sqrt{12} and ........ are like surds. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Two surds are called like surds if their radicands are the same after simplifying them to their simplest form.First,simplify $\sqrt{12}$:$\sqrt{12

Vedclass · Accepted Answer

$\sqrt{48}$

Vedclass · Answer

(C) Two surds are called like surds if their radicands are the same after simplifying them to their simplest form.First,simplify $\sqrt{12}$:$\sqrt{12} = \sqrt{4 	imes 3} = 2\sqrt{3}$.Now,check the options:$A) \sqrt{24} = \sqrt{4 	imes 6} = 2\sqrt{6}$$B) \sqrt{36} = 6$ (This is a rational number,not a surd)$C) \sqrt{48} = \sqrt{16 	imes 3} = 4\sqrt{3}$$D) \sqrt{60} = \sqrt{4 	imes 15} = 2\sqrt{15}$Since $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{48} = 4\sqrt{3}$,they share the same irrational factor $\sqrt{3}$.Therefore,$\sqrt{12}$ and $\sqrt{48}$ are like surds.

$\sqrt{12}$ and ........ are like surds.

Similar Questions

Show that any positive odd integer is of the form $4q+1$ or $4q+3,$ where $q$ is some integer.

Without actually performing the long division,find if $\frac{987}{10500}$ will have a terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

Prove that the product of any two consecutive positive integers is divisible by $2$.

The last digit of $2^{3} \times 5^{125}$ is . . . . . . .

If $n$ is an odd integer,then show that $n^{2}-1$ is divisible by $8$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module