Algebra Concepts and Techniques for Solving SSC Questions

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Basic and Rich Algebraic Concepts for elegant solutions of SSC CGL problems

Algebra concepts and techniques for quick solution of SSC Algebra questions

Algebra concepts and techniques to solve tough SSC questions in a few simple steps

Algebra Concepts and Techniques for Solvinng SSC Questions is a comprehensive collection of techniques for SSC algebra problem solving in a few steps.

These are invaluable for solving Algebra questions in SSC CGL or SSC CHSL.

Sections are,

  1. General algebra problem solving concepts
  2. Frequently used algebra concept formulas in solving competitive algebra questions
  3. Advanced algebra concepts and techniques for solving hard algebra problems in simple steps.

You may click here to skip the following section on General algebra problem solving concepts. Our suggestion is not to skip, as the concepts form the foundation of techniques.

General algebra problem solving concepts

Algebra is one of the most useful disciplines in Maths as a whole. It is used in Number system problems, Trigonometric problems and other topic areas. Its use is not limited to real number domain, it is used as well in dealing with problems involving imaginary numbers.

Its widespread use originates from its inherent abstraction in defining and dealing with its component elements.

Component elements in Algebra

The smallest component elements in algebra are the variables and the constant coefficients. These two components combine to form algebraic terms and the terms combine to form the algebraic expressions. The combining process happens through the use of usual mathematical operators.

When an expression has an equality operator we get an equation.

Let us present an example.

$ax^2 + bx + c$ is an algebraic expression whereas, $ax^2 + bx + c = 0$ is a frequently encountered algebraic equation.

In this equation, we define $x$ to be the variable and $a$ and $b$ to be the constant coefficients of the terms $ax^2$ and $bx$, the third $c$ being an independent coefficient and term.

When we define $x$ to be the variable, it means, we don't know its exact value, its value can vary within a range or sometimes the variable can take up any range of values, or for that matter may not have any definable value even. A variable in an expression may have real values or imaginary values.

The nature of the expression is determined by the maximum power of its variables. In this specific example as above, the expression is a quadratic one in single variable $x$. If power of all variables is unity, the equation is classified as a linear equation.

All arithmetic rules apply to algebraic expression manipulation

This is inherent in the definition of the discipline and its component elements.

The basic concepts of addition, subtraction, multiplication and division along with the concepts of fraction manipulation, factorization, LCM, HCF, Indices are all applicable to algebraic expressions and equations.

Similarly the other rules such as associative rules and the such that are elaborated in our article on Number System are also applicable to Algebraic manipulations.

Mathematical modeling

Introduction of the concept of variables, value of which are not known in the beginning of a large scale real life analysis, further leads to formation of relationships between important and relevant variables expressed in the form of expressions, equations and inequalities and then on to analysis of these to finally arrive at useful results.

This whole process is done by forming first the mathematical model representing close to real life situation and then analyzing the model using plethora of tools, techniques and strategies that are well known and not so well known.

Mathematical Modeling of most of the important real life problems though is infeasible.

An example of a simple model

Problem example 1

Khokon bought 5 oranges and 5 lemons at a total cost of Rs. 80, and 7 days later bought 3 oranges and 4 lemons at a total cost Rs. 50. If the prices remained unchanged in the two purchases, what was the per piece price of orange in Rs.?

Standard Solution

Let us assume per piece price of orange be $a$ and that of lemon be $b$.

These assumptions enable us to express the first purchase transaction in the form of the equation,

$5\times{a} + 5\times{b} = 80$.

And the second transaction as,

$3\times{a} + 4\times{b} = 50$.

This is the simple mathematical model representing the two transactions.

The process of solution for values of two variables in two linear equations is well known and easily understood. We need to eliminate one of the variables by first transforming one equation and then subtracting the transformed one from the other.

We will transform, let us say the first equation of our problem, with the target to eliminate variable $b$ (as we need to know the value of the other variable $a$ to get the answer). To do this we use the useful property of an equation that,

You can multiply, divide, add, and subtract any value to (or from) the two expressions on both sides of an equation without changing the nature of the equation.

Here, as we want to make the coefficient of variable $b$ equal to $4$ (so that $b$ cancels out when we subtract one equation from the other after transformation), we multiply the first equation (effectively each term of the first expression) by $\displaystyle\frac{4}{5}$. This transforms the first equation to,

$4a + 4b = 64$.

Subtracting the second equation from this transformed equation we have the desired answer of the per piece price of oranges in Rs. as,

$a = 14$.

General rule of solvability of a system of linear equations

Linear equations means all variables in a set of equations have only power of unity.

