If -2-1-414-3-1-732-5-2-236 and -6-2-449- find the approximate value of 2-3-2-3-2-3-2-3-3-1-3-1

2+√(3)2 √(3)+2 √(3)2+√(3)+√(3) 1√(3)+1 = (2+√( 3 ) )^ 2+(2 √( 3 ) )^ 2 (2 √( 3 ) )(2+√( 3 ) ) + (√( 3 ) 1)^ 2 (√( 3 ) )^ 2 (1)^ 2 = 4+2√( 3 ) ...

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Byju's Answer

Standard VIII

Mathematics

Finding Square Root of a Number

If √2=1.414...

Question

If $\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236$ and $\sqrt{6} = 2.449$ , find the approximate value of $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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Solution

$\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
$= \frac{(2 + \sqrt{3})^{2} + (2 - \sqrt{3})^{2}}{(2 - \sqrt{3}) (2 + \sqrt{3})} + \frac{(\sqrt{3} - 1)^{2}}{(\sqrt{3})^{2} - (1)^{2}}$
$= \frac{4 + 2 \sqrt{3} + 3 + 4 - 2 \sqrt{3} + 3}{(2)^{2} - (\sqrt{3})^{2}} + \frac{(3) - 2 \sqrt{3} + 1}{3 - 1}$
$= \frac{14}{1} + \frac{4 - 2 \sqrt{3}}{2}$
$= \frac{28 + 4 - 2 \sqrt{3}}{2}$
$= \frac{32 - 2 \sqrt{3}}{2}$
$= 16 - \sqrt{3}$
$= 16 - 1.732$
$\approx 14.26$

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Similar questions

Q. If

\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236

and

\sqrt{6} = 2.449,

find the value of

\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

Q. If

\sqrt{5} = 2.236

and

\sqrt{6} = 2.449

, find the value of

\frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}}

Given , $\sqrt{2} = 1.414, \sqrt{3} = 1.732$ and $\sqrt{6} = 2.449$ , find the value of $\frac{1}{\sqrt{3} - \sqrt{2} - 1}$ correct to 3 place of decimal

Find the value to three places of decimals of each of the following. It is given that

\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236

and

\sqrt{10} = 3.162

\frac{2 + \sqrt{3}}{3}

Q. Question 3
If

\sqrt{2} = 1.414

and

\sqrt{3} = 1.732

, then find the value of

\frac{4}{3 \sqrt{3} - 2 \sqrt{2}} + \frac{3}{3 \sqrt{3} + 2 \sqrt{2}}

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If √2=1.414,√3=1.732,√5=2.236 and √6=2.449, find the approximate value of 2+√32−√3+2−√32+√3+√3−1√3+1

2+√32−√3+2−√32+√3+√3−1√3+1=(2+√3)2+(2−√3)2(2−√3)(2+√3)+(√3−1)2(√3)2−(1)2=4+2√3+3+4−2√3+3(2)2−(√3)2+(3)−2√3+13−1=141+4−2√32=28+4−2√32=32−2√32=16−√3=16−1.732≈14.26

If $\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236$ and $\sqrt{6} = 2.449$ , find the approximate value of $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$