Square And Qube

$$\sqrt {4{a^2} - 4a + 1} + 3a$$
$$ = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2a} \right)}^2} - 2 \times 1 \times 2a} + 3a$$
$$ = \sqrt {{{\left( {1 - 2a} \right)}^2}} + 3a$$
$$ = \left( {1 - 2a} \right) + 3a$$
$$ = \left( {1 + a} \right)$$
$$ = \left( {1 + 0.1039} \right)$$
$$ = 1.1039$$

$$\eqalign{ &×= \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 - 1} }} \times \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 + 1} }}$$
$$\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{ {3 - 1} }}$$
$$\,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2}$$
$$\,\,\,\,\,\, = 2 + \sqrt 3 $$
$$y = \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 + 1} }} \times \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 - 1} }}$$
$$\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{ {3 - 1} }}$$
$$\,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2}$$
$$\,\,\,\,\,\, = 2 - \sqrt 3 $$
$$\therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14$$

Money collected = (59.29×100) paise = 5929 paise.
∴ Number of members = $$\sqrt {5929} $$   = 77

$$\sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} $$
$$ = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} $$
$$ = \sqrt {49 - 45} $$
$$ = \sqrt 4 $$
$$ = 2$$

$$\frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \,$$
$$ = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }}$$
$$ = \frac{{5 - 20 + 50}}{{2\sqrt 5 }}$$
$$ = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }}$$
$$ = \frac{{35\sqrt 5 }}{{10}}$$
$$ = \frac{{7 \times 2.236}}{2}$$
$$ = 7 \times 1.118$$
$$ = 7.826$$

$${\text{Give}}\,{\text{Expression}}$$
$$= \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} $$
$$ = 5$$

$${\text{Let}}\sqrt {0.0169 \times x} = 1.3$$
$${\text{Then}},\,0.0169x = {\left( {1.3} \right)^2} = 1.69$$
$$ ⇒ x = \frac{{1.69}}{{0.0169}} = 100$$

$${\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right)^2}$$
$$ = {\left( {\sqrt 3 } \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - 2 \times \sqrt 3 \times \frac{1}{{\sqrt 3 }}$$
$$ = 3 + \frac{1}{3} - 2$$
$$ = 1 + \frac{1}{3}$$
$$ = \frac{4}{3}$$

A number ending in 8 can never be a perfect square.

$$\,\,\,\,\,\,\,\,2|\overline 6 \overline {40} \overline {09} \,(\,\,253$$
$$\,\,\,\,\,\,\,\,\,\,\,|4$$
$$\,\,\,\,\,\,\,\,\,\,\,| - - - - - $$
$$\,\,\,\,45\,|240$$
$$\,\,\,\,\,\,\,\,\,\,\,|225$$
$$\,\,\,\,\,\,\,\,\,\,\,| - - - - - $$
$$503\,\,|\,\,\,\,\,\,1509$$
$$\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,1509$$
$$\,\,\,\,\,\,\,\,\,\,\,| - - - - - $$
$$\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,x$$
$$\,\,\,\,\,\,\,\,\,\,\,| - - - - - $$
$$\therefore \sqrt {64009} = 253$$