## Use transformation to square of sum and surd rationalization concepts to solve tough surd problems in a few simple steps

We have found students to feel Algebra problems a bit more problematic than the math problems on other topics. There are good reasons for this unease with Algebra, one being the abstraction and the other being ability to build layer after layer of mis-direction. But when Algebra is mixed up with Surds, the situation sometimes seems to be out of bounds.

You will be able to understand the solutions better or even to improve some of them if you go through the surd solving tutorials before going ahead,

**Basic and rich concepts on fractions and decimals part 1**

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

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

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

The concepts and methods explained in these tutorials should be enough to solve most of the difficult surd problem that you would encounter.

Coming to the point we will take up solving problems straightaway.

### Problem 1.

If $x = \displaystyle\frac{\sqrt{3}}{2}$, then the value of $\displaystyle\frac{\sqrt{1 + x}}{1 + \sqrt{1 + x}} + \displaystyle\frac{\sqrt{1 - x}}{1 - \sqrt{1 - x}}$ is,

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

**Solution:**

**First stage Problem analysis:**

Looking at the target expression we decide, unless we express $(1 + x)$ and $(1 - x)$ * as square of sums*, we won't be able to come out of the square root on these two main expressions and can't reach the solution in a long time to come.

By substituting the value of $x$ we get,

$\sqrt{1 + x} = \sqrt{1 + \displaystyle\frac{\sqrt{3}}{2}}$

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

**First stage input transformation:**

The target is more defined now. * We need to express the numerator of the surd sum under the root as a square* which at first glance seems to be not possible.

We remember our experience of dealing with such two-term surd expression where, multiplying the expression by 2 we made the expression a square in our **Algebra Solution Set 13.**

Taking the cue, we multiply the numerator and denominator both by 2,

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

$=\sqrt{\displaystyle\frac{(\sqrt{3} + 1)^2}{4}}$

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

This is our **main breakthough. **And it is a case of applying** Transformation to square of sum technique **in surds.

Similarly we get,

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

**Second stage:**

Putting these simple expressions of $\sqrt{1 + x}$ and $\sqrt{1 - x}$ in the target expression, we get,

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

Simplifying,

$E =\displaystyle\frac{\sqrt{3} + 1}{3 + \sqrt{3}} + \displaystyle\frac{\sqrt{3} - 1}{3 - \sqrt{3}}$

Factoring out $\sqrt{3}$ from the denominators of the two terms, turns each denominator factor equal to the numerator and it results in,

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

This is application of the useful * Simplification technique of factoring out* that helps to simplify surd expressions so many times.

**Answer:** Option c: $\displaystyle\frac{2}{\sqrt{3}}$.

#### Key concepts used:

- By
on the target end state expression and**End State Analysis**we decide that,**the free resources of choice values****transforming**both $(1+x)$ and $(1-x)$**into squares**to bring them out of the square roots in the target expression**is a must.** - Substituting the value of $x$ in $\sqrt{1 + x}$, the situation at first glance seems to be not favorable. Now we apply
**the important technique of transforming the surd term to twice the term by multiplying the expression by 2**. Objective is to use this transformed twice the original surd term as the middle term of a square of sum of two terms, that is, the $2ab$ in $(a + b)^2$. - Using intermediate
as soon as we transform the surd term, we recognize it to be a square of a two term surd expression. This is**input transformation technique****Pattern recognition skill**in action resulting in use of. Pattern recognition is one of the most important**Pattern recognition principle****problem solving skills**. *Now applying principle of induction, we say that the other expression $(1 -x)$ also will be simplified similarly except for the sign.*- We use the
, and substitute the two simple resulting expressions in the target expression and simplify.**Substitution technique** - In the last stage also we apply a special
to finally home in to a very simple result. This technique is very useful especially in surds problems.**Simplification by factoring out technique**

### Problem 2.

If $a = \sqrt{6} - \sqrt{5}$, $b = \sqrt{5} - 2$ and $c = 2 - \sqrt{3}$ then the correct comparative relationship between $a$, $b$ and $c$ is,

- $b \lt a \lt c$
- $a \lt b \lt c$
- $b \lt c \lt a$
- $a \lt c \lt b$

**Problem analysis **

This is a problem of surd expression comparison, where the two term surd expressions involve subtractions.

We need to analyze the expressions more closely, transform them if required and find useful patterns. At this point of time we are not clear about all the steps that we have to take. Our main objective now is to find useful significant patterns in the three equations and see how we can use the patterns.

#### First stage input transformation and pattern identification

In the three surd expressions there are three surd terms, $\sqrt{6}$, $\sqrt{5}$ and $\sqrt{3}$ and one integer $2$. But, if you look closely, $2$ can be expressed as $\sqrt{4}$. With this nugget of new knowledge, let us represent the values of $a$, $b$ and $c$ in a **more regular form** in terms of four surd terms,

$a = \sqrt{6} - \sqrt{5}$,

$b = \sqrt{5} - \sqrt{4}$,

$c = \sqrt{4} - \sqrt{3}$.

