Question

Given √3=1.732app.,find to three places of decimal the value of 1+2√3/2-√3

Answer 1

1 + 2√3 / (2-√3).

Rationalising the denominator 1 + 2√3(2+√3) /( 2-√3)(2+√3)

You get 1 + 4√3 + 6 = 7 + 4√3

= 7 + 4 x 1.732

= 7 + 6.928 = 13.928

Answer 2

Answer:

$\frac{1+2\sqrt{3} }{2-\sqrt{3} }=16.66$

Step-by-step explanation:

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

Simplify to three decimal places.

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

Rationalize the denominator by $2+\sqrt{3}$ we have:
$\frac{1+2\sqrt{3} }{2-\sqrt{3} }=\frac{1+2\sqrt{3} }{2-\sqrt{3} }*\frac{2+\sqrt{3} }{2+\sqrt{3} }$

        $=\frac{2+\sqrt{3}+4\sqrt{3}+6 }{4-3} \\=8+5\sqrt{3}\\=8+5(1.732)\\=16.66$

Hence, $\frac{1+2\sqrt{3} }{2-\sqrt{3} }=16.66$

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