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How to solve difficult Surd Algebra problems in a few simple steps 4

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How to solve complex surds problems step by step by analytical techniques

Solving complex surds problems needs strategic use of suitable surds problem solving techniques. How to do this is explained clearly by solving a few carefully selected surds problems.

Complex surds invariably involve algebra. Broadly the topic of discussion here is thus Efficient Surds Algebra problem solving techniques.

Sections are,

  1. Carefully selected solved problems to highlight how to solve surds problems step by step quickly. Why and how to use the techniques are explained as clearly as possible.
  2. Surds Guide list of all links to Concept tutorials on surds, question sets and solution sets on surds and articles designed to show how to solve specific types of surds problem quickly.

Recommended reading

Before going ahead, you will be able to understand the solutions better if you go through the following concept and technique tutorials on what are surds and how to solve surds,

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 in minimum time, most of the difficult surd problem that you would face in various competitive exams.

Coming to the point, let us now take up solving problems straightaway.


Solved example problems

Problem 1. Double square root surds and Surd term factoring

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

Solution: First stage Problem analysis: Simplifying Double square root of surds

First realization: Unless we express $(1 + x)$ and $(1 - x)$ as square of sums and simplify the double square root of surds, we won't be able to proceed further.

Substituting the value of $x$,

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

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

First stage transformation of given relation:

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.

Reusing the technique, 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 breakthrough by simplifying double square root of surds.

Similarly,

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

Solution: Second stage simplification of surds: Surd term factoring technique

Substituting these simple values of $\sqrt{1 + x}$ and $\sqrt{1 - x}$ in the target expression,

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

Simplifying,

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

Now identify the key pattern that $\sqrt{3}$ can be factored out of both the terms of each denominator.

This results in each denominator factor equal to the numerator canceling each other out,

$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 simple but very useful Surd term factoring technique that helps to simplify surd expressions so many times.

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

Key concepts used:

  • Identifying that double square root simplification must be done first. This must be the first key action.
  • Creating forcibly a coefficient of 2 in the surd term, which is the middle term of the three term expansion of square of sum of two terms—$2ab$ in $(a + b)^2$. The basic concept used is—the surd term must be the middle term of the three term expansion of square of sum and it must have coefficient 2.
  • In the last stage, the key pattern identification of $\sqrt{3}$ that can be factored out and subsequent application of Surd term factoring technique solved the problem.

To have comprehensive knowledge on simplifying double square root surds and surd term factoring techniques, read the tutorial,

How to solve surds part 2 - double square root surd simplification and surd term factoring techniques.


Problem 2. Surd expression comparison and Inequality concepts

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,

  1. $b \lt a \lt c$
  2. $a \lt b \lt c$
  3. $b \lt c \lt a$
  4. $a \lt c \lt b$

Solution: First stage: Problem analysis to Identify key patterns and Primary action

This is a problem of surd expression comparison, where the two term surd expressions that are to be compared are in the form of subtractions $\sqrt{p+1}-\sqrt{p}$.

This makes direct comparison of the expression not feasible.

First express $2$ as $\sqrt{4}$ to make all four terms $\sqrt{6}$, $\sqrt{5}$, $\sqrt{4}$ and $\sqrt{3}$ in the three expressions of uniformly surd form. The three surd expressions are now,

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

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

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

You can easily see the first key pattern,

The difference of squares of the terms in each surd expression is 1.

And now identify the second key pattern that if these three expressions were addition of two surd terms, $\sqrt{6}+\sqrt{5}$, $\sqrt{5}+\sqrt{4}$, and $\sqrt{4}+\sqrt{3}$ instead, you could have easily compared the the three. The concept involved is,

Surd expressions say, $(\sqrt{6}+\sqrt{5})$ and $(\sqrt{5}+\sqrt{4})$ can easily be compared with each other as one term $\sqrt{5}$ is common and both the expressions are additive. As $\sqrt{6} \gt \sqrt{4}$ the first expression is larger than the second. It is that simple.

And you can easily,

Convert the subtractive surd expressions to additive by surd rationalization technique of multiplying and dividing each expression by the form of expression $\sqrt{p+1}+\sqrt{p}$ and then inverting each.