The solvability rule says,

You can solve and get values of all of the $n$ variables if you have at least $n$ number of unique equations in $n$ variables.

In simpler terms, to solve for values of three variables you need at least three unique linear equations in terms of the three variables, for two you need at least two equations and so on.

Unique equations means, you must not be able to derive any equation from any other in the set of equations being evaluated. For example,

$4a + 12b = 24$ and $a + 3b = 6$ are not two unique equations as, just by multiplying the second equation by $4$ you get the first.

Focal point is quick Algebra problem solving

Though it starts simple, algebra is large and complex. We won't cover any more of equations, inequalities, binomial theorems and such other concepts of algebra.

Our main focus in this session will be how to use the most basic concepts and the rich derived concepts along with powerful general problem solving strategies and techniques to solve algebraic problems involving expression manipulations in least number of steps and time.

In other words, we will focus on quick elegant solutions of hard algebra problems mainly for competitive tests.

Nevertheless the concepts that follow should be highly effective in nurturing the problem solving abilities in the student's mind by focusing on reaching the solution quickly using analytical abilities in contrast to going through long series of manipulative routine steps wasting valuable time and energy. This applies to school students as well.


Causes behind difficulties with Algebra

First : Though the topic of Algebra is based on a small set of concepts, dealing with abstract symbolic variables poses first level of difficulty in forming the problem definition in comparison to problems involving numbers.

Second: Furthermore, based on the small number of building blocks of concepts, Algebraic expressions can be made to appear very complex posing the second level of difficulty to the student in understanding and solving such problems.

Third: Identification of useful patterns in complex algebraic expressions is an essential skill in solving Algebra problems, and, not everyone is comfortable in recognizing useful patterns which is not easily visible. This forms the third level of difficulty with problems on Algebra in general.

Some of these important and useful patterns may systematically be compiled into a set of rich concepts on top of the basic concepts, for solving such problems easily in future.

Fourth: Knowledge of basic and rich concepts is though not enough - one needs to have basic problem solving skills using powerful general problem solving strategies and techniques.

Fifth: Lastly, it is a fact of life and Algebra that even after intensive skill enhancing practices, you may face a totally new type of problem in the final performing stage. You need to be able to bridge the final gap, as we say, to be in full control of any domain, in this case, Algebra.


Concept layers

Like any subject or activity area, the important and useful concepts fall in two layers - the basic concept layer that everyone knows, and the rich concept layer that is derived from the basic concepts and actual problem solving experience.

When a new rich concept is discovered in the process of solving a new type of problem in a few steps elegantly, it is added to the rich concept layer incrementally and continually. Thus this rich concept layer is an extensible open ended layer of concepts.

The problem solving concept and process layer uses these subject concept layers to solve any problem in the domain efficiently.

Remember, it is not problem solving any which way to us - it is efficient problem solving at low cost (in this case time) .


Frequently used algebra concept formulas in solving competitive algebra questions

The basic operations involved in Algebra are none other than all the basic arithmetic operations, but on abstract symbolic variables and expressions, not on numbers.

The more important relationships in Algebra forming the basic concept layer of Algebra are the following.

$(a + b)^2 = a^2 + 2ab + b^2$, in the form of square of sum of two variables.

$(a - b)^2 = a^2 - 2ab + b^2$, subtractive square of sum of two variables.

$a^2 - b^2 = (a + b)(a - b)$, this is one of the most useful algebraic relationships.

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, in the form of cube of sum of two variables.

$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$, this cube form is used frequently and it is better to remember it as a basic concept.

$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, negative counterpart of cube of sum of two variables.

$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$, which again is useful.

Derived from the cubes of sums we get the next set of basic relationships that are frequently used.

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, a very useful two-factor expansion of sum of cubes to remember.

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, equally useful two-factor expansion of subtractive sum of cubes.

Similarly it pays to know the similar expressions in three variables.

$(a + b + c)^2 $

$\hspace{1mm} = a^2 + b^2 + c^2 + 2(ab + bc + ca)$,

$(a + b + c)^3 $

$\hspace{1mm} = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c +a)$,

$a^3 + b^3 + c^3 - 3abc $

$\hspace{1mm} = (a + b + c) \times{}$

$\hspace{10mm} (a^2 + b^2 + c^2 - ab - bc - ca)$.

At the end, we would present Sreedhar Acharya's formula for finding the two roots of the genral quadratic equation,

$ax^2 + bx + c = 0$, where $a$, $b$ and $c$ are the two coefficients and the constant and $x$ is the variable.

