Rationalization and Simplification of algebraic fraction | kullabs.com
Notes, Exercises, Videos, Tests and Things to Remember on Rationalization and Simplification of algebraic fraction

Please scroll down to get to the study materials.

Note on Rationalization and Simplification of algebraic fraction

• Note
• Things to remember
• Videos
• Exercise
• Quiz

The process of converting an irrational number to a rational number by multiplying it with a suitable rationalizing factors is called the rationalization of an irrational number.

Rationalization of the denominator of binomial form:

To rationalize the denominator of a binomial form, we multiply and divide the given binomial irrational number by the conjugate the irrational number in the denominator of the given expression.

Conjugate irrational number:

The binomial irrational numbers which differ only in sign ( + or - ) between the terms connecting them are called conjugate irrational numbers. The conjugate irrational numbers are a rational factor of each other. When we multiply them we get the rational number.
For example: $$(\sqrt{7} + 2) (\sqrt{7} - 2)$$ $$= 7 - 4 \: = 3$$. So $$\sqrt{7} + 2$$ is conjugate of$$\sqrt{7} - 2$$.

An equation in which unknown quantity occurs under the rational or root sign is called radical equation. To solve the radical equations, different methods are used. But the usual method is to eliminate the radicals by the process of "squaring both sides" and then solving the resulting equation.

Simplification of algebraic fraction

Three steps are necessary to simplify the algebraic fractions.

Step 1: Factorize both numerator and denominator and reduce them to lowest term.
Step 2: Take LCM of denominators.
Step 3: Simplify the numerator and get a simple form of a fraction.

• The process of converting an irrational number to a rational number by multiplying it with a suitable rationalizing factors is called the rationalization of an irrational number.
• An equation in which unknown quantity occurs under the rational or root sign is called radical equation.
• Three steps are necessary to simplify the algebraic fractions.

Step 1: Factorize both numerator and denominator and reduce them to lowest term.
Step 2: Take LCM of denominators.
Step 3: Simplify the numerator and get a simple form of a fraction.

.

Very Short Questions

Solution:

\begin{align*} &=\frac{x^3}{x-1} -\frac{x}{x-1}+ \frac{1}{x+1} - \frac{x^2}{x+1}\\ &= \frac{(x^3-x)(x+1)+(x-1)(1-x^2)}{(x-1)(x+1)}\\&=\frac{x^4+x^3-x^2-x+x-x^3-1+x^2}{(x-1)(x+1)}\\&=\frac{x^4-1}{(x-1)(x+1)}\\&=\frac{(x^2+1)(x^2-1)}{x^2-1} \\ &=x^2+1\: \:_\text {Ans.} \end{align*}

\begin{align*} &= \frac {1}{8(1- \sqrt {a})} - \frac {1}{8(1 + \sqrt {a})} +\frac {2\sqrt {a}}{8(1- \sqrt {a})} \\&= \frac {1 + \sqrt {a} - (1 - \sqrt {a})}{8(1 -a\sqrt {a} (1 + \sqrt {a})} + \frac {2\sqrt {a}}{8((1 -a)} \\ &= \frac {1 + \sqrt {a} -1 + \sqrt {a} }{8(1 -a)} +\frac {2\sqrt {a}}{8(1- \sqrt {a})} \\&= \frac {2\sqrt {a} + 2\sqrt {a}}{8 (1 -a)} \\& = \frac {4\sqrt {a}}{8(1- \sqrt {a})} \\&= \frac {\sqrt {a}}{2(1 -a)} \\ \end{align*}

Solution:

