## Unless you simplify a square root of surds expression, you won't be able to solve the surd problem

In this *second concept session on how to solve surds*, you will learn *three important surds simplification techniques,*

**Square root of surds:**How to simplify square root of surds that are actually surds expressions under a second square root. You have to free the inner surd expression by transforming it to a whole square surd expression.**Surd term factoring:**How to simplify surds expressions by taking a surd factor out of a term in a two term surd expression. Example: $3+\sqrt{3}=\sqrt{3}(\sqrt{3}+1)$, and,**Surd coefficient comparison:**How to solve a surds equation by comparing coefficients of the surd and non-surd terms on two sides of the equation.

Suitable surds example problems and test level surd problems are used for showing how the surd simplification techniques work. These techniques are indispensable for solving surds problems in any relevant competitive exam such as SSC CGL Tier II.

**Desirable prerequisite:** Knowledge on *surds rationalization technique*. If you don't know about it, read through our article: * How to solve surds 1, Rationalization*.

### How to convert a square root of surds to a whole square surd expression

Square root of surds appear in two forms. First three examples of first type are easier to simplify. These are,

$\sqrt{10+2\sqrt{21}}$

$\sqrt{3+2\sqrt{2}}$

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

In many surd simplification problems such square root surds appear. Unless you free up the surd expression from the surrounding square root there is no way you can proceed further.

The obvious way to do this is,

To express the two term surd expression under square root as a square of another two-term surd expression.

#### Example problem 1.

Simplify, $\sqrt{10+2\sqrt{21}}$.

**Solution:**

$\sqrt{10+2\sqrt{21}}$

$=\sqrt{(\sqrt{7})^2 + 2\times{\sqrt{7}}\times{\sqrt{3}} + (\sqrt{3})^2}$

$=\sqrt{(\sqrt{7}+\sqrt{3})^2}$

$=\sqrt{7}+\sqrt{3}$.

The surd under square root is free of enclosing square root and can easily be operated in simplifying the larger problem in which it is a part.

In the three term expanded form of $(a+b)^2=a^2 + 2ab+b^2$ equivalent to $\sqrt{10+2\sqrt{21}}$, values of $a$ and $b$ were determined so that the surds expression under square root could be converted to a whole square.

#### The Key pattern

The surd term $2\sqrt{21}$ in $10+2\sqrt{21}$ must be the middle term in its equivalent whole square expression of the form,

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

**Note:** *Why must the surd term be the middle term?* Simply because, squares of surds and their sum are rational numbers leaving the middle term of the three term expansion, $a^2 + 2ab+b^2=(a+b)^2$ as the only surd term.

Breakthrough is provided by the middle term of the form of $2ab$.

**Technique:** Ignoring the coefficient 2, break up rest of the surd middle term into two factors. At least one of these two will be a surd so that sum of squares of the two equals the rational numeric term of the original two term surd expression under square root.

In this case, middle term excluding coefficient 2 is $\sqrt{21}$.

It has two surd factors, $\sqrt{7}$ and $\sqrt{3}$ and sum of squares of the two is 10 as expected.

This is easy.

#### Example problem 2

The second example problem is also easy.

Simplify, $\sqrt{3+2\sqrt{2}}$.

**Solution:**

$\sqrt{3+2\sqrt{2}}$

$=\sqrt{(\sqrt{2})^2 +2\times{\sqrt{2}}\times{1}+1^2}$

$=\sqrt{(\sqrt{2}+1)^2}$.

Let us simplify a third surd example problem of same type.

#### Example problem 3.

Simplify, $\sqrt{9+4\sqrt{5}}$.

**Solution:**

This example problem is a little different. The middle term coefficient after excluding 2 is $2\sqrt{5}$. So one term must be 2 and the second $\sqrt{5}$, squaring each and adding you get 9, satisfying the given expression.

So,

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

$=\sqrt{(\sqrt{5})^2 + 2\times{\sqrt{5}}\times{2} + 2^2}$.

$=\sqrt{(\sqrt{5}+2)^2}$.

**Note:** *Here the the convention of writing the larger term of the surd expression first is followed.*

With a little bit of practice you should be able to simplify such square of surds or double square root surd expressions without any difficulty.