The two roots are governed by the formula,

$\displaystyle\frac{-b \pm{\sqrt{b^2 - 4ac}}}{2a}$.

If $b^2 \lt {4ac}$, then the roots of $x$ are imaginary.


Advanced algebra concepts and techniques for solving hard algebra problems in simple steps

Algebra Concept and Technique 1. Three variable zero sum principle

The rich concept of three variable zero sum principle is,

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$ which is a very useful concept.


Proof of three variable zero sum principle:

$a + b + c = 0$,

Or, $(a+b) =-c$.

Raise both sides to their cubes,

$(a + b)^3 =-c^3$

Or, $a^3+b^3+c^3+3ab(a+b)=0$.

Substitute, $a+b=-c$,

$a^3 + b^3 + c^3 = 3abc$, a compact and highly useful result.

To identify this highly useful result we have given the concept a name of Three variable zero sum principle and included it in our rich concept layer.


You may refer to details of application of this valuable rich concept in our second quick algebra simplification session.

Algebra Concept and Technique 2: Chained equation treatment technique

An example of a chained equation,

$a^x = b^y = c^z$.

In this form manipulation of the bases indices in this case are difficult.

Technique is to equate the three parts of the equation to a dummy variable, say $q$,

$a^x = b^y = c^z=q$.

Now the chained equation in this form can be expressed as three separate equations,

$a^x=q$,

$b^y = q$, and,

$c^z=q$.

allows independent pairwise comparison and useful result derivation.

The new equality is introduced in the form,

$a^x = b^y = c^z = q$, from which we can form independent equations, $a^x = q$, $b^y = q$ and $c^z = q$, all in terms of $q$.

You may refer the effective use of this technique in our first quick algebra simplification session.

Algebra Concept and Technique 3: Base equalization technique

We have found application of this profound technique in widely different areas, of Maths and in real life problems as well.

You may explore its use in indices problems, in our first quick algebra simplification session, fraction problems and in diverse areas.

Algebra Concept and Technique 4: Principle of inverses

This is another very useful concept set. You may refer to its detailed treatment in our session on inverses.

Briefly one of the useful results of principle of inverses in Algebra is,

If $x + \displaystyle\frac{1}{x} = n$, where $n$ usually is a positive integer or zero, we can always derive similar expressions in sum of inverses for powers 2, 3 and beyond. The basic advantage with this type of expressions of inverses results from the variable $x$ disappearing when the two inverses are multiplied together, $x\times{\displaystyle\frac{1}{x}} = 1$.

Example problem:

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

Solution:

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

Squaring we get,

$x^2 + \displaystyle\frac{1}{x^2} + 2 = 4$,

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

So, $x^3 + \displaystyle\frac{1}{x^3}$

$\hspace{5mm}= \left(x + \displaystyle\frac{1}{x}\right)\left(x^2 + - 1 + \displaystyle\frac{1}{x^2}\right)$

$\hspace{5mm}= 2$.

The principle of inverses can also be applied in solving real life problems.

Algebra Concept and Technique 5: Substitution technique or principle of representative

When in complex expressions we observe some of the component expressions appear unchanged throughout the problem, our first action will always be to replace these more complex component expressions by single variables.

This immediately simplifies the clutter and visual complexity of the problem, and brings into focus commonly used results that were not visible earlier due to the clutter.

This again is a general problem solving principle and is applied unknowingly in many diverse real life problem areas.

You may refer to the use of this powerful technique in our second quick algebra simplification session as well as the third quick algebra simplification session.

Algebra Concept and Technique 6: Principle of collection of like terms

This is a very useful technique applied especially in Algebra with great effectiveness. It simply says,

In a complex algebraic expression with many individual terms, collect together the terms in small groups, so that each group can take up a new meaningful existence, thus simplifying the whole expression in one glorious sweep.

We have already highlighted the use of this concept in the first quick algebra simplification session as well as the second quick algebra simplification session.

Algebra Concept and Technique 7: Input transformation technique

Though this sounds simple, many tricky problems can be elegantly solved by transforming the given expression in well known symmetric forms that bear direct relations to the end state expression.

You may refer to the effective use of this technique in our second quick algebra simplification session as well as the third quick algebra simplification session.

Algebra Concept and Technique 8: Principle of zero sum of square terms

This principle is based on fundamental mathematical principles. It says,

If,

$a^2 + b^2 + c^2 = 0$, and $a$, $b$ and $c$ are real, then, $a=b=c=0$.

On many occasions, use of this basic principle is the only way you can solve the problem quickly.

