## Enhanced input transformation and end state analysis solve the problem in a few simple steps

The problems involving Algebra are taught in schools and also form an important part of most of the competitive job tests such as SSC CGL, Bank POs etc.

We'll not repeat the usual **reasons behind the difficulties** faced by students in solving Algebra problems and the **set of basic **and **rich concepts** that are **invaluable** in reaching **elegant solutions** for seemingly difficult Algebra problems.

In this session **in the first problem,**

We have appliedenhanced input transformation techniquealong withEnd state analysisto find a new elegant path to the solution.

And in the second problem,

We have shown how

, can be effectively applied.Many ways technique that improves problem solving skills

For comparison, we will not present any commonly encountered long winded laborious solutions from this point on.

You may watch the video below.

**For best results,** you should attempt to solve each of these problems and measure your time and number of steps to the solution before you go ahead with the explanation of the solution.

### Problem 1.

If $x = \sqrt{\displaystyle\frac{\sqrt{5} + 1}{\sqrt{5} - 1}}$, find $5x^2 - 5x -1$.

- 4
- 3
- 0
- 5

**Solution: ****First stage Problem analysis:**

As the given input is under double square root, it is our first job to rescue the surd expression involved from its prison of the outer square root. Once we free up the surd expression from its square root blockade, we will be able to apply further simplification techniques on it.

#### First stage action: freeing the surd expression in input from its outer square root

Examining the expression under square root, having enough experience in dealing with surds, the way forward comes easy to us, we need to apply * Surd rationalization technique.* The denominator will turn to, $a^2 - b^2 = 5 - 1 = 5$ and the numerator, $(a + b)^2 = (\sqrt{5} + 1)^2$,

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

*Upto this point it is all routine and conventional.* No new approach was visible. But now there are at least two ways to the solution. You may try to find out yourself.

#### Second stage first approach - conventional

$x = \displaystyle\frac{(\sqrt{5} + 1)}{2}$,

Or, $x^2 = \displaystyle\frac{(\sqrt{5} + 1)^2}{4}=\displaystyle\frac{6 + 2\sqrt{5}}{4} = \displaystyle\frac{3 + \sqrt{5}}{2}$.

Substituting the values of $x^2$ and $x$ in target expression,

$5x^2 - 5x -1$

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

$=\displaystyle\frac{1}{2}\times{\left(15 + 5\sqrt{5} - 5\sqrt{5} - 5 - 2\right)}$

$=\displaystyle\frac{8}{2}$

$=4$.

This is the way we are used to solve our math problems, not much thinking, safe and routine procedures.

Usually this conventional procedural way is longer, time consuming, boring with no scope of creativity and excitement and error-prone.

#### Elegant solution

Especially these MCQ test problems have ample scope for finding hidden elegant solutions because of a strong pressure to solve a problem in a minute's time. Without the elegant solution, this may not be possible. When you follow the problem solver's path or the elegant path, you would usually do all your reasoning and minor calculations in your mind rather than on paper.

This way is much faster and less error-prone because **firstly you are dealing with concepts and reasoning more** and **secondly routine operations on numbers and symbols less.**

There is generally a great emphasis on lightning fast mental math speed for success in Competitive exams. We remove this emphasis. We just say, as we have said so many times in our discourses,

For success in competitive exams, the student should have fairly good mental math speed. And we qualify it here with, "anyone with a good resolve can develop that fairly good mental math speed."

With only high speed mental math skills you may not go far, because at the core of problem solving lies your deductive reasoning ability, pattern identification and use ability and ability to use basic and rich concepts.

With these skills the problem gets simplified so much that routine calculation becomes a minor issue.

#### Deductive reasoning in the search for elegant method

Leaving the conventional path of substitution, we reason out,

- the choice values being simple, the $\sqrt{5}$ must be eliminated in the target expression, and more importantly,
- our experience based strategy is always trying to transform the given expression, using end state analysis approach, as alike to the end expression as possible. In this approach, as the given expression is smaller, labour is less and if elegant solution exists, we reach it faster.
- from the simple answer values, applying our deductive reasoning, we become more or less sure of significant gains along this path.

