**Roots**

The symbol n $\sqrt[n]{b}$ is read the nth root of b, where n is called the index, b is called the base, and √ is called the radical. $\sqrt[n]{b}$ denotes that number which raised to the nth power yields b. In other words, a is the nth root of b if a^{n} = b . For example, $\sqrt{9}$ = 3 because 3^{2} = 9 , and $\sqrt[3]{-8}$ = -2 because (-2)^{3} = -8. Even roots occur in pairs: both a positive root and a negative root. For example, $\sqrt[4]{16}$ = 2 since 2^{4} = 16 , and $\sqrt[4]{16}$ = -2 since (-2)^{4} = 16 . Odd roots occur alone and have the same sign as the base: $\sqrt[3]{-27}$ =-3 since (-3)^{3} = -27 . If given an even root, you are to assume it is the positive root. However, if you introduce even roots by solving an equation, then you __must __consider both the positive and negative roots:

x^{2} = 9

$\sqrt{{x}^{2}}=\pm \sqrt{9}$

$x=\pm 3$

Square roots and cube roots can be simplified by removing perfect squares and perfect cubes, respectively. For example,

$\sqrt{8}=\sqrt{4\xb72}=\sqrt{4}\sqrt{2}=2\sqrt{2}$

$\sqrt[3]{54}=\sqrt[3]{27\xb72}=\sqrt[3]{27}\sqrt[3]{2}=3\sqrt[3]{2}$

Even roots of negative numbers do not appear on the GRE. For example, you will not see expressions of the form √-4 ; expressions of this type are called complex numbers.

There are only two rules for roots that you need to know for the GRE:

$\sqrt[n]{xy}=\sqrt[n]{x}\sqrt[n]{y}$

**For example**,

$\sqrt{3x}$ = $\sqrt{3}$$\sqrt{x}$

$\sqrt[n]{\frac{x}{y}}$ = $\frac{\sqrt[n]{x}}{\sqrt[n]{y}}$

**For example**,

$\sqrt[3]{\frac{x}{8}}$ = $\frac{\sqrt[3]{x}}{\sqrt[3]{8}}$ = $\frac{\sqrt[3]{x}}{2}$

**Caution:** $\sqrt[n]{x+y}\ne \sqrt[n]{x}+\sqrt[n]{y}$

For example, $\sqrt{x+5}\ne \sqrt{x}+\sqrt{5}$ Also, $\sqrt{{x}^{2}+{y}^{2}}\ne x+y$ This common mistake occurs because it is similar to the following valid property: $\sqrt{{\left(x+y\right)}^{2}}=x+y$ (If *x + y* can be negative, then it must be written with the absolute value symbol: *|x + y|* ). Note, in the valid formula, it’s the whole term, *x + y*, that is squared, not the individual *x* and *y*.

To add two roots, both the index and the base must be the same. For example, $\sqrt[3]{2}+\sqrt[4]{2}$ cannot be added because the indices are different, nor can $\sqrt{2}+\sqrt{3}$ be added because the bases are different. However, $\sqrt[3]{2}+\sqrt[3]{2}=2\sqrt[3]{2}$ In this case, the roots can be added because both the indices and bases are the same. Sometimes radicals with different bases can actually be added once they have been simplified to look alike. For example,

$\sqrt{28}+\sqrt{7}=\sqrt{4\xb77}+\sqrt{7}=\sqrt{4}\sqrt{7}+\sqrt{7}$

$\mathrm{=\; 2}\sqrt{7}+\sqrt{7}=3\sqrt{7}$

You need to know the approximations of the following roots:

$\sqrt{2}\approx 1.4$

$\sqrt{5}\approx 2.2$

**Example 1:**

**Column A**

*x*

**Column B**

*y*

where x^{2} = 4

y^{3} = -8

*y*

^{3}= -8 yields one cube root,

*y*= –2. However

*, x*

^{2}= 4 yields two square roots,

*x*= ±2. Now, if

*x*= 2, then Column A is larger; but if

*x*= –2, then the columns are equal. This is a double case and the answer is (D).

**Example 2:**

If *x* < 0 and y is 5 more than the square of* x*, which one of the following expresses *x* in terms of *y*?

(A) $x=\sqrt{y-5}$

(B) $x=-\sqrt{y-5}$

(C) $x=\sqrt{y+5}$

(D) $x=\sqrt{{y}^{2}-5}$

(E) $x=-\sqrt{{y}^{2}-5}$

*y*is 5 more than the square of

*x*” into an equation yields:

$y={x}^{2}+5$

y – 5 = x

^{2}

$\pm \sqrt{y-5}=x$

Since we are given that

*x*< 0, we take the negative root, $-\sqrt{y-5}=x$

The answer is (B).