We presented an example of effective use of this valuable rich concept in our second quick Algebra simplification session.

Algebra Concept and Technique 9: Use of a new factor technique

By this technique, an existing expression is multiplied with a new factor thus simplifying the whole expression.

For example, when we encounter the expression, $a^2 - ab + b^2$ we might like to multiply it by $a + b$ to simplify the product to $a^3 + b^3$.

You may refer to the perfect use of this technique in our third algebra simplification session.

Algebra Concept and Technique 10: Use of delayed evaluation technique

In simplification of algebraic expression a very useful technique to use is the delayed evaluation technique. It says,

Delay evaluation of especially long time-consuming products till the last stage possible.

More often than not, at the last stage, the time-consuming product itself may not be required to be done.

Example problem:

If $a \gt 1$ and $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$, then the value of $a^4 - \displaystyle\frac{1}{a^4}$ is,

  1. $\frac{57895}{20736}$
  2. $\frac{57985}{20736}$
  3. $\frac{59825}{20736}$
  4. $\frac{58975}{20736}$

Final hurdle

We know $\left(a + \displaystyle\frac{1}{a}\right) = \displaystyle\frac{25}{12}$ and $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$.

So multiplying the two,

$\left(a^2 - \displaystyle\frac{1}{a^2}\right)=\displaystyle\frac{25\times{7}}{12^2}$.

Again we know,

$\left(a^2 + \displaystyle\frac{1}{a^2}\right)=\displaystyle\frac{337}{12^2}$.

Multiplying the two we get finally,

$\left(a^4 - \displaystyle\frac{1}{a^4}\right) = \displaystyle\frac{337\times{25}\times{7}}{12^4}$

Applying free resource use principle

In the beginning itself we have looked at the large choice values and decided to use whatever technique we know to avoid large multiplication. That is why we kept the factors, 7, 25 and 337 undisturbed without multiplying them.

First we test the numerators of choice values for factor 25 and quickly eliminate option a and b.

Next we test the two values of numerators of options c and d, that is, 59825 and 58975 for divisibility by 7. Either you apply the 7 divisibility technique or direct mental division here. We preferred the second approach and found 59825 not a multiple of 7 but 58975 to be a perfect multiple of 7. So the choice is option d.

Notice that we have ignored calculating $12^4$ altogether because all the denominators of four choices are same. If you actually calculate $12^4$, it will be equal to $20736$.

Answer: Option d: $\displaystyle\frac{58975}{20736}$.

Application of powerful free resource use principle made this quick elegant solution possible.

You may refer to the detailed treatment in How to solve difficult algebra problems in a few simple steps 5.


For effectively using these concepts in solving any algebra problem elegantly, it is necessary to use the powerful general problem solving strategies and techniques such as End state analysis, Pattern recognition, Solving a simpler problem, Problem breakdown technique and the likes along with the subject concepts.

Algebra Concept and Technique 11. Simplification technique

Though practically all concepts and techniques in Algebra are meant for simplification of complex algebraic expressions, by this generic name we state a really simple but often ignored step in any simplification process. It states simply,

If you find an opportunity to simplify a complex expression, simplify it immediately without any delay.

This seems obvious, but it simplifies the expression preparing the ground for applying more sophisticated techniques. We will highlight the usefulness of the concept through an example problem. Our recommendation, never miss a step of simplification by using the most basic algebraic concepts.

Example Problem 1.

If $x + \displaystyle\frac{1}{x} = 5$, then the value of $\displaystyle\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}$ is,

  1. $\frac{47}{21}$
  2. $\frac{41}{23}$
  3. $\frac{43}{23}$
  4. $\frac{45}{21}$

Solution: first stage objective

Though the given expression is an inverse sum, looking at the assymmetric fraction of large target expressions we decide not to evaluate inverse sums in other suitable powers of $x$.

Rather, we focus our attention on the large fraction in the target expression for simplifying. This is our first objective.


Note: It is important to note that we have reached this first objective by examining the end state or the target expression comparing it with the beginning state or the given expression by applying the most frequently used End State analysis approach, a powerful general problem solving concept.


When we compare the denominator with the numerator, we identify that the numerator has an expression same as the denominator, that is, $x^4 +1$, though a bit hidden away, the first term in the beginning and the second term at the end of the expression. But we have detected its presence by using our pattern recognition skill, an essential skill in any type of problem solving.

We take this opportunity to immediately simplify by using the most basic concept of dividing part of the numerator by the denominator. This is simplification technique at work.

Thus we have the target expression,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1} = 1 + \displaystyle\frac{E_1}{E_2}$, the fractional parts are given names.

