Exponents and roots are the basic topics that GRE tests from your grade school knowledge. So, one thing which is certain is that on the final day, you will see questions based on these topics. Hence, solving lots of GRE arithmetic practice sets on exponents and roots can prove to be highly beneficial.

Exponents and roots, though two different topics, yet are closely related to one another. The questions based on these topics are meant to trick you but if your concept is clear then surely you will get all the correct answers. An exponent along with root appears to be a bit tricky but if you know the loop-holes and cheat tricks, you will have the answers in split seconds. These will help you to easily master this section in the GRE Quantitative Reasoning section.

**What are exponents?**

The repeated multiplication of a number with itself is represented using exponents. For example; 7Ã—7Ã—7Ã—7Ã—7Ã—7Ã—7Ã—7= 78

Where, 7 is known as the base and 8 is called as exponent; hence, it is read as â€˜ 7 to the eighth power’ or â€˜7 raised to the eighth power’.

When the exponent is equal to two, it is known as squaring. If exponent is equal to three, then it is known as cube.

If a negative number is raised to n power, the result can be positive or negative depending on whether n is even or odd. If the exponent of a negative number is even then the resultant value will be positive, otherwise, it is negative.

**Negative Exponents or Zero**

Exponents can be negative or zero as well and the process of representing these exponents are:

\(\small p^{0} = 1\),where p is any non-zero number

If p=0, that is, 0

^{o}is an undefined number.

For all non-zero numbers with negative exponents, \(\smallÂ p^{-1} = \frac{1}{p} \; and \; p^{-n} = \left ( \frac{1}{p} \right )^{n}\)

It is important to note that, \(\small p \times p^{-1} = p \times \frac{1}{p} = 1\)

**So, what is a square root?**

The inverse of exponent is root. The square root of a non-negative number p is a number â€˜a’, such that p^{2}= a

For example, 2 and -2 are square roots of 4, 3 and -3 both are square roots of 9, -4 and 4 both are square roots of 16 and so on.

The symbol used for representing square root isâˆš, which is known as radical sign.

**Higher order roots**

The order of square root is 2. For higher order roots of any positive number, the representation is done in a similar way. Cube root is of order 3 whereas, fourth root is of order 4.

Let’s solve a question and brush up our concepts:

**Question:** \(\small (49 + 5\sqrt{2})^{\frac{1}{2}} + (49 – 5\sqrt{2})^{\frac{1}{2}}\)
**Solution: **

**step 1:** Square the given expression, so that roots can be removed. But remember to un-square the result in end to get the right answer.

**step 2:** Hence we get the expression as,Â \(\small (49 + 5\sqrt{2})^{\frac{1}{2}} + (49 – 5\sqrt{2})^{\frac{1}{2}} =

((49 +5\sqrt{2})^{\frac{1}{2}})^{2} + ((49 – 5\sqrt{2})^{\frac{1}{2}})^{2} + 2(49 + 5\sqrt{2})^{\frac{1}{2}}(49 – 5\sqrt{2})^{\frac{1}{2}}\)

**step 3:** \(\small 49 + 5\sqrt{2} + 2(49 + 5\sqrt{2})^{\frac{1}{2}}(49 – 5\sqrt{2})^{\frac{1}{2}} + 49 – 5\sqrt{2}\)

**step 3:** \(\small 98 + 2((49 + 5\sqrt{2})(49 – 5\sqrt{2}))^{\frac{1}{2}}\)

**step 4:** \(\small 98 + 2(49^{2} – (5\sqrt{2})^{2})^{\frac{1}{2}}\)

**step 5:** \(\small 98 + 2 (2401 – 50)^{\frac{1}{2}}\)

**step 6:** \(\small 98 + 2 \sqrt{2351}\)

**step 7:**We need to un-square it, so, answer will beÂ \(\small \sqrt{98 + 2\sqrt{2351}}\)

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