Observe that the method in all three cases depends on stripping off a coefficient 2 from the surd middle term.

What do we do if such a coefficient of 2 doesn't exist at all in the surd middle term of the two term surd expression under square root!

#### Example problem 4: Surd middle term of a square root of surd expression without factor 2

Let us take a very common example of a two term surd expression $(2 + \sqrt{3})$. This is a frequently occurring surd expression. Sometimes it is put under a square root,

$\sqrt{2+\sqrt{3}}$.

How would you eliminate the enclosing square root in this case?

**The method follows from two step deductive reasoning,**

- If the surd expression under root is to be freed of square root, it must be expressed as $a^2 +2ab +b^2$.
- If
**2 is absent in the surd middle term**, we must introduce 2 artificially by multiplying and dividing the surd expression under square root itself by 2, and then try to express the numerator surd expression as a whole square.

Let us apply the new method to the fourth example problem.

The surd expression to be simplified is,

$\sqrt{2+\sqrt{3}}$.

Multiplying and dividing the surd expression under square root by 2 you get,

$\displaystyle\frac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}$

$=\displaystyle\frac{1}{\sqrt{2}}\sqrt{(\sqrt{3}+1)^2}$

$=\displaystyle\frac{1}{\sqrt{2}}(\sqrt{3}+1)$.

Simplifying only this type of expression is a little tricky, but with practice and right techniques, simplifying any form of square root of surds should be easy.

Now you will learn how Surd term factoring technique is applied.

### Surd term factoring technique

#### Examples

Any of the following two form of surd expressions may appear in surd problems with no other action on them possible,

$2 + \sqrt{6}$

$3 + \sqrt{15}$.

If you look closely, you will find each of the integers under square root consists of two prime factors. If one of these two factors also matches with the second numeric term, there is a strong case of surd term factoring.

Let us see how.

The first example,

$2+\sqrt{6}=\sqrt{2}(\sqrt{3}+\sqrt{2})$.

You have taken the factor $\sqrt{2}$ not only out of the surd term, but

also out of the numeric term.

Often you will find that the transformed surd expression either appears both in numerator and denominator or its complementary surd expression thus simplifying the problem to a great extent.

In the second example,

$3+\sqrt{15}=\sqrt{3}(\sqrt{5}+\sqrt{3})$.

In both these cases you have taken the common factor out of both the two terms in the surd expression.

Sometimes you take a factor out of the single surd term. Example,

$5+\sqrt{24}=5+2\sqrt{6}=(\sqrt{3}+\sqrt{2})^2$.

By taking out factor 4 from 24 under square root, the expression is so transformed that it could be *expressed as a square of sum expression.*

Occasionally you might be able to identify a common surd term hidden in the numeric term. An example is,

$3+\sqrt{3}=\sqrt{3}(\sqrt{3}+1)$

With this background, let us now solve a few actual test level surd problems where square root surds expressions appear as well as you need to use surd term factoring technique.

In the fourth problem you will learn how to apply the *coefficient comparison and equalization technique* to solve a surds equation.

### Test level problems: Simplifying square root of surds, surd term factoring and coefficient comparison technique

#### Problem 1.

The value of $\sqrt{5+2\sqrt{6}} - \displaystyle\frac{1}{\sqrt{5+2\sqrt{6}}}$ is,

- $\sqrt{5}-1$
- $1+\sqrt{5}$
- $2\sqrt{2}$
- $\sqrt{2}$

#### Solution 1. Problem 1.

Let us first simplify the subtractive sum of inverses,

$E=\sqrt{5+2\sqrt{6}} - \displaystyle\frac{1}{\sqrt{5+2\sqrt{6}}}$

$=\displaystyle\frac{4+2\sqrt{6}}{\sqrt{5+2\sqrt{6}}}$, simply combining the two terms,

$=\displaystyle\frac{4+2\sqrt{6}}{\sqrt{(\sqrt{3}+\sqrt{2})^2}}$

$=\displaystyle\frac{2(2+\sqrt{6})}{\sqrt{3}+\sqrt{2}}$.