This is a significant improvement and will make further simplification much easier.

From a complex fraction of $\displaystyle\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}$, the fraction to be evaluated has been simplified to $\displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1}$ in which the critical improvement has been elimination of the $x^4$ term and the numeric digit $1$ from the numerator with a reduction of number of terms from 5 to 3.

We'll see shortly in the same example how use of this simple step prepared the ground for application of more sophisticated algebraic techniques towards the final solution.

Recommendation

Never miss an opportunity to simplify an expression using the most basic mathematical concepts. That step will enable you to take more powerful steps towards solution later. This is as true in Algebra as in real life problem solving.

Algebra Concept and Technique 12. Many ways technique

This is a general problem solving and learning technique and we will use this technique in Algebra now with great positive results. The problem example used is the same and we will start from where we left in explaining the previous technique.

Here we will use the technique stated as,

If the most obvious approach does not seem to produce immediate results, explore any other approach available and use the approach to see if it breaks the immediate bottleneck.

In short, don't be locked in the habit of using the most obvious approach, explore to see if any other approach is available and use it.

Again, this concept seems to be obvious, but when applied its value would be clear to you.

Application of the technique

We have simplified the target expression in the previous problem as,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1}$ with the given expression as,

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

In most cases, we would use the given inverse sum along the lines of principle of inverses and will consider evaluating the required inverse sums of squares or cubes. But in this case examining the large fraction of the target expression we do not find any immediate use of the more frequently used form of principle of inverses.

Being aware of the many ways technique, we choose instead to transform the given expression in its second form,

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

Or, $x^2 - 5x + 1 = 0$.

We will use this second form of input expression to simplify the target expression invoking Many ways technique.

Just as you try to solve a problem in many ways to improve your problem solving skills and learn to solve problems efficiently, you may need to use a given resource also in more than one way.

Here instead of using an inverse sum in its usual manner we are now using it in expanded second form.

Now we will invoke another powerful algebraic concept where we will use this second form of the given expression to simplify the target expression greatly.

Algebra Concept and Technique 13. Continued factor extraction technique for simplification

When there is no other way, we apply this powerful algebraic technique for simplifying quite intimidating expressions with the help of a simpler input expression of zero value.

Essentially, we extract the input expression as a factor stage by stage from the higher order target expression and at each stage, after extraction as a factor in a part of the target expression, we substitute zero value for the input expression to eliminate that part altogether.

Then again we take up second stage extraction of the input expression as a factor in the remaining portion of the target expression.

At every stage the target expression thus gets simplified in big steps.

Let's see how this is done with the numerator of the fraction in the target expression,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1} = 1 + \displaystyle\frac{E_1}{E_2}$, where $E_1$ and $E_2$ are the numerator and denominator respectively,

and the given expression expanded in the second form of a zero valued expression,

$x^2 - 5x + 1 = 0$.

We will extract the LHS of the equation as a factor and eliminate all terms involved in the factored part of the expression.

Applying the technique on the numerator,

$E_1 = 3x^3 + 5x^2 + 3x = 3x(x^2 - 5x + 1) + 15x^2 + 5x^2$

$= 20x^2$.

In the first step (or in any stage) we have extracted or formed the LHS of the given expression as a factor, absorbing the highest order term of the given expression with the other terms automatically formed.

In this case, when we formed the LHS $(x^2 - 5x + 1)$ as a factor from the target expression absorbing the highest order term $3x^3$, we find the third term 1 of the factor equivalent to $3x$ already existing in the target expression. Thus we needed to compensate only for the second term of $-5x$ equivalent to $+15x^2$ outside the brackets in the main part of the rest of the expression.

At any stage, absorbing the highest order (with highest power) term in the factor by the first term of the factor, writing down the factored expression and compensating for the rest of the terms with opposite signs outside the factored expression are the main steps to carry out in this technique.

As the value of the factor is zero, the whole lot of terms involved and absorbed in the factored expression are reduced to zero and are eliminated at one stroke. The result becomes a simple $15x^2 + 5x^2 = 20x^2$, a very significant simplification.

This is one stage factor extraction only but it shows the power of simplification in no uncertain terms.

Just to satisfy academic curiosity, we will show the final steps to the solution below which you may skip.


Final steps to the solution:

So, the target expression is transformed to,

$E = 1 + \displaystyle\frac{20x^2}{x^4 + 1}$

$=1 + \displaystyle\frac{20}{x^2 + \displaystyle\frac{1}{x^2}}$

At last we have got our inverse expression in the target.

