## Surd problems are not so difficult to solve

We have given a brief introduction to surds in our article * Fractions and decimals part 1*. You may like to go through it.

In this article, * we will concentrate mostly on how to solve surd problems* with examples taken at the level of SSC CGL. We will cover the most frequently used simplification of surd problems by surd term rationalization.

### What surds are

Surds are a group of irrational numbers that are expressed as,

$\sqrt{\text{Product of Prime number factors}}$

**Examples of surds** are,

$\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ and so on.

There can be more than one prime factors within square roots, but for the surd property to be retained, **there must be at least one prime factor within the square roots that is unequal to any other factor.**

For example, $\sqrt{12}$ is a surd with three factors 2, 2 and 3, only one of which 3 is a prime unequal to the other two factors of 2 within the square root.

In $\sqrt{15}$, two unequal factors are both prime.

**Irrational numbers** are a large set of numbers that have three distinctive properties,

- An irrational number must have a decimal part that is non-terminating and non-repeating. An example is, 0.01001000100001.....
- An irrational number cannot be expressed as a fraction of two integer numbers $\displaystyle\frac{p}{q}$, where $p$ and $q$ are integers, while any rational number can be expressed as a fraction.
- Normal arithmetic operations cannot be carried out on irrational numbers.

In spite of these general difficulties of math on irrational numbers, its subset surds have gained prominence in school math and competitive math because of three reasons,

- Surds are expressed in compact form and involve only a square root. If squared, the difficult surd is no more.
and possible between two surds, because the operations take place within the square root.**Two of the basic arithmetic operations, namely multiplication and division are allowed**- A
make surds a popular branch for judging problem solving ability through math problem solving.**number of quick problem solving special properties or methods to simplify surd expressions**

Using suitable examples in this session we will explain how to solve surd problems easily and quickly. We will cover the most basic method for surd problem solving, namely eliminating surd denominator by surd rationalization.

### Multiplication, Division and Indices on surds

An example of **surd multiplication** is,

$\sqrt{3}\times{\sqrt{7}}=\sqrt{21}$.

The multiplication happens within the square root, resulting in the product 21, but within square root and the * result remains a surd*.

An example of **surd division** is,

$\displaystyle\frac{\sqrt{2}}{\sqrt{5}}=\sqrt{0.4}$, which is again a surd.

Though after the operations, number of terms reduces, the result still remains a surd.

**Indices on surds** is just an extension of existing indices concepts because, surd term has a power of $\displaystyle\frac{1}{2}$,

$\left(\sqrt{3}\right)^3=3^{\frac{3}{2}}$, still a surd.

**Surd comparison** is an important class of surd problems that we will deal with in details in a later session.

### Surd rationalization by eliminating surd suitably

A large number of surd problems involve at least one surd fraction term in a surd expression. Following is an example,

#### Problem example 1.

$\displaystyle\frac{1}{\sqrt{2}+1}+3-\sqrt{2}$.

As such, by normal means simplification seems to be difficult and awkward. In this general situation,

We resort to eliminate the surd in the denominator.

This is the general problem solving strategy of any simplification problem involving complicated fraction terms. The general approach towards such simplification is,

Simplify, and if possible eliminate the denominator of a fraction term completely.

#### How to simplify surd expression such as $\displaystyle\frac{1}{\sqrt{2}+1}$

The * first important method of surd simplification* rely on the algebraic expression,

$(a+b)(a-b)=a^2-b^2$.

As the individual terms gets squared, surds terms are eliminated from the result.

Using this method on the expression, we * multiply and divide the fraction term by the surd expression complementary to the denominator*, in this case by $\sqrt{2}-1$.

**Note:** While deciding on this multiplying factor, care is taken to keep the larger term positive. For example we don't use $1 - \sqrt{2}$, but use generally $\sqrt{2}-1$ as $\sqrt{2} \gt 1$ and will ensure a positive result unless otherwise affected.

Thus,

$\displaystyle\frac{1}{\sqrt{2}+1}$

$=\displaystyle\frac{1}{\sqrt{2}+1}\times{\displaystyle\frac{\sqrt{2}-1}{\sqrt{2}-1}}$

$=\displaystyle\frac{\sqrt{2}-1}{2-1}$

$=\sqrt{2}-1$.