#### First Significant pattern in three equations

With this conversion now we notice the first significant pattern in the three equation as, $\sqrt{5}$ is common to first and second equation but it is being subtracted in the first and is subtracted from in the second equation - in opposite sides of the minus sign.

The same is true for the second and the third equations where the term $\sqrt{4}$ is common between the two.

#### How to use the significant pattern of commonality

Here we remember the * fundamental property of subtraction* - it is

*. That means, $(5 - 3) \neq (3 - 5)$.*

**not a Commutative operation**Because of this **inability to use the common term fruitfully for comparison of two expressions involving subtraction,** we identify the **need of similar complementary sum of terms involving addition**, where **addition is a commutative property.**

For example, let us consider two expressions,

$x + \sqrt{5}$, and

$\sqrt{5} + y$.

Because of **commutative property of addition** we can rewrite the first equation as,

$\sqrt{5} + x$.

Now **we can compare the two expressions and can say which one is larger depending on relative values of $x$ and $y$. **

For example, if $x \gt y$, the first expression involving $x$ will be greater than the second, $\sqrt{5}$ being common to both, and vice versa.

**So the next question inevitably is - ***can we transform the given surd expressions in subtraction of terms form to similar surd expressions in sum of terms form? If we could, we would then be able to easily compare all the three values with each other because of the commonality of terms between first and second and second and third expressions.*

#### Second significant pattern in each of the three two term surd expressions - Surd Rationalization property

The difference of squares of the terms in each of the two term expressions is 1.

We call this as * Surd rationalization property.* Whenever we observe this property in any two term surd expression, we may think of inversion and rationalization.

#### Input transformation using Surd rationalization property

Whenever a two term surd expression satisfies this property, if we invert and rationalize, the denominator turn to 1 and disppears leaving us with an addition complementary two term expression.

**For example,** if we invert $\sqrt{6} - \sqrt{5}$ and rationalize, it is transformed to $\sqrt{6} + \sqrt{5}$,

$a = \sqrt{6} - \sqrt{5}$,

Or, $\displaystyle\frac{1}{a} = \displaystyle\frac{1}{\sqrt{6} - \sqrt{5}}$

$= \displaystyle\frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2}$

$=\sqrt{6} + \sqrt{5}$.

Now if we invert and rationalize the other two equations we would get three sum of surds with same terms and will be able to compare all the three with each other, because **result of comparing inverted variables is equivalent to converse of comparing orginal variables.**

**Relationship between comparison of inverted variables and comparison of original variables - use of inequality property**

The basic property of comparison of inverted and original variables states,

If inverted variable $a$ is greater than inverted variable $b$, then $a \lt b$, the nature of comparison result is also inverted.

For example,

If $\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b}$, it follows from **inequality properties,** $a \lt b$. This is a **basic inequality property.**

**Final stage of inversion, rationalization and comparison**

So we invert the variables and rationalize each getting,

$\displaystyle\frac{1}{a} = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2}$

$=\sqrt{6} + \sqrt{5} $

Or, $\displaystyle\frac{1}{a}= 2\sqrt{3} + \sqrt{5}$,

$\displaystyle\frac{1}{b} = \frac{\sqrt{5} + \sqrt{4}}{(\sqrt{5})^2 - (\sqrt{4})^2}$

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

Or, $\displaystyle\frac{1}{b}= 2 + \sqrt{5}$,

$\displaystyle\frac{1}{c} = \frac{\sqrt{4} + \sqrt{3}}{(\sqrt{4})^2 - (\sqrt{3})^2}$

$=\sqrt{4} + \sqrt{3} $

Or, $\displaystyle\frac{1}{c}= 2 + \sqrt{3}$.

As $\sqrt{5} \gt \sqrt{3}$,

$\displaystyle\frac{1}{b} \gt \displaystyle\frac{1}{c}$.

Similarly as, $2\sqrt{3} \gt 2$,

$\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b}$.

Joining the two relations, we have a relationship of inverses,

$\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b} \gt \displaystyle\frac{1}{c}$.

The property of greater than or lesser than relationship or inequality relationship is,

If we invert the variables, the relationship also inverts, that is, greater than changes to lesser than and vice versa.

So we have our desired comparative relationship between $a$, $b$ and $c$ as,

$a \lt b \lt c$.

**Answer:** Option b: $a \lt b \lt c$.

#### Key concepts used:

- Noting the
of the difference of squares of each pair of terms as 1, we decided to apply**Surd rationalization property**but to do so we first inverted the variables.**Surd Rationalization technique,** - Simplifying the results of rationalization produced promising expressions.
- Comparing two pairs of expressions we could form clear comparative relation of values of the inverted variables.
- Joining the two comparative relations, a three term comparative relation between inversed variables is formed.
- As inverting the variables results in inversion of the relationship also according to
**inequality property,**we obtained our desired comparative relation between the three surd expressions.

**Note:** When surd expressions in subtraction of terms form are to be compared, we need to find ways to transform the expressions to sum of terms form generally by inversion and rationalization.