We evaluate the inverse in squares in no time,

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

Or, $x^2 + \displaystyle\frac{1}{x^2} = 25 -2 = 23$.

This time we have invoked the Many ways technique in its most used form by transforming the given expression by the techniques embodied in principle of inverses.

Thus we get,

$E = 1 + \displaystyle\frac{20}{23}$

$=\displaystyle\frac{43}{23}$

Answer: Option c: $\displaystyle\frac{43}{23}$.


This is an example of one stage factoring out. Occasionally longer expressions require two or more stages of factor extraction, each stage being simplified greatly because of zero value of the factor.

You may refer to the detailed multiple solutions to this problem in our session on How to solve difficult Algebra problems in a few simple steps 6.

Update on 25-5-2019:

While creating video on the article mentioned above, we solved this problem in just a few steps that should take 20 to 30 seconds of time. We present the solution below to highlight how strategy and objective driven key pattern identification can result in very quick elegant solutions to problems.

You may watch the video here.

Elegant solution to the problem

If $x + \displaystyle\frac{1}{x} = 5$, then the value of $\displaystyle\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}$ is,

  1. $\displaystyle\frac{47}{21}$
  2. $\displaystyle\frac{41}{23}$
  3. $\displaystyle\frac{43}{23}$
  4. $\displaystyle\frac{45}{21}$

Elegant solution: First stage: Follow the algebraic expression simplification strategy of simplifying the target expression first

Algebraic expression simplification strategy says,

Simplify the target expression as far as possible by itself, before even thinking of substituting given values.

With this goal, we identify the presence of $x^4+1$ in the numerator (combine first and last term). This is first key pattern identification.

The target expression is simplified to,

$1+\displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1}$.

Elegant solution: Second stage: With goal of converting target in terms of given expression, find the second key pattern

When you fix the right goal, chances of finding shortest solution also increases.

At this second stage, our goal is clear:

How best to convert the much simpler second term of the target in terms of given expression of sum of inverses, $\left(x+\displaystyle\frac{1}{x}\right)$.

With this goal, combined with continued drive towards simplifying the target expression by itself, we discover that if the numerator and denominator of the second term are divided both by $x^2$, we meet two objectives with one action. The result,

$1+\displaystyle\frac{3\left(x+\displaystyle\frac{1}{x}\right) + 5}{x^2 + \displaystyle\frac{1}{x^2}}$.

The numerator value is 20 and the denominator value is, $(5^2-2)=23$, as,

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

The target expression value,

$1+\displaystyle\frac{20}{23}=\displaystyle\frac{43}{23}$.

Answer: Option c: $\displaystyle\frac{43}{23}$.

The solution should take maximum of 30 seconds and without any writing.

Simplification technique is more general while algebraic expression simplification strategy is specific. It is always to be tried as a habit, before you substitute given values to the target expression.

Algebra Concept and Technique 14. Reuse technique

It is a simple but effective technique that saves time for solution. It states,

When you are at a specific evaluation stage, instead of starting from the beginning point, use all the evaluated results that can be used for saving maximum time.

We will highlight the use of this simple but useful technique through a fragment of a problem.

Example problem 2.

We are given, $a \gt 1$ and $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$, and we are to evaluate the value of $a^4 - \displaystyle\frac{1}{a^4}$.

Getting the inverse sum in square is easy,

$a + \displaystyle\frac{1}{a} = 2\frac{1}{12} = \frac{25}{12}$,

Or, $a^2 + \displaystyle\frac{1}{a^2} = \frac{25^2}{12^2} - 2 = \frac{337}{144}$.

At this point we take stock and understand clearly that if we get the value of $a^2 - \displaystyle\frac{1}{a^2}$, we can easily calculate the value of the target expression. And to evaluate, $a^2 - \displaystyle\frac{1}{a^2}$, we need to get the value of $a - \displaystyle\frac{1}{a}$, as value of $a + \displaystyle\frac{1}{a}$ is already available.

Thus we form a subtraction of square of inverses directly from the last available result of $a^2 + \displaystyle\frac{1}{a^2} = \frac{337}{144}$,

$a^2 - 2 + \displaystyle\frac{1}{a^2} = \displaystyle\frac{337}{144} - 2$

Or, $\left(a - \displaystyle\frac{1}{a}\right)^2 = \displaystyle\frac{49}{144}$,

Or, $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$, as $a \gt 1$.

We have reused the value of already evaluated expression, $a^2 + \displaystyle\frac{1}{a^2}$.

Instead, we could have started from the beginning with the given expression, $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$ and squaring and subtracting 4, and then taking the square root we would have had our desired value of $\left(a - \displaystyle\frac{1}{a}\right)$.