Taking factor $\sqrt{2}$ out of the two terms in the numerator,

$E=\displaystyle\frac{2\sqrt{2}(\sqrt{3}+\sqrt{2})}{\sqrt{3}+\sqrt{2}}$

$=2\sqrt{2}$.

**Answer:** Option c: $2\sqrt{2}$.

You have not only taken the surd expression in the denominator out of its enclosing square roots, but also identified the pattern of convenient surd factoring possibility on the numerator by taking the factor $\sqrt{2}$ both out of $2$ and $\sqrt{6}$.

**This type of factorization is not readily visible** and a closer analytical look is required to identify the possibility. This technique of * Surd term factoring *makes many times simplification very easy.

**In this solution, surd rationalization is not used at all.**

#### Alternate solution: First simplifying square root of surds and then applying surd rationalization

The same problem you will solve now by simplifying both the instances of square root of surds first,

$E=\sqrt{5+2\sqrt{6}} - \displaystyle\frac{1}{\sqrt{5+2\sqrt{6}}}$

$=\sqrt{(\sqrt{3}+\sqrt{2})^2}-\displaystyle\frac{1}{\sqrt{(\sqrt{3}+\sqrt{2})^2}}$, simplifying both square root of surds,

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

$=(\sqrt{3}+\sqrt{2})-(\sqrt{3}-\sqrt{2})$, multiplying and dividing second term with $(\sqrt{3}-\sqrt{2})$, surd rationalization,

$=2\sqrt{2}$.

This solution takes fewer number of steps and is faster.

A difficult surd problem can often be solved in more than one way. You have to adopt the simplest path.

#### Problem 2.

The value of $\sqrt{\displaystyle\frac{(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2})}{5+\sqrt{24}}}$ is,

- $\sqrt{6}-\sqrt{2}$
- $2-\sqrt{6}$
- $\sqrt{6}-2$
- $\sqrt{6}+\sqrt{2}$

#### Solution 2. Problem 2.

The two factors, $(\sqrt{12}-\sqrt{8})$ in the numerator and $(5+\sqrt{24})$ in the denominator catch your attention first. Having terms rich with factors, these two surd expressions need immediate simplification by surd term factoring.

Follow the never-failing strategy,

While solving any difficult surd problem,

first simplifythe surd expressions on which surd term factoring technique can easily be applied.

Simplifying first factor in the numerator by Surd term factoring,

$\sqrt{12}-\sqrt{8}=2(\sqrt{3}-\sqrt{2})$.

Multiply this result with the second factor using $(a+b)(a-b)=a^2-b^2$, and you get in the numerator just $2$.

Now turn your attention to the denominator expression.

Again applying surd term factoring, take 4 out of square root term 24,

$5+\sqrt{24}=5+2\sqrt{6}=(\sqrt{3}+\sqrt{2})^2$, applying square of surds simplification.

The given expression is simplified to,

$E=\displaystyle\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.

Now rationalize the denominator by multiplying and dividing with $(\sqrt{3}-\sqrt{2})$,

$E=\sqrt{2}(\sqrt{3}-\sqrt{2})=\sqrt{6}-2$.

**Answer.** Option c: $\sqrt{6}-2$.

To solve this problem, you have applied all three surd problem solving techniques you have learned till now,

- Surd rationalization
- Simplifying square of surds, and
- Surd term factoring.

Though the third technique seems to be lightweight compared to the other two, *ability to see the possibility of surd term factoring often becomes crucial to arrive at the solution.*

Let us now solve the third test level difficult surd problem.

#### Problem** 3.**

The value of $\displaystyle\frac{1}{\sqrt{12-\sqrt{140}}}-\displaystyle\frac{1}{\sqrt{8-\sqrt{60}}}-\displaystyle\frac{2}{\sqrt{10+\sqrt{84}}}$ is,

- 3
- 0
- 1
- 2

#### Solution 3: Problem analysis and solving execution

Whenever you meet a surd expression under a square root, you know for sure the surd expression under the square root must be converted to a square of sum. This problem has three such expressions. In each take out first the factors from each of the surd terms applying surd term factoring technique (and as per the stratety just stated).