### Problem 3.

If $x=\displaystyle\frac{4\sqrt{15}}{\sqrt{5} + \sqrt{3}}$, then the value of $\displaystyle\frac{x + \sqrt{20}}{x - \sqrt{20}} + \displaystyle\frac{x + \sqrt{12}}{x - \sqrt{12}}$ is,

- $\sqrt{3}$
- 2
- 1
- $\sqrt{5}$

#### Solution: **First stage: input simplification by rationalization**

$x=\displaystyle\frac{4\sqrt{15}}{\sqrt{5} + \sqrt{3}} = 2\sqrt{15}(\sqrt{5} - \sqrt{3})$

$=10\sqrt{3} - 6\sqrt{5}$

#### Second stage: target expression simplification by using input expression

Putting this value of $x$ in the target expression,

$E = \displaystyle\frac{10\sqrt{3} - 6\sqrt{5} + 2\sqrt{5}}{10\sqrt{3} - 6\sqrt{5} - 2\sqrt{5}}$

$\hspace{10mm} + \displaystyle\frac{10\sqrt{3} - 6\sqrt{5} + 2\sqrt{3}}{10\sqrt{3} - 6\sqrt{5} - 2\sqrt{3}}$

$=\displaystyle\frac{10\sqrt{3} - 4\sqrt{5}}{10\sqrt{3} - 8\sqrt{5}} + \displaystyle\frac{12\sqrt{3} - 6\sqrt{5}}{8\sqrt{3} - 6\sqrt{5}}$

$=\displaystyle\frac{5\sqrt{3} - 2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}}$

$\hspace{10mm} + \displaystyle\frac{6\sqrt{3} - 3\sqrt{5}}{4\sqrt{3} - 3\sqrt{5}}$.

#### Third stage: applying Numerator simplification technique

In the first term we notice the term $5\sqrt{3}$ common between the numerator and denominator and in the second term also we find similar pattern of the term $3\sqrt{5}$ common between the numerator and denominator. Furthermore, the other term also is in same surds $\sqrt{5}$ and $\sqrt{3}$ respectively. This is **useful pattern identification.**

When we find such a pattern, the common term is eliminated by applying the ** numerator simplification technique, **more specifically, by subtracting 1 to the fraction term (or adding 1 to the fraction term when the common term is of opposite signs between the numerator and denominator) and compensating the subtraction by addition of another 1. As the second term also is in terms of the same surd, the numerator is simplified to a single term expression.

**Example**

To simplify, $\displaystyle\frac{x + 3p}{x + 2p}$, we subtract and add 1 to the fraction,

$\displaystyle\frac{x + 3p}{x + 2p} - 1 + 1 =\displaystyle\frac{p}{x + 2p} + 1$.

$= \displaystyle\frac{2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} + \displaystyle\frac{2\sqrt{3}}{4\sqrt{3} - 3\sqrt{5}} + 2$.

Applying the **numerator simplification technique** on the last result of target expression simplification,

$E=\displaystyle\frac{5\sqrt{3} - 2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} - 1 + 1$

$\hspace{10mm} + \displaystyle\frac{6\sqrt{3} - 3\sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} - 1 + 1$

$=\displaystyle\frac{2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} + \displaystyle\frac{2\sqrt{3}}{4\sqrt{3} - 3\sqrt{5}} + 2$

This is significant simplification, but the last barrier remains to be crossed.

#### Fourth stage: Simplification by factoring out technique

Again we examine the numerator and denominator closely and detect the possibility of taking out the term $\sqrt{5}$ as a factor from the first denominator and $\sqrt{3}$ as a factor from the second denominator, thus canceling the factors and simplifiying greatly,

$E=\displaystyle\frac{2}{\sqrt{15} - 4} + \displaystyle\frac{2}{4 - \sqrt{15}} + 2$

$=2$, a simple result.

**Answer:** Option b: 2.

#### Key concepts, techniques and steps:

- First step is
in as simple form as possible. We use**input transformation****surd rationalization technique.** - Second step is to substitute the value of $x$ in the large target expression and
**carry out the first level simplification.** - In the third step,
**noticing same term appearing in numerator and denominator,**we applyby subtracting 1 and adding 1**numerator simplification technique****to cancel the similar terms**from the numerator and**make it a single term expression.** - In the fourth step, we apply the
of each of the surd fraction terms and as a result the denominator and numerator of the two terms become same except for one minus sign. Thus the two terms cancel out leaving the plus 2 as the final result.**factoring out technique individually on the numerator and denominator**

#### Recommendation:

Even after applying so many techniques and strategies on this specific problem, it finally takes a bit more time than other problems.

In fact, the conventional solution is so long we have not presented it here.

If you are not habituated in applying the rich algebraic concepts and problem solving techniques, we would recommend, do not attempt this humongous problem in the final test, if at all it appears in the final test paper.

Most importantly though, the learning of techniques and strategies in solving such a problem is significant. At the least you can gain rich experience from the proces of solving such difficult problems during practice sessions. After all it is always true that,

The more difficult is the problem you solve the more things you learn.

The list of * Difficult algebra problem solving in a few steps quickly* is available at,

*.*

**Quick algebra**To go through the extended resource of **powerful concepts and methods** to solve difficult algebra problems, you may click on,