This is the identification of the most important key action that would solve the problem immediately.

Solution: Second stage: Surd expression comparison by Surd rationalization and Inequality concepts

So we invert the variables and rationalize each,

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

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

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

As $\sqrt{5} \gt \sqrt{3}$ with the second term equal between $\sqrt{5} + \sqrt{4}$ and $\sqrt{4} + \sqrt{3}$,

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

Similarly as, $\sqrt{6} \gt \sqrt{4}$ with second term equal between $\sqrt{6} + \sqrt{5}$ and $\sqrt{5} + \sqrt{4}$,

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

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

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

Now you have to use the Inequality concept,

If we invert the variables in an inequality, the relationship also gets inverted, that is, greater than relation gets changed to lesser than relation and vice versa.

Finally, you have the 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:

  • Identifying that the difference of squares of each pair of terms is 1, it was decided to apply Surd Rationalization technique, but to do so the variables were first inverted.
  • Surd rationalization applied on each of the three. Comparing two pairs of additive expressions, clear comparative relation of the two values of the inverted variables could easily be formed.
  • Joining the two comparative relations, a three term comparative relation between inverted variables is formed.
  • Inequality concepts: As inverting the variables results in inversion of the relationship also, the desired comparative relation between the three surd variables obtained.

To know more on how to compare surd expressions, read the article,

How to solve surds part 3 - surd expression comparison and ranking.

Problem 3. Surd rationalization, Componendo, Target simplification first and Surd term factoring

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,

  1. $\sqrt{3}$
  2. 2
  3. 1
  4. $\sqrt{5}$

Solution: First stage: input simplification by rationalization

By applying surd rationalization on given value of $x$,

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

The given value of $x$ not being simple, we'll follow the very important strategy of target expression simplification first before substituting given value of $x$.

Solution: Second stage: Numerator simplification of both terms of the target expression

Both the terms in the target expression are of the componendo dividendo pattern (or signature of componendo dividendo),

In each fraction term, the two terms in numerator and denominator are same with only one pair opposite in sign.

With such a form of an algebraic fraction, it is easily possible to eliminate one of the numerator terms by adding or subtracting 1.

Question is which of the two numerator terms to eliminate?

Again we take a strategic decision to eliminate the more complex value of $x$. This would achieve maximum amount of simplification of the two terms of the target expression. This will be the dividendo operation of componendo dividendo.

Subtracting 1 from each of the two terms and adding compensating 2,

$E = \displaystyle\frac{x + \sqrt{20}}{x - \sqrt{20}} -1 + \displaystyle\frac{x + \sqrt{12}}{x - \sqrt{12}} -1 +2$

$=\displaystyle\frac{2\sqrt{20}}{x - \sqrt{20}} + \displaystyle\frac{2\sqrt{12}}{x - \sqrt{12}} +2$

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

We have now substituted the value of $x$ and simplified $\sqrt{20}$ as $2\sqrt{5}$ and $\sqrt{12}$ as $2\sqrt{3}$.

Further simplifying by cancelling out $2\sqrt{5}$ and $2\sqrt{3}$ between numerator and denominator of the first and the second terms respectively,

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

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

How very convenient! The first two terms cancel out leaving just 2 as the final result,

$E=2$.

In fact, following this strategic approach the apparently complex problem can even be solved wholly in mind.

Answer: Option b: 2.

Key concepts, techniques and steps:

  • First step is input transformation eliminating denominator. We use here surd rationalization technique.
  • Second step is to adopt the strategy of target expression simplification by itself before substitution of complex given value of $x$.
  • In this step, noticing each of the two fraction terms of the target expression conforming to the powerful componendo dividendo pattern, we decide to eliminate the more complex term $x$ from both the numerators by subtracting 1 from each fraction term and adding compensating 2. This is numerator simplification.
  • Lastly, after substitution of value of $x$, we apply the factoring out technique individually on the numerator and denominator of each of the surd fraction terms and as a result the two terms become same but opposite in sign. Thus the two terms cancel out, leaving the 2 as the final result.