The difference between the two approaches would be,

In the first approach of using Reuse technique, we have calculated $\displaystyle\frac{337}{144} - 2$ which took little time, while in the second approach we had to first form the square and then calculate $\displaystyle\frac{625}{144} - 4$.

The second step would have taken 5 to 10 seconds extra which in itself may seem to be small, but in competitive tests may turn out to be the difference between success and failure.

Apart from success in tests,

Our main philosophy being Efficient Problem Solving, we won't like to miss any opportunity in finding a shorter way to the solution.

That's where lies the importance of this apparently obvious technique to be practiced as a habit in all situations.

In fact,

All innovations reuse.

It is a foundational technique.

You may refer to detailed use of this technique in our session on How to solve difficult Algebra problems in a few simple steps 5.

Algebra Concept and Technique 15. Surd rationalization property

Though surds form a part of number system, it is closely intertwined with Algebra in forming Surd Algebra problems that are usually a bit more complex than usual. That's why we include the basic surd concepts as rich algebraic concepts here.

We have already seen how Surd rationalization works in our first part of the basic and rich algebra concepts. Now we will go through two very important concepts. The first is the Surd rationalization property.

Input transformation by Surd Rationalization Property

In dealing with surds, we examine a two term surd expression always to detect if it satisfies the Surd Rationalization Property which states,

If the difference of the square of the two terms in the surd expression is 1, when we inverse the expression and rationalize, its denominator will become 1, and the denominator will thus be eliminated very conveniently, leaving us with a second complementary surd expression.

For example, if the starting expression were $2 + \sqrt{3}$ that satisfies Surd rationalization property, after inverting and rationalizing we get as a result, its complementary expression, $2 - \sqrt{3}$.

As an example we will take up a third example problem.

Example problem 3.

If $x = 3 + 2\sqrt{2}$, then the value of $\displaystyle\frac{x^6 + x^4 + x^2 + 1}{x^3}$ is,

  1. 192
  2. 216
  3. 204
  4. 198

Problem analysis and solution

Without waiting, we divide the numerator of the target expression by the denominator, because we notice the possibility of generating inverse expressions that we are comfortable with. This also is a case of using Simplification technique that we discussed earlier.

$\displaystyle\frac{x^6 + x^4 + x^2 + 1}{x^3}$

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

Our intention is to find a useful value of the sum of inverse of $x$ and applying the techniques of principle of inverses evaluate the target expression with ease.

Observing the surd rationalizing property present in the given expression $x = 3 + 2\sqrt{2}$, immediately we take the steps to invert $x$ and rationalize using rationalization technique,

$\displaystyle\frac{1}{x} = \displaystyle\frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = 3 - 2\sqrt{2}$, a very convenient expression.

With this result we get the value of sum of inverses,

$x + \displaystyle\frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$

With this sum of inverses we would easily be able to evaluate sum of inverse squares and then the sum of inverse cubes to evaluate finally the target expression.

After simplifying the target expression in the form of sum of inverses, we needed a convenient value of such a sum of inverses as input. The given input expression though was in a different from, it had the Surd rationalization property.

Detecting this property, evaluating first the inverse by inverting and rationalizing and then summing the inverse of $x$ with the direct $x$ produced the desired sum of inverses in convenient form.

Recommendation: Wherever you detect presence of surd rationalization property in a two term surd expression you may take up inversion and then rationalization.

You may refer to the detailed use of this useful concept in our sessions on How to solve difficult Algebra problems in a few simple steps 5 and How to solve difficult Surd Algebra problems in a few simple steps 4.

Algebra Concept and Technique 16. Transformation to square sum technique

This forms two of the most important techniques in surd expression simplification.

The objective that these pair of techniques fulfill is to transform a two term surd expression into a square of two term surd expression.

Wherever the original expression is under square roots, you have to apply one of these two techniques to bring the expression out of the square root for further simplification steps. In all such problems involving square root of surd expressions, application of one of these two techniques is indispensable.

First technique of transformation of a two term surd expression as a square of a second two term surd expression:

We will highlight the technique using a problem example.

Example problem 4.

Given $x = 3 + 2\sqrt{2}$, evaluate, $\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}$.

First stage pattern identification

Examining the given expression we notice that the surd term $2\sqrt{2}$ has a coefficient of 2. This is the first essential property for expressing the surd expression as a square of sum. This will form the second term $2ab$ in the general square expression $(a + b)^2 = a^2 + 2ab + b^2$.

