Rationalise the denominator in each of the following:
$\frac{1}{7-4 \sqrt{3}}$
$\frac{1}{7-4 \sqrt{3}}$
(A) To rationalise the denominator of $\frac{1}{7-4 \sqrt{3}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $7+4 \sqrt{3}$.
$\frac{1}{7-4 \sqrt{3}} \times \frac{7+4 \sqrt{3}}{7+4 \sqrt{3}} = \frac{7+4 \sqrt{3}}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{7+4 \sqrt{3}}{49 - (16 \times 3)}$
$= \frac{7+4 \sqrt{3}}{49 - 48}$
$= \frac{7+4 \sqrt{3}}{1}$
$= 7+4 \sqrt{3}$
$\frac{1}{7-4 \sqrt{3}} \times \frac{7+4 \sqrt{3}}{7+4 \sqrt{3}} = \frac{7+4 \sqrt{3}}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{7+4 \sqrt{3}}{49 - (16 \times 3)}$
$= \frac{7+4 \sqrt{3}}{49 - 48}$
$= \frac{7+4 \sqrt{3}}{1}$
$= 7+4 \sqrt{3}$
Explore More
Similar Questions
If $\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ find the value of $a$ and $b$.
Medium
View SolutionState whether the following statements are true or false. Justify your answer.
$(i)$ The square of an irrational number is always rational.
$(ii)$ $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.
$(i)$ The square of an irrational number is always rational.
$(ii)$ $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.
Medium
View SolutionInsert a rational number and an irrational number between the following:
$0.15$ and $0.16$
$0.15$ and $0.16$
Medium
View SolutionRationalise the denominator of the following: $\frac{\sqrt{40}}{\sqrt{3}}$
Easy
View SolutionState whether the following statement is true or false:
$(-1)^{11} = -1$
$(-1)^{11} = -1$
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo