0$)
Statement: Simplify $\sqrt{72} - \sqrt{32} + \sqrt{50}$
Solution:
$$\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$$$$\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$$$$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$$
$$6\sqrt{2} - 4\sqrt{2} + 5\sqrt{2} = (6 - 4 + 5)\sqrt{2} = 7\sqrt{2}$$
Answer: $7\sqrt{2}$
Statement: Rationalize the denominator: $\frac{2}{3\sqrt{2}}$
$$\frac{2}{3\sqrt{2}} = \frac{2}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{3 \cdot 2} = \frac{2\sqrt{2}}{6} = \frac{\sqrt{2}}{3}$$
Answer: $\frac{\sqrt{2}}{3}$
Statement: Simplify $\frac{1}{\sqrt{7} - 2}$
$$\frac{1}{\sqrt{7} - 2} = \frac{1}{\sqrt{7} - 2} \cdot \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$$
Answer: $\frac{\sqrt{7} + 2}{3}$
Statement: Simplify $\sqrt{5 + 2\sqrt{6}}$
Represent as $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$
$a + b = 5, 2\sqrt{ab} = 2\sqrt{6}$, so $ab = 6$
$$t^2 - 5t + 6 = 0$$$$(t - 2)(t - 3) = 0$$
$a = 2, b = 3$ (or vice versa)
$$\sqrt{5 + 2\sqrt{6}} = \sqrt{(\sqrt{2} + \sqrt{3})^2} = \sqrt{2} + \sqrt{3}$$
Answer: $\sqrt{2} + \sqrt{3}$
Statement: Compute $\sqrt[3]{27} \cdot \sqrt[3]{8} + \sqrt{144} - \sqrt[4]{16}$
$$\sqrt[3]{27} = 3$$$$\sqrt[3]{8} = 2$$$$\sqrt{144} = 12$$$$\sqrt[4]{16} = 2$$
$$3 \cdot 2 + 12 - 2 = 6 + 12 - 2 = 16$$
Answer: 16
Wrong: $\sqrt{x^2} = x$ for all $x$
Correct: $\sqrt{x^2} = |x|$
For example, $\sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|$, and not $-3$
Wrong: $\sqrt{9 + 16} = \sqrt{9} + \sqrt{16} = 3 + 4 = 7$
Correct: $\sqrt{9 + 16} = \sqrt{25} = 5$
This is a serious mistake!
Wrong: $2\sqrt{3} = \sqrt{2 \cdot 3} = \sqrt{6}$
Correct: $2\sqrt{3} = \sqrt{4 \cdot 3} = \sqrt{12}$
Wrong: Is the conjugate of $(a + \sqrt{b})$ equal to $(a - \sqrt{b})$ or $(a + \sqrt{c})$?
Correct: The conjugate of $(a + \sqrt{b})$ is exactly $(a - \sqrt{b})$
$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$
Wrong: $\sqrt[4]{x^4} = x$ for all $x$
Correct: $\sqrt[4]{x^4} = |x|$, and for an even-degree root the radicand must be non-negative
### Tip 1: Memorize the basic roots$$\sqrt{4}=2, \sqrt{9}=3, \sqrt{16}=4, \sqrt{25}=5, \sqrt{36}=6$$$$\sqrt[3]{8}=2, \sqrt[3]{27}=3, \sqrt[3]{64}=4, \sqrt[3]{125}=5$$
### Tip 2: Simplify ALL radicals in the answerIf a radical of the form $\sqrt{8}$ remains in the answer, it means the solution is not finished. Simplify it to $2\sqrt{2}$.
### Tip 3: For cube roots remember$$\sqrt[3]{-8}=-2, \quad \sqrt[3]{-27}=-3$$The cube root of a negative number is negative!
### Tip 4: Try to represent a nested radical as the square of a sum or differenceSuch expressions often have elegant simplifications.
### Tip 5: Rationalize the denominatorOn the exam it is customary to write the answer without radicals in the denominator.
### Tip 6: Check your answers by direct computationIf you obtained $\sqrt{7} + 2$ as the answer to $\frac{1}{\sqrt{7}-2}$, check:$$(\sqrt{7} + 2)(\sqrt{7} - 2) = 7 - 4 = 3$$$$\frac{1}{\sqrt{7}-2} = \frac{\sqrt{7}+2}{3}$$ — correct!
All course topics · Practice problems
Please enable JavaScript to use the practice app.