Problem 4. How to solve surds equations

What are the real roots of the equation, $\sqrt{3x+4} +\sqrt{3x-6}=10$?

  1. $\displaystyle\frac{15}{4}$
  2. $\displaystyle\frac{35}{4}$
  3. $\displaystyle\frac{25}{4}$
  4. $\displaystyle\frac{5}{4}$

Solution: Problem analysis and strategy decision

The essential action that we have to take for solving a surds equation of this form is,

To raise the equation to its square.

Analyzing the two equations, you find that raising the equation to its square in this form would create unnecessary complexity.

This is where using the result of $(a+b)(a-b)=a^2-b^2$, multiply both sides of the equation by $(\sqrt{3x+4} -\sqrt{3x-6})$, reducing the number of terms under square root to just 1,

$3x+4 -(3x-6)=10=10(\sqrt{3x+4} -\sqrt{3x-6})$,

Or, $\sqrt{3x+4} -\sqrt{3x-6}=1$.

Add it up with the given equation to eliminate the second term,

$2\sqrt{3x+4} =11$

Square both sides of the equation,

$4(3x+4)=121$,

Or, $3x=105$,

Or, $x=\displaystyle\frac{35}{4}$.

Answer: Option b: $\displaystyle\frac{35}{4}$.

We will solve at the end a completely different type of surds problem.

Problem 5. Surds problem with surds equation: Comparison of coefficients of similar variables

If $\displaystyle\frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}=A + \sqrt{B}$ then $B-A$ is,

  1. $-13$
  2. $3\sqrt{3}-7$
  3. $13$
  4. $\sqrt{13}$

Solution 5: Problem analysis and solving execution

First convert the denominator surds expression under square root to a square of sum to simplify the double square root of surds,

$7 + 4\sqrt{3}=(2+\sqrt{3})^2$.

Rationalizing the transformed denominator $2+\sqrt{3}$, you get the given expession as,

$(4+3\sqrt{3})(2-\sqrt{3})=A + \sqrt{B}$,

Or, $-1+2\sqrt{3}=A+\sqrt{B}$.

As $\sqrt{B}$ is the surd term it must be equal to the SIMILAR irrational surd term on the LHS. Similarly rational term $A$ must also be equal to $-1$.

This is because,

A surd term being an irrational number with a non-terminating non-repeating decimal component, it cannot be added numerically to a rational number giving a result that you can express with certainty.

Note: Surd form is $\sqrt{n}$ where $n$ has at least one factor as a prime number. Numerically it always has component of a non-terminating non-repeating decimal component and cannot be expressed as a fraction of the rational number form, $\displaystyle\frac{p}{q}$, where both $p$ and $q$ are integers.

This is what we call Coefficient comparison and equalization for similar variables that don't mix together. It follows a fundamental algebraic principle,

In an equation, coefficients of similar type of variables on both sides of the equation must be equal.

Thus,

$A=-1$, and

$2\sqrt{3}=\sqrt{B}$,

Or, $B=12$, and

$B-A=13$.

Answer: Option c: $13$.

To solve this last problem also you needed three methods,

  1. First, conversion to square of sum surd expression for simplifying the double square root surds,
  2. Second, surd rationalization, and
  3. Third, Coefficient comparison and equalization for similar variables.

By applying these three methods along with Surd term factoring whenever needed, most of the difficult surd problems can be solved.

You will find the many different types of surds algebra problems solved in the solution sets with links below.

End note

Strategic analytical approach generally results in much faster solution than any other approach.

Overall, the solution of these example problems are rich in learning potential and generally,

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


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


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,

Suresolv Algebra Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II and Other Competitive exams.

The guide list of articles includes ALL articles on Algebra in Suresolv and is up-to-date.

Guided help on Surds, fractions and indices in Suresolv

To get the best results out of the extensive range of articles of tutorials, questions and solutions on Surds, fractions and indices in Suresolv, follow the guide,

Suresolv Surds, fractions, Indices reading and practice guide for competitive exams.

The guide list of articles include ALL articles on Surds, fractions and indices in Suresolv and is up-to-date.


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