Add $4 \sqrt{6}-8 \sqrt{10}$ and $3 \sqrt{6}+12 \sqrt{10}$.
(N/A) To add the given expressions,we group the like terms:
$(4 \sqrt{6}-8 \sqrt{10}) + (3 \sqrt{6}+12 \sqrt{10})$
$= (4 \sqrt{6} + 3 \sqrt{6}) + (-8 \sqrt{10} + 12 \sqrt{10})$
$= (4 + 3) \sqrt{6} + (-8 + 12) \sqrt{10}$
$= 7 \sqrt{6} + 4 \sqrt{10}$
$(4 \sqrt{6}-8 \sqrt{10}) + (3 \sqrt{6}+12 \sqrt{10})$
$= (4 \sqrt{6} + 3 \sqrt{6}) + (-8 \sqrt{10} + 12 \sqrt{10})$
$= (4 + 3) \sqrt{6} + (-8 + 12) \sqrt{10}$
$= 7 \sqrt{6} + 4 \sqrt{10}$
Explore More
Similar Questions
If $a = \frac{\sqrt{5}}{8}$ and $\frac{8}{a} = b \sqrt{5}$,then find the value of $b$.
Medium
View SolutionFill in the blank to make the following statement true:
The square root of $121$ is ..........
The square root of $121$ is ..........
Easy
View SolutionFind the value of $a$ :
$\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11}$
$\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11}$
Difficult
View SolutionRationalise the denominator of the following:
$\frac{16}{\sqrt{41}-5}$
$\frac{16}{\sqrt{41}-5}$
Difficult
View SolutionFor each question,select the proper option from four options given,to make the statement true: $0.3\overline{6}=$
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