This is the * complementary form of the surd fraction* and is the most frequently used expression form in surd problems.

Thus, knowing that the surd fraction can be replaced by its complementary surd expression, we carry out this part automatically in mind. This process we call as Surd rationalization and we consider it as a *Quick math problem solving method as well as a part of Mental maths*.

With awareness about how we can eliminate a surd denominator expression, let us continue solving Problem example 1.

#### Solution 1. Problem example 1.

$\displaystyle\frac{1}{\sqrt{2}+1}+3-\sqrt{2}$

$=\sqrt{2}-1+3-\sqrt{2}$

$=2$.

A simple result because after rationalization of denominator, the surd created in the numerator cancels out the already existing surd in the second term.

We will show a more extensive use of this method in a second problem example taken at the SSC CGL level.

#### Problem example 2.

Value of $\displaystyle\frac{1}{3-\sqrt{8}}-\displaystyle\frac{1}{\sqrt{8}-\sqrt{7}}+\displaystyle\frac{1}{\sqrt{7}-\sqrt{6}}$

$\hspace{30mm}-\displaystyle\frac{1}{\sqrt{6}-\sqrt{5}}+\displaystyle\frac{1}{\sqrt{5}-2}$ is,

- 2
- 3
- 4
- 5

#### Solution 2. Problem example 2.

Each of the denominator has one unique property in addition to being a two term subtractive surd expression—the difference between the two terms, when squared, is 1.

This is a very convenient situation, as after we rationalize a denominator, say, $\sqrt{p}-\sqrt{q}$, by multiplying both numerator and denomintor by its complementary, $\sqrt{p}+\sqrt{q}$, the denominator is transformed to,

$\left(\sqrt{p}\right)^2 - \left(\sqrt{q}\right)^2=p-q=1$.

Simply speaking each denominator will be eliminated.

For convenience of mental processing we will always assume $3=\sqrt{9}$ and $2=\sqrt{4}$—a uniform look aids mental processing.

What will happen to the numerators?

Let us jot down the expression of numerators,

$(\sqrt{9}+\sqrt{8}) - (\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})$

$\hspace{30mm}-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+\sqrt{4})$

All terms except the first and the last are canceled out.

So final result is,

$\sqrt{9}+\sqrt{4}=5$.

**Answer:** Option d: 5.

This is the most frequently used approach to simplify a surd expression.

We will explain other ways to solve surd problems in later such sessions.

### Concept tutorials on Fractions, Surds, decimals and related topics

**Fractions and Surds concepts part 1**

**Base equalization technique is indispensable for solving fraction problems**

**How to solve surds part 1, Rationalization**

**How to solve surds part 2, Double square root surds and surd term factoring**

**How to solve surds part 3, Surd expression comparison and ranking**

#### Question sets and Solution Sets in Fractions and related topics

**SSC CGL level Solution set 75 on fractions decimals indices 7**

**SSC CGL level Question set 75 on fractions decimals indices 7**

**SSC CGL level Solution Set 73 on Surds and Indices 7**

**SSC CGL level Question Set 73 on Surds and Indices 7**

**SSC CGL level Solution Set 70 on fractions and surds 6**

**SSC CGL level Question Set 70 on fractions and surds 6**

**SSC CGL level Question Set 61 on fractions indices surds 5**

**SSC CGL level Solution Set 61 on fractions indices surds 5**

**SSC CGL level Question Set 60 on fractions indices surds 4**

**SSC CGL level Solution Set 60 on fractions indices surds 4**

**SSC CGL level Question Set 59 on fractions square roots and surds 3**

**SSC CGL level Solution Set 59 on fractions square roots and surds 3**

**SSC CGL level Question Set 47 on fractions decimals and surds 2**

**SSC CGL level Solution Set 47 on fractions decimals and surds 2**

**SSC CGL level Question Set 17 on fractions decimals and surds 1**

**SSC CGL Level Solution set 17 on fractions decimals and surds 1**