Applying surd term factoring first and square of surds simplification technique next, on the three denominators you get,

$12 - \sqrt{140}=12-2\sqrt{35}=(\sqrt{7}-\sqrt{5})^2$,

$8-\sqrt{60}=8-2\sqrt{15}=(\sqrt{5}-\sqrt{3})^2$, and

$10 +\sqrt{84}=10+2\sqrt{21}=(\sqrt{7}+\sqrt{3})^2$.

Sum of surds in the brackets are then the three denominators (after taking out of the square roots).

Now rationalize to form the simplified expression eliminating the denominators in the process,

$\displaystyle\frac{1}{\sqrt{12-\sqrt{140}}}-\displaystyle\frac{1}{\sqrt{8-\sqrt{60}}}-\displaystyle\frac{2}{\sqrt{10+\sqrt{84}}}$

$=\displaystyle\frac{1}{\sqrt{7}-\sqrt{5}}-\displaystyle\frac{1}{\sqrt{5}-\sqrt{3}}-\displaystyle\frac{2}{\sqrt{7}+\sqrt{3}}$

$=\displaystyle\frac{\sqrt{7}+\sqrt{5}}{2}-\displaystyle\frac{\sqrt{5}+\sqrt{3}}{2}-\displaystyle\frac{\sqrt{7}-\sqrt{3}}{2}$

$=0$, all terms cancel out.

**Answer:** Option b: 0.

To solve this third problem also you have applied all the three techniques you have learned,

- Surd term factoring
- Simplification of square of surds, and
- Surd rationalization.

Now you will learn how to apply the fourth surd problem solving technique of * Surd coefficient comparison and equalization of similar variables* between two sides of an equation.

### Coefficient comparison and equalization for similar variables

#### Problem** 4.**

If $\displaystyle\frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}=A + \sqrt{B}$ then $B-A$ is,

- $-13$
- $3\sqrt{3}-7$
- $13$
- $\sqrt{13}$

#### Solution** 4: Problem analysis and solving execution**

First convert the denominator surd expression under square root to a square of sum on the way to simplifying the square of surds,

$7 + 4\sqrt{3}=(2+\sqrt{3})^2$.

Rationalizing the transformed denominator $2+\sqrt{3}$, you get the given expession as,

$(4+3\sqrt{3})(2-\sqrt{3})=A + \sqrt{B}$,

Or, $-1+2\sqrt{3}=A+\sqrt{B}$.

As $\sqrt{B}$ is the surd term it must be equal to the SIMILAR irrational surd term on the LHS. Similarly rational term $A$ must also be equal to $-1$.

This is because,

A surd term being an irrational number with a non-terminating non-repeating decimal component, it cannot be added numerically to a rational number giving a result that you can express with certainty.

**Note:** Surd form is $\sqrt{n}$ where $n$ has at least one integer factor that is not a square. Numerically it always has component of a *non-terminating non-repeating decimal component* and cannot expressed as a fraction of the rational number form, $\displaystyle\frac{p}{q}$, where both $p$ and $q$ are integers.

This is what we call * Coefficient comparison and equalization for similar variables* that don't mix together. It follows a

**fundamental algebraic principle,**In an equation, coefficients of similar type of variables on both sides of the equation must be equal.

Thus,

$A=-1$, and

$2\sqrt{3}=\sqrt{B}$,

Or, $B=12$, and

$B-A=13$.

**Answer:** Option c: 13.

To solve this last problem also you needed three methods,

- First, conversion to square of sum surd expression using simplifying square of surds technique
- Second, surd rationalization, and
- Third, Coefficient comparison and equalization for similar variables.

By applying these three methods along with Surd term factoring whenever needed, most of the difficult surd problems can be solved.

You would be surprised to know that,

For solving difficult problems in Algebra as well as Trigonometry, you would have to apply these surd simplification techniques you have just learned.

In the next concept tutorial session on how to solve surds, you will learn how to solve the more involved problems of **Surd expression comparison and ranking.**

### 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, Surds, Indices and related topics

**SSC CGL level Solved Question set 93 on Surds and Indices 10**

**SSC CGL level Solved Question set 92 on Surds and Indices 9**

**SSC CGL level Solved Question set 91 on Surds and Indices 8**

**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**