This time we have included "Assurance" in the steps to the solution
Abstract: Algebraic rich concepts and techniques along with general problem solving strategies enable elegant solution to the difficult SSL CGL Algebra problem in this tenth session of algebra problem solving in a few simple steps. But out of two elegant solutions this time, which one should we select? The associated degree of certainty provides us with an usable criterion for choosing the method of solution. In brief, Assurance in method of solution will be considered as an important criterion for selection of problem solving method.
Students in schools or in competitive tests such as SSC CGL generally find Algebra problems difficult.
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.
The list of Difficult algebra problem solving in a few steps quickly is available at, Quick algebra.
In this session we will take up the important aspect of assured path to the solution in addition to few steps to the solution as before. In other words, we will explain the process of solving the chosen problem in a few steps, but in addition we will also consider, in all similar problem solving which method will lead you to the solution with more certainty.
How much confidence do you have on the method of solving the problem that you have chosen - degree of assurance
When you start solving a not so easy algebra problem under time pressure in a major competitive exam such as SSC CGL, the most important factor working towards success will undoubtedly be the problem solving strategy you will adopt after brief analysis of the problem.
If you choose to take up an uncertain path, not only you may need to backtrack after a few steps, the inherent uncertainty will make you unsure, less confident and slower in thinking. It is kind of double loss situation.
Along a tentative uncertain exploratory path even if you reach the solution, valuable time will surely be wasted in the process.
Our recommendation is,
- In exam-hall decision making you must be fast, sure and confident with no hesitation at all at any point, including choice of problem solving strategy for each problem.
- In the process, you might make one or two mistakes, but make those few mistakes quickly and confidently to take up a new problem without any delay.
- Better it is to leave a few problems if you feel any bit unsure, because in any case you are not expected to solve all MCQ based questions accurately in any test, and
- Best it is to choose the problem solving strategy which will surely lead you to the solution in a few steps. This is what we call Assured Solution in a few steps.
Speed is certainly important, but degree of assurance of a problem solving strategy is no less important.
In this session while solving the problem we have chosen this time,
We will show you two methods to solution, both elegant and fast. In one we will use, useful pattern recognition technique and in the other the technique of coefficient comparison of like terms. While we compare the two, you can assess yourself the pros and cons of the methods from your own point of view.
It is of utmost importance that you follow your own inclinations with acquired wisdom.
We urge you to try solving the problem on your own and note the time you took before going through our solution processes.
Watch the video below,
The factors of $(a^2 +4b^2 + 4b - 4ab -2a -8)$ are,
First Elegant Solution by useful pattern recognition technique - problem analysis and the process of problem solving:
As always we will analyze first the End state, (in this case the given expression that will be produced as an output of the product of a choice) by following End state analysis approach, because that is the most preferred first step in any problem solving activity.
We also know, in majority of the difficult competitive test math problems, useful pattern recognition plays one of the most important roles in solving the problems quickly and surely.
Having the skill to detect useful patterns, it should then take you very little time to detect the half hidden three terms in the given expression,
$a^2 + 4b^2 - 4ab = (a-2b)^2$.
Using this result by our deductive reasoning mechanism, the immediate next question asked is,
- Which product out of the four choices can generate this square of sum expression?
You latch on to the first choice product very quickly, yes it generates $(a-2b)^2$, as both the factors start with $(a-2b)$. In any of the other three products, both factors do not have $(a-2b)$ as part the two factors.
By this result you are fairly certain that the first choice represents the solution. But to be absolutely sure a second question needs to be answered,
- What about the rest of the terms of the product? In $(a^2 +4b^2 + 4b - 4ab -2a -8)$, the three terms, $a^2 + 4b^2 - 4ab$ have already been taken care of by the square of sum $(a-2b)^2$ leaving the three terms, $4b -2a -8$.
We can easily enumerate these three terms in the product of the first choice mentally,
$=(a-2b)^2 + [2(a-2b) -4(a-2b) - 8]$
$=(a-2b)^2 + [-2(a-2b) - 8] $
$=(a-2b)^2 + [4b -2a -8]$.
These three terms also match exactly with the rest of the three terms in the given expression after taking care of common $(a -2b)^2$. You have your solution then in no time at all.
Observe that we are thinking in terms of $(a-2b)$. This is direct application of abstraction technique when we take an expression $(a-2b)$ as a variable so that deductions are significantly simplified with minimal chances of error.
The important theme is,
Deal with larger chunks as long as possible and not the smaller many chunks of detail from the beginning of the problem solving process.
This is a different form of delayed evaluation technique we have used earlier in some of our discussions. Here this opportunity of using less detail and larger chunk of an unbroken expression $(a-2b)$ is created by directly using abstraction technique. We use the expression as a single variable mentally. This technique we have used earlier as Reverse substitution technique.
Let us now see how the second method of solution compares against the first.
Second elegant solution by coefficient comparison technique - problem solving process
By problem analysis it is evident that the given expression will be equal to one of the four products in the choices and thus we can classify the problem as fit case of using coefficient comparison technique.
So this time we will follow the pretty solid and assured step by step method of comparing coefficients of like terms between the given expression and the product expressions in the choices. While comparing the coefficients, we should be able to form the coefficients in the products mentally without actually expanding the products.
Coefficient comparison technique
The coefficient comparison technique is stated as,
The coefficients of like terms on the LHS and RHS of an equation must be same for the equality to hold.
We use here the method of mentally calculating each coefficient of a term in the expanded target expression from the two products by the concept of coefficient formation in a product of sums. This concept is brought out briefly in the following example.
Example of coefficient formation in a product of two expressions as factors
The coefficients of $x^4$, $x^3$, $x^2$, $x$ and $x^0$ in,
$(ax^2 + bx + c)(px^2 + qx + d)$
will be respectively,
Coefficient of $x^4$ = $ap$,
Coefficient of $x^3$ = $aq + bp$
Coefficient of $x^2$ = $ad + cp + bq$
Coefficient of $x$ = $bd + cq$
Coefficient of $x^0$ = $cd$, a total of 9 terms.
Coming back to our problem, first we compare coefficient of $a^2$
Result: in all four choices coefficient of $a^2$ is 1. Null result. No decision can be taken.
Second we compare the coefficient of $b^2$.
Result: in first three expressions coefficient is 4., but in the fourth choice, the coefficient of $b^2$ is -4. Thus the fourth choice is eliminated.
Third we compare coefficient of $b$.
In this step we are evaluating the first three choices only, the fourth choice having been eliminated in the last step.
In the first choice, coefficient of $b$ is given by, $8b - 4b = 4b$, that is 4, same as in the given expression. But in second and third choices, the coefficient of $b$ is -4. contadicting term formation of the given expression. These choices must then be considered as invalid leaving the first one as the only feasible choice.
To confirm, we test the coefficients of the rest of the terms in the same way but only against the product of the first choice and we find all of them to be same in the given expression as also in the product of the first choice.
Answer: Option a: $(a-2b-4)(a-2b+2)$.
In this case again we find the problem could be solved in more than one way. This is a case of Many ways technique, practice of which enhances the power of your problem solving skill in general.
We have shown two elegant methods of solution each with its own requirement of skill-set.
The first solution leads you quickly to the solution if you have the skill of recognizing useful patterns in target expressions. Of course the useful pattern has to be there in the first place. This method though is fast and elegant, is a bit uncertain as you don't know for sure whether you will be able to discover a useful pattern right at the start.
In the second method we recognize the form of the problem to be perfectly suitable for application of the standard assured method of coefficient comparison of like terms of LHS and RHS and having the skill-set of forming coefficients of variable terms in a product of expression quickly, we straightway resort to apply the method. By this method also we reach the solution quickly. Additionally, however complex or long the expressions were, the method is such as would surely lead you to the solution. The degree of assurance of the method is high.
There is a third possible method of solving the problem by direct expansion of a product in a choice and comparing with the given expression. While expanding, 9 terms will be generated which will need simplification. To do it for all four choices will take time and we would never adopt this approach. But In our problem, as the answer is in the first choice, even by this inherently inefficient method you may reach the solution as quickly as in other two methods. But how would you know that the first choice is the answer? That is the element of uncertainty and because of this uncertainty we would not consider this third approach as an acceptable method.
Finally then both the acceptable methods that we have shown are elegant and fast but which one you would choose will depend on whether you are able to detect the pattern quickly or whether you can compare the coefficients quickly. It is a question of which of the skill-sets you have.
In any problem, detecting and using friendly pattern leads you to the solution quickest. But if you fail to find such a pattern in a few seconds, you should know that the second method of coefficient comparison of like terms will lead you to the solution equally fast but with no uncertainty. This high degree of certainty inbuilt in this technique makes it a preferred method if you can deduce the coefficients quickly.
Be comfortable in using both the techniques so that according to the nature of the problem you can choose the most suitable one for elegant and also assured solution in a few steps.
To go through other powerful concepts and methods to solve difficult algebra problems, you may click on,