Based on this deductive reasoning, **we decide to transform the already transformed input expression further rather than substitute** the value of $x$ in the target expression.

#### Further transformation of given expression towards target expression

$x = \displaystyle\frac{(\sqrt{5} + 1)}{2}$,

Or, $x^2 = \displaystyle\frac{(\sqrt{5} + 1)^2}{4}$,

Or, $x^2 = \displaystyle\frac{3 + \sqrt{5}}{2}$,

Or, $x^2 - \displaystyle\frac{\sqrt{5} + 1}{2} - 1 = x^2 - x - 1=0$.

**Third stage of input transformation**

Now we can see the goal clearly and resort to the final third stage of input transformation,

$x^2 - x - 1=0$,

Or, $5x^2 - 5x - 1=4$.

**Answer:** Option a: 4.

This technique is a powerful one and we name it as **Enhanced input transformation technique**.

When the situation is suitable for applying this powerful technique, we don't just stop after first stage input transformation. We continue transforming the input expression to make it as nearly same as the target expression. In the process two or more stages of transformation may need to be done, but the process will always be guided by a comparison between the transformed input and the target expression.

**Comment:** Though it took a bit of time to explain, it finally has been a 50 seconds problem at the maximum because of the * deductive reasoning with a purpose* and

*carried out*

**use of end state analysis approach of continuously comparing input expression with the target expression****all in your mind.**

#### Key concepts used:

- Looking at the input expression and
**finding a double square root**we take up the only possible recourse of**freeing the surd expression from its outer square root**byTill this point it is conventional.**Surd rationalization technique.** - The conventional path to the solution is to substitute the simplified value of variable $x$ in the target quadratic expression in $x$ and go through routine procedure.
- By the use of
comparing the target end state expression with the given expression, and also applying**End State Analysis**taking into account of the simple choice values in the answer, we take the alternate path of further transforming the already simplified input expression.**Free resource use principle** - At this second level of input transformation, as expected, the surd $\sqrt{5}$ is quickly eliminated leaving us with a quadratic equation only in $x$.
- Now we are so near to the solution that the third minor input transformation comes quick and easy.
- Together these three stages form what we call
Necessarily, to use this powerful technique, you need to compare the transformed input with the target expression using**Enhanced Input transformation technique.**before and after every transformation.**End state analysis** - Decisions were guided by
and**deductive reasoning**and**pattern analysis**at the base level.**identification skills**

### Problem 2.

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

- 6
- 5
- 8
- 4

**Problem analysis and elegant solution**

Usually we are given sum of inverses in the input and happily go on exploiting its virtues. In contrast the sum of inverses is the target expression here, and not the given expression. This is just an observation with no signifucant impact in our choice of solution path.

For choosing the elegant path to the solution our main focus will always be the given input expression and if its specialities do not call for immediate action, we compare it with the target expression trying to find similarities.

#### First clue in the given expression

With just a look at the given expression we identify it as a fit case for applying the greatly useful **componendo and dividendo technique.**

#### Input transformation

In this technique we just add 1 to two sides of the equation in the first step and subtract 1 in the second stage. In the third stage we divide one result with the other thus simplifying the equation greatly,

$\displaystyle\frac{x^2 - x + 1}{x^2 + x + 1} = \displaystyle\frac{2}{3}$,

Or, $\displaystyle\frac{x^2 - x + 1}{x^2 + x + 1} + 1 = \displaystyle\frac{2}{3} + 1$,

Or, $2\displaystyle\frac{x^2 + 1}{x^2 + x + 1} = \displaystyle\frac{5}{3}$,

This is the first step. *In the second step* **we subtract input expression from 1**,

$1 - \displaystyle\frac{x^2 - x + 1}{x^2 + x + 1} = 1 - \displaystyle\frac{2}{3}$,

Or, $2\displaystyle\frac{x}{x^2 + x + 1} = \displaystyle\frac{1}{3}$.