The second essential property is, the sum of squares of two target terms in the square of sum must be equal to the first term in integer form. In our case it is 3 and it satisfies this second condition as, $3 = (\sqrt{2})^2 + 1^2$.

Unless the given two term surd expression satisfies these two conditions, transforming it into a square of two term surd expression won't be possible.

In our problem then we can directly have our transformation,

$x = 3 + 2\sqrt{2} = (\sqrt{2})^2 + 2\times{1}\times{\sqrt{2}} + 1^2$

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

So, $\sqrt{x} = \sqrt{2} + 1$.

Now only we can think of evaluating the target expression.

We need to form a sum of inverses and we detect presence of surd rationalization property in $\sqrt{x}$. Thus we invert, rationalize and then subtract,

$\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}$

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

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

This first form appears in surd expressions frequently.

Second technique of transformation of a two term surd expression as a square of a second two term surd expression:

This second form appears less frequently and so is harder to detect. It needs first a minor transformation of the given expression to convert the surd term to a term with a coefficient of 2 and then this transformed expression can be treated with the first technique as detailed above.

Let us highlight the use of this technique through a problem example.

Example problem 5.

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,

  1. $2$
  2. $2 - \sqrt{3}$
  3. $\frac{2}{\sqrt{3}}$
  4. $1$

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 for 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,

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

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

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

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

Similarly we get,

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

With this main breakthrough the rest of the steps don't pose much of difficulty.

For detailed treatment, you may refer to our session on Algebra Solution Set 13 and How to solve difficult Surd Algebra problems in a few simple steps 4.

Algebra Concept and Technique 17. Numerator simplification technique

Most times our objective in simplifying algebraic expressions is to simplify either numerator or denominator or both of the expressions. By Numerator simplification technique we address a specific case of simplifying the numerator with great positive results.

The two conditions for application of this very useful technique occurs when in a fraction of two term algebraic expressions,

  1. Both the numerator and the denominator have one common term, and
  2. The other terms left out are formed by same variable or surd but with different coefficients.

When we detect such a pattern we simply subtract a 1 from the fraction expression and compensate it by addition of another 1. The subtraction operation cancels out the common terms and transforms the numerator to a single term expression, a great improvement.

We will highlight the use of this technique through a problem fragment.

Example problem 6.

Simplify, $\displaystyle\frac{5\sqrt{3} - 2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} + \displaystyle\frac{6\sqrt{3} - 3\sqrt{5}}{4\sqrt{3} - 3\sqrt{5}}$.

Useful Pattern recognition

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. Thus the first condition is satisfied.

Furthermore, the other terms also are in same surds $\sqrt{5}$ and $\sqrt{3}$ respectively, but with different coefficients and thus the situation satisfies the second condition for applying the numerator simplification technique.

When we find such a pattern, the common terms are eliminated by applying the numerator simplification technique, more specifically, by subtracting 1 from the fraction term (or adding 1 to the fraction term when the common terms are of opposite signs between the numerator and denominator) and compensating the subtraction by addition of another 1.

As the second terms also are in terms of the same surd (or variable), the numerator is simplified to a single term expression.

Applying the numerator simplification technique on the target expression we get,

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

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

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

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

For detailed use of this technique you may refer to our session on How to solve difficult Surd Algebra problems in a few simple steps 4.

Algebra Concept and Technique 18. Simplification by factoring out technique

This is similar but not the same as the Continued extraction of factor technique. When we apply this technique, we detect a factor common to all terms of an expression and take it out of the expression thus simplifying it.

This again is a naturally obvious technique which should invariably be applied whenever opportunity arises. The important point to emphasize is,

In general, the common factor is not so easily visible.

This is the reason for our inclusion of this obvious simplification approach as an important rich algebraic concept. You have to use your pattern identification skills consciously to find out the hidden common factor.

We will highlight the use of this technique through a fragment of a problem. This is taken from the point where we left in highlighting the numerator simplification technique.

Example problem 7.

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

Identifying hidden pattern

Examining the two denominators closely we 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,

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

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

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

$=2$, a really simple result.

For detailed use you may refer to our session on How to solve difficult Surd Algebra problems in a few simple steps 4.


In itself these concepts and techniques are practical ways to simplify otherwise complex algebraic expressions quickly in a few steps, but one must recognize the fact that these powerful concepts and techniques can't be used in isolation.

For effectively using these concepts in solving any algebra problem elegantly in a few steps, it is necssary to use the powerful general problem solving strategies and techniques such as End state analysis, Many ways technique, Pattern recognition, Problem breakdown technique and the likes along with the subject concepts.



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