Question

find the values correct up to 3 decimal places. if √2=1.412 √3 =1.732 √5= 2.2360 √6 = 2.4495& √ 10 = 3.162(a) 3–√5/3+2√5 (b)1+√2/3–2√2​

Answer 1

......... ...... ........

Answer 2

Answer:

Given: Values of √2 = 1.412 , √3 = 1.732 , √5 = 2.2360 , √6 = 2.4495 &

          √10 = 3.162

To find: value of $\frac{3-\sqrt{5}}{3+2\sqrt{5}}\:\:and\:\:\frac{1+\sqrt{2}}{3-2\sqrt{2}}$

Consider,

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

$\implies\frac{3-\sqrt{5}}{3+2\sqrt{5}}\times\frac{3-2\sqrt{5}}{3-2\sqrt{5}}$

$\implies\frac{9+10-6\sqrt{5}-3\sqrt{5}}{9-4\times5}$

$\implies\frac{19-9\sqrt{5}}{-11}$

$\implies\frac{9\sqrt{5}-19}{11}$

$\implies\frac{9\times2.2360-19}{11}$

$\implies0.102$

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

$\implies\frac{1+\sqrt{2}}{3-2\sqrt{2}}\times\frac{3+2\sqrt{2}}{3+2\sqrt{2}}$

$\implies\frac{3+4+3\sqrt{2}+2\sqrt{2}}{9-4\times2}$

$\implies\frac{7+5\sqrt{2}}{1}$

$\implies7+5\times1.412$

$\implies13.412$