Question

arrange in assending order 3√2, 2√8, 4, √50, 4√3​

Answer 1

${\large{ \underline{\pmb{ \sf{RequirEd \: solution...}}}}}$

⋆ Understanding the question: This question says that we have to arrange the following surds/numbers in ascending order. They are 3√2, 2√8, 4, √50, 4√3.

⋆ Using concept: To solve this question we have to convert all these numbers into square root form.

⋆ Now let's solve this question!

~ According to the question

First number:

${\sf{:\implies 3 \sqrt{2} = \sqrt{{3}^{2}} \times 2 }}$

${\sf{:\implies \sqrt{9} \times 2 }}$

${\sf{:\implies \sqrt{18} }}$

Second number:

$\sf \implies 2 \sqrt{8} = \sqrt{{2}^{2} \times 8}$

$\sf \implies \sqrt{4 \times 8}$

$\sf \implies \sqrt{32}$

Third number:

$\sf \implies 4 = \sqrt{16}$

Fourth number:

• It is already in square root form.

Fifth number:

${\sf{:\implies 4 \sqrt{3} = \sqrt{{4}^{2}} \times 3 }}$

${\sf{:\implies \sqrt{8} \times 3 }}$

${\sf{:\implies \sqrt{24} }}$

~ Now, We can easily compare all the numbers, As they are in square root form,

The numbers are:

• √18
• √18√32
• √18√32√16
• √18√32√16√50
• √24

~ Ascending order of these square roots is,

$\sf \implies \sqrt{16} < \sqrt{18} < \sqrt{24} < \sqrt{32} < \sqrt{50}$

~ Ascending order of the actual numbers,

$\sf \implies 4 < 3 \sqrt{2} < 4 \sqrt{3} < 2 \sqrt{8} < \sqrt{50}$

${\large{\pmb{\sf{\underline{Additional \: KnowlEdge...}}}}}$

• Rational numbers: The numbers which can be written in p/q form. where q ≠ 0 i.e., q is not equal to zero. For example - 7/8, 9/8, 567/738.

• Irrational number: Irrational number are the totally opposite to rational numbers. They cannot be expressed in the form of p/q.
• The best examples for irrational numbers is ${\sf{\pi}}$ and ${\frak{e}}$