Now we take the ratio of the two results,

$\displaystyle\frac{x^2 + 1}{x} = 5$,

Or, $x + \displaystyle\frac{1}{x} = 5$

**Answer:** Option b: 5.

#### Alternate method 1

We will show you an alternate method of solution that is **driven more by the form of the target expression** than the form of the given expression. The target expression is brought in at the very first stage,

$\displaystyle\frac{x^2 - x + 1}{x^2 + x + 1} = \displaystyle\frac{2}{3}$,

Or, dividing the numerator and denominator both by $x$, we get,

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

Now we can take two paths. Let us take the conventional path and **cross-multiply** the two sides of the equation which in this case is simple,

$3\left(x + \displaystyle\frac{1}{x}\right) - 3 = 2\left(x + \displaystyle\frac{1}{x}\right)+ 2$.

Transposing the terms,

$x + \displaystyle\frac{1}{x} = 5$, our solution.

#### Alternate method 2

Instead of the cross-multiplication and transposition, you could have imagined in your mind that $\left(x + \displaystyle\frac{1}{x}\right) = p$, a dummy variable and used the componendo and dividendo technique in a flash,

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

Or, $\displaystyle\frac{p - 1}{p + 1} = \displaystyle\frac{2}{3}$,

Or, $p = x + \displaystyle\frac{1}{x} = 5$.

No cross-multiplication and rearrangement of terms needed. If you are conversant with this simple but powerful technique, Componendo dividendo is always fast and accurate.

#### Key concepts and techniques used in the first method:

- Deciding that applying
**Componendo dividendo technique**on the input expression is the first and most important action to be taken because of the form of the expression being perfectly suited for applying this powerful simplification technique. - We apply this technique and easily derive the simple result all in our mind. For your benefit we have shown the steps explicitly.
- The result of application of this technique is so simple, with just a look at the target expression, we get to the solution in just one more simple step.
- In the final count it turned out to be a 25 second problem at most.

#### Key concepts and techniques used in the second method:

- Directly bringing in the target expression by simply dividing the numerator and denominator by $x$.
- Cross multiplication of the four terms on two sides of the equation.
- Simplification.
- These last two steps are very basic and age-old staple diet techniques used by millions of children over ages.
- These are routine with no thinking involved except mechanically correct following of the procedural steps.

#### Key concepts and techniques used in the third method:

The first step is same as the second method by bringing in the target expression at the very beginning,

- by dividing the numerator and denominator by $x$
in both numerator and numerator,**forming the target expression** - using
**abstraction**andof a dummy variable for the target expression in mind,**reverse way substitution** - application of the
now on the transformed equation produces the final result in one shot.**componendo dividendo technique**

**Important Note**

Though the problem looked innocuous and simple, it could be solved in **no less than three different ways, all distinct.** This is what we call, a great example of use of **Many ways Technique for improving problem solving skill. **The **more you find more than one fairly efficient paths to the solution of a problem**, it would be obvious that your ability or skill

- to
**see**more than one path to the solution is quite well-developed, - to
**use**various powerful techniques and strategies to solve the problem in its topic area is also very well-developed, and - you have an inherent drive to
**explore new**pathways which is the basis of innovative new thinking.

But **two conditions of usefulness are applicable here,**

**the solutions should all be good**, not just the laborious run-of-the-mill-conventional-inefficient solution, and- you must have the
**skill to compare**and evaluate the worth of a path to the solution.

### Summarization

Through * analytical treatment of the solution process of a few selected sums* it was shown how the basic and rich

*concept sets of Algebra together*with

*powerful general problem solving strategies*enable

*elegant solution of the problems in a few simple steps even for seemingly difficult algebra problems.*

*A very important aspect brought out is the exposition of the Many ways problem solving skill building technique, continued use of which enhances your power of innovation, no less.*

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

*.*

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