\begin{align*} &= \frac {1}{x- 1} - \frac {3}{x} + \frac {3} {x+1}-\frac{1}{x+2} = \frac{x-3(x-1)}{x(x-1)}+ \frac{3}{x+1}-\frac{1}{x+2}\\&= \frac{x-3x+3}{x(x-1)}+\frac{3}{x+1}-\frac{1}{x+2} =\frac{(3-2x)(x+1)+3x(x-1)}{x(x-1)(x+1)}-\frac{1}{x+2}\\&=\frac{-2x^2-2x+3x+3+3x^2-3x}{x(x^2-1)}-\frac{1}{x+2}\\&=\frac{x^2-2x+3}{x(x^2-1)}-\frac{1}{x+2}= \frac{(x^2-2x+3)(x+2)-x(x^2-1)}{x(x^2-1)(x+2)} \\&=\frac{x^3+2x^2-2x^2-4x+3x+6-x^3+x}{x(x^2-1)(x+2)}=\frac{6}{x(x^2-1)(x+2)}\: \:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \frac{1}{(x-1)(x-2)}-\frac{3}{(x-2)(x-3)}+\frac{2}{(x-3)(x-1)}\\&=\frac{x-3-3(x-1)+2(x-2)}{(x-1)(x-2)(x-3)}=\frac{x-3-3x+3+2x-4}{(x-1)(x-2)(x-3)}\\&=\frac{-4}{(x-1)(x-2)(x-3)}=\frac{4}{(x-1)(x-2)(3-x)}\: \:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \frac{1}{a- 3} - \frac {1}{a-1} + \frac {1} {a+3}-\frac{1}{a+1}\\&=\frac{1}{a-3}+\frac{1}{1+3}-[\frac{1}{a-1}+\frac{1}{a+1}]=\frac{a+3+a-3}{(a-3)(a+3)}-\frac{a+1+a-1}{(a+1)(a-1)}\\&=\frac{2a}{(a-3)(a+3)}-\frac{2a}{(a+1)(a-1)}=\frac{2a}{a^2-9}-\frac{2a}{a^2-1}\\&=\frac{2a(a^2-1)-2a(a^2-9)}{(a^2-9)(a^2-1)}=\frac{2a^3-2a-2a^3+18a}{(a^2-9)(a^2-1)}=\frac{16a}{(a^2-9)(a^2-1)} \: \:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \frac{1}{x- 1} - \frac {2}{2x+1} + \frac {1} {x+1}-\frac{2}{2x-1}=\frac{x+1+x-1}{(x-1)(x+1)}-[\frac{2}{2x+1}+\frac{2}{2x-1}] \\&= \frac{2x}{(x-1)(x+1)}-[\frac{2(2x-1)+2(2x+1)}{(2x+1)(2x-1)}] =\frac{2x}{(x-1)(x+1)}-\frac{(4x-2+4x+2)}{(2x+1)(2x-1)}\\&=\frac{2x}{(x-1)(x+1)}-\frac{8x}{(2x+1)(2x-1)} = \frac{2x}{x^2-1} - \frac{8x}{4x^2-1}\\&=\frac{2x(4x^2-1)-8x(x^2-1)}{(x^2-1)(4x^2-1)}\\&=\frac{8x^3-2x-8x^3+8x}{(x^2-1)(4x^2-1)}\\&=\frac{6x}{(x^2-1)(4x^2-1)}\:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \frac{1}{x- 5} - \frac {1}{x-3} + \frac {1} {x+5}-\frac{1}{x+3}=\frac{x+5+x-5}{(x-5)(x+5)}-[\frac{1}{x-3}+\frac{1}{x+3}]\\&=\frac{2x}{x^2-25}-\frac{x+3+x-3}{(x-3)(x+3)}=\frac{2x}{x^2-25}-\frac{2x}{x^2-9}=\frac{2x(x^2-9)-2x(x^2-25)}{(x^2-25)(x^2-9)}\\&=\frac{2x^3-18x-2x^3+50x}{(x^2-25)(x^2-9)}= \frac{32x}{(x^2-25) (x^2-9)} \: \:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \frac {1}{x+8} - \frac {1}{x-4} + \frac {1} {x-8}-\frac{1}{x+4}=\frac{1}{x+8}+\frac{1}{x-8}-\frac{1}{x-4}-\frac{1}{x+4}\\&=\frac{x-8+x+8}{(x+8)(x-8)}-[\frac{1}{x-4}+\frac{1}{x+4}]= \frac{2x}{(x+8)(x-8)}-[\frac{x+4+x-4}{(x-4)(x+4)}\\&=\frac{2x}{x^2-8^2}-\frac{2x}{x^2-4^2}=\frac{2x}{x^2-64}-\frac{2x}{x^2-16}=\frac{2x(x^2-16)-2x(x^2-64)}{(x^2-16)(x^2-64)}\\&=\frac{2x^3-32x-2x^3+128x}{(x^2-16)(x^2-64)}=\frac{96x}{(x^2-16)(x^2-64)}\:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \frac{2}{a^2-2a-24} + \frac {2}{a^2-3a-18} + \frac {1} {a^2+7a+12}\\&=\frac{2}{a^2-6a+4a-24}+\frac{2}{a^2-6a+3a-18}-\frac{1}{a^2+3a+4a+12}\\&=\frac{2}{a(a-6)+4(a-6)}+\frac{2}{a(a-6)+3(a-6)}-\frac{1}{a(a+3)+4(a+3)}\\&=\frac{2}{(a-6)(a+4)}+\frac{2}{(a+6)(a+3)}-\frac{1}{(a+3)(a+4)}\\&=\frac{2(a+3)+2(a+4)-1(a-6)}{(a+3)(a+4)(a-6)}\\&=\frac{2a+6+2a+8-a+6}{(a+3)(a+4)(a-6)}\\&=\frac{3a+20}{(a+3)(a+4)(a-6)} \: \:_\text {Ans.} \end{align*}

0%

irrational
rational
decimal
none
• If the product of two surds is a rational numbers, each of them is called the _____ factor of the other.

fraction
all
decimal
rationalising

dividing
subtracting
multiplying

2(sqrt{3})
2(sqrt{5})
2(sqrt{7})
2(sqrt{1})
• (sqrt{5}) - (sqrt{3}) is called the conjugate of _____ or vice versa.

both
(sqrt{5}) + (sqrt{3})
(sqrt{5}) - (sqrt{3})
none
• You scored /5

Forum Time Replies Report

8 ³√4x-7=13

Solve

5^x 2 3×5^x------------------------5^x ×3-5^x-1