Practice MCQ Questions and Answer on Square Root and Cube Root

Solution:

 $$\eqalign{ & a = \frac{{\sqrt 3 }}{2}{\text{ (given)}} \cr & \therefore \sqrt {1 + a} + \sqrt {1 - a} \cr & = \sqrt {1 + \frac{{\sqrt 3 }}{2}} + \sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2 + \sqrt 3 }}{2}} + \sqrt {\frac{{2 - \sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2\left( {2 + \sqrt 3 } \right)}}{4}} + \sqrt {\frac{{2\left( {2 - \sqrt 3 } \right)}}{4}} \cr & = \sqrt {\frac{{4 + 2\sqrt 3 }}{4}} + \sqrt {\frac{{4 - 2\sqrt 3 }}{4}} \cr} $$ $$ = \sqrt {\frac{{3 + 1 + 2 \times \sqrt 3 \times 1}}{2}} + $$     $$\sqrt {\frac{{3 + 1 - 2 \times \sqrt 3 \times 1}}{2}} $$     \[\because \left\{ \begin{gathered} {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} - 2.\sqrt 3 .1 = {\left( {\sqrt 3 - 1} \right)^2} \hfill \\ {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} + 2.\sqrt 3 .1 = {\left( {\sqrt 3 + 1} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a + b} \right)^2} \hfill \\ \end{gathered} \right\}\] $$\eqalign{ & = \sqrt {\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{2}} + \sqrt {\frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}} \cr & = \frac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{2} \cr & = \frac{{2\sqrt 3 }}{2} \cr & = \sqrt 3 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \Leftrightarrow \sqrt {x + \frac{x}{y}} = x\sqrt {\frac{x}{y}} \cr & \Leftrightarrow x + \frac{x}{y} = {x^2}.\frac{x}{y} \cr & \Leftrightarrow \frac{{xy + x}}{y} = \frac{{{x^3}}}{y} \cr & \Leftrightarrow xy + x = {x^3} \cr & \Leftrightarrow y + