Principle of inverses and targeted input transformation help to solve in a few steps
Students in school or in competitive tests like 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 highlight how to reach the elegant solution in quick time in contrast with the inefficient way to solve such problems in specific and in general through a carefully selected problem.
In this problem,
Converting the denominator in the form of sum of inverses immediately simplified the problem and once again showed the utility of principle of inverses. Additionally we showed another method of solution not just for comparison, but to highlight the indispensable problem solving skill enhancement technique, the Many ways technique.
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,
Problem 1.
If $a(2 + \sqrt{3}) = b(2 - \sqrt{3}) = 1$, then the value of $\displaystyle\frac{1}{a^2 + 1} + \displaystyle\frac{1}{b^2 + 1} $ is,
- $1$
- $4$
- $-5$
- $9$
Elegant Solution: First stage Problem analysis and input transformation by Surd rationalization technique:
The given condition is a three-part continued equation in two variables and the target expression is in terms of these two variables. The first task is then to separately form two equations for these two variables $a$ and $b$ and simplify. Then only we will think of further actions.
$a(2 + \sqrt{3}) = 1$,
Or, $a= \displaystyle\frac{1}{2 + \sqrt{3}}=2 - \sqrt{3}$, by applying Surd rationalization technique multiplying numerator and denominator by $2 - \sqrt{3}$.
Similarly, $b(2 - \sqrt{3})=1$,
Or, $b= \displaystyle\frac{1}{2 - \sqrt{3}}=2 + \sqrt{3}$, again by surd rationalization but this time using multiplying factor $2 + \sqrt{3}$.
Different ways to solve appear after this point.
Second stage analysis and transforming target to sum of inverses:
With values of $a$ and $b$ being in ideal form for using sum of their inverses, we look for sum of inverses in the target expression as well. But first let us show why we consider the values of $a$ and $b$ to be ideal for using as sum of inverses. This is pattern analysis and recognition skill at work.
$a + \displaystyle\frac{1}{a} = 2 - \sqrt{3} + \displaystyle\frac{1}{2 - \sqrt{3}}$
$\hspace{15mm}=2 - \sqrt{3} + 2 + \sqrt{3} = 4$
In the same way,
$b + \displaystyle\frac{1}{b} = 4$
Now we will transform the target expression introducing sum of inverses.
Transforming target expression
$\displaystyle\frac{1}{a^2 + 1} + \displaystyle\frac{1}{b^2 + 1} $
$=\displaystyle\frac{1}{a\left(a + \displaystyle\frac{1}{a}\right)} + \displaystyle\frac{1}{b\left(b + \displaystyle\frac{1}{b}\right)}$
$=\displaystyle\frac{1}{4}\left(\displaystyle\frac{1}{a} + \displaystyle\frac{1}{b}\right)$
$=\displaystyle\frac{1}{4}(2 + \sqrt{3} + 2 - \sqrt{3})$
$=1$
Answer: Option a: $1$.
Key concepts and techniques used:
- Separating out the two variables into two independent equations of surd values from the given three-part continued equation.
- Rationalizing and simplifying the values of $a$ and $b$ using surd rationalization technique to eliminate the inverse form.
- Now comparing the target expression with the simplified form of input variable values taking the decision to use principle of inverses.
- Sum of inverses of both the input variables resulted in same value 4.
- Now transforming the target expression in the form of sum of inverses to get the result in one shot.
- This solution is conceptually advanced but can be carried out in a very short time wholly in the mind.
Second solution of Problem 1
As mentioned earlier up to the point of separating out the two variables into two surd expressions and simplifying their values, the path is same.
$a(2 + \sqrt{3}) = 1$,
Or, $a= \displaystyle\frac{1}{2 + \sqrt{3}}=2 - \sqrt{3}$.
$b(2 - \sqrt{3}) = 1$,
Or, $b= \displaystyle\frac{1}{2 - \sqrt{3}}=2 + \sqrt{3}$.
Now only we will tread a more conventional and less conceptual path.
Using the input values to directly deduce the denominators of the target expression
$a=(2 - \sqrt{3})$,
Or, $a^2 + 1 = (2 - \sqrt{3})^2 + 1 = 8 - 4\sqrt{3}$
Similarly,
$b=(2 + \sqrt{3})$,
Or, $b^2 + 1 = (2 + \sqrt{3})^2 + 1 = 8 + 4\sqrt{3}$
A more efficient path
Here again you may choose a more efficient path of applying the most simple Simplifying technique that says, simplify an expression as soon as you get an opportunity to simplify and then only carry on further,
$a^2 + 1 = 8 - 4\sqrt{3}=4(2 - \sqrt{3})$,
Or, $\displaystyle\frac{1}{a^2 + 1}=\displaystyle\frac{1}{4(2 - \sqrt{3})}=\displaystyle\frac{2 + \sqrt{3}}{4}$,
And similarly,
$b^2 + 1 = 8 + 4\sqrt{3}=4(2 + \sqrt{3})$,
Or, $\displaystyle\frac{1}{b^2 + 1}=\displaystyle\frac{1}{4(2 + \sqrt{3})}=\displaystyle\frac{2 - \sqrt{3}}{4}$.
Summing up we get the result as 1.
Wholly deduction oriented conventional path to the solution
Substituting the values of $a^2 + 1$ and $b^2 + 1$ in target expression,
$\displaystyle\frac{1}{a^2 + 1} + \displaystyle\frac{1}{b^2 + 1}$
$=\displaystyle\frac{1}{8 - 4\sqrt{3}} + \displaystyle\frac{1}{8 + 4\sqrt{3}}$
$=\displaystyle\frac{8 + 4\sqrt{3} + 8 - 4\sqrt{3}}{(8 - 4\sqrt{3})(8 + 4\sqrt{3})}$
$=\displaystyle\frac{16}{(64 - 16\times{3})}$
$=1$.
Our recommendation will always be to choose the more conceptual path to the solution with least amount of deductions. This is the fastest way to the solution while the examining authorities actually try to test whether you can think in abstract conceptual terms which is the basis of efficient and innovative problem solutions.
Summarization
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.
Regarding choice of method, choose always the method that uses basic and rich subject concepts and general problem solving techniques. This approach only can take you to the highest level of performance excellence.
Guided help on Algebra in Suresolv
To get the best results out of the extensive range of articles of tutorials, questions and solutions on Algebra in Suresolv, follow the guide,
The guide list of articles includes ALL articles on Algebra in Suresolv and is up-to-date.