How to solve Surds part 2, double square root surds and surd term factoring | SureSolv

How to solve Surds part 2, double square root surds and surd term factoring

Unless you transform a two-term surd expression under a square root, you can't solve the surd problem

how-to-solve-surds-part-2-double-square-root-surds

We have given a brief introduction to surds in our article Fractions and decimals part 1and How to solve surds 1, Rationalization. You may like to go through these before going ahead further in this session.

In this session we will learn about three additional surd techniques to solve surd problems,

  1. Double square root surds: How to solve surd problems, where a two-term surd expression appears under a second square root. Unless we free up the surd expression under square root we can't deal in any way with the double square root situation—use of conversion to square of sum surd expression.
  2. Surd term factoring: How to solve surd problems by taking a factor out of a surd term
  3. Surd coefficient comparison: How to solve a surd problem by comparing and equalizing coefficients of surd and non-surd terms on two sides of an equation—use of Coefficient equalization of similar variables.

Through solving of chosen SSC CGL level problems we will show how to apply these three different techniques to solve apparently difficult surd problems effectively.

Double square root surd expression and Conversion to square of sum surd expression technique

Three examples of double square root surd expressions are,

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

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

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

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

The obvious and simple way to do this is,

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

Sometimes it is easy, other times transforming to a square of sum a bit tricky.

Let us convert the three examples.

Example 1.

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

$=\sqrt{7 + 2\times{\sqrt{7}}\times{\sqrt{3}} + 3}$

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

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

The surd under square root is now free of enclosing square root but in changed form.

We have used the algebraic expression,

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

and attempted to determine the values of $a$ and $b$ in the surd expression so that the two term expression under square root can be converted to a square of two term surd expression.

Coefficient 2 identifies the middle term: The main help is provided by the middle term of the form of $2ab$. Isolating the coefficient 2, the rest should be broken up into two factors, both or one of which should be a surd, so that after squaring each of these and summing up, result becomes the second numeric term of the original two term surd expression.

In this case, middle term without coefficient 2 is $\sqrt{21}$. It consists of a unique pair of two surd factors, $\sqrt{7}$ and $\sqrt{3}$. After squaring each and taking their sum, we get the result as 10 as expected.

This is is easy to transform.

Example 2

The second example is,

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

$=\sqrt{2 +2\times{\sqrt{2}}\times{1}+1}$

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

This is also easy.

Let us take the third example,

Example 3.

The third example is,

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

This example 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 we get 9, satisfying the given expression.

So,

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

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

We follow the convention of writing the larger term of the surd expression first.

So with a little bit of practice you should be able to deal with such double square root surd expressions.

Notice that the method in all three cases depends on isolating the coefficient 2 of the middle term. What do we do if such a coefficient of 2 is not present in the two term surd expression under square root!

Example 4 of middle term of a double square root surd expression without factor 2

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

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

How should we eliminate the enclosing square root in this case?

The method follows from two step deductive reasoning,

  1. If the surd under root is to be freed of square root, it must be expressed as $a^2 +2ab +b^2$.
  2. If 2 is absent in any term, we must supply 2 artificially by multiplying and dividing the surd expression under square root itself by 2, and then on start analyzing the expression.

Let us apply this new method to the fourth double square root surd expression.

Our expression was,

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

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

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

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

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

Only this type of expression is a little tricky, but with practice, dealing with any form should become easy.

Before taking up application of these methods in solving actual problems let us show in what form of surd expressions, Surd factoring technique can be applied.

Surd term factoring technique

Examples

The following two examples of surd expressions might appear normal expressions with no other action on them possible.

$2 + \sqrt{6}$

$3 + \sqrt{15}$.

If you look closely you will find the surd term under the 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,

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

We have taken out 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, in this case, $(\sqrt{3}-\sqrt{2})$ appears as another factor, so that the expression immediately gets simplified to a great extent.

Similarly in the second example,

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

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

With this background, let us now solve a few SSC CGL level surd problems where double square root surd expressions appear as well as we need to use surd term factoring technique.

Solving double square root surd problems and use Surd term factoring when needed

Problem example 1.

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

  1. $\sqrt{5}-1$
  2. $1+\sqrt{5}$
  3. $2\sqrt{2}$
  4. $\sqrt{2}$

Solution 1. Problem example 1.

Let us first simplify the subtractive sum of inverses,

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

$=\displaystyle\frac{4+2\sqrt{6}}{\sqrt{5+2\sqrt{6}}}$

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

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

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: $2\sqrt{2}$.

We 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 case 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 is called Surd term factoring which many times makes simplification easy. In this example, need to rationalize the denominator didn't arise at all.

Problem example 2.

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

  1. $\sqrt{6}-\sqrt{2}$
  2. $2-\sqrt{6}$
  3. $\sqrt{6}-2$
  4. $\sqrt{6}+\sqrt{2}$

Solution 2. Problem example 2.

First we need to simplify the three factor expression and only then think of double square root technique if the situation demands.

Simplifying first factor in the numerator by Surd term factoring,

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

Multiplying this with the second factor using $(a+b)(a-b)=a^2-b^2$, we get in the numerator, the result of just $2$.

Now we need to give attention to the denominator expression. Taking 4 out of square root term 24,

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

So the given expression is transformed to,

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

Now we need to rationalize the denominator,

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

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

To solve this problem, we needed to use all three surd problem solving techniques we have learned till now,

  1. Surd rationalization
  2. Conversion to square of sum surd expression, and
  3. Surd term factoring.

Though the third technique seems to be lightweight in comparison to the other two, sometimes ability to see the factoring possibility becomes crucial. That was the case in the first problem as well as in the second problem.

Let us now solve a third SSC CGL level surd problem.

Problem example 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

  1. 3
  2. 0
  3. 1
  4. 2

Solution 3: Problem analysis and solving execution

Whenever we meet a surd expression under a square root, we know the surd expression under the square root must be converted to a square of sum. This problem has three such expressions. In each we take out first the factors from the surd term.

Taking these one by one,

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

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

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

Surd sums in the brackets are then the three denominators (after taking out of the square roots). We rationalize to form the simplified expression,

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

$=0$, all terms cancel out.

Answer: Option b: 0.

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

  1. Surd term factoring
  2. Conversion to square of sum surd expression, and
  3. Surd rationalization.

We will now end this session with application of the fourth surd solving technique, namely, Surd coefficient comparison and equalization of similar variables between two sides of an equation.

Coefficient equalization of similar variables

Problem example 4.

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

  1. $-13$
  2. $3\sqrt{3}-7$
  3. $13$
  4. $\sqrt{13}$

Solution 4: Problem analysis and solving execution

First we convert the denominator surd expression under square root to a square of sum,

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

Rationalizing the transformed denominator $2+\sqrt{3}$, we 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 surd term on the LHS (irrational and rational can't be added, though we represent an addition we can't arrive at any result by carrying out the addition).

This is what we call Coefficient equalization of similar variables that don't mix together. This is a fundamental algebraic principle.

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 we needed three methods,

  1. First, conversion to square of sum surd expression,
  2. Second, surd rationalization, and
  3. Third, Coefficient equalization of similar variables.

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

In the next tutorial session on how to solve surds, we will take up the more involved problems of surd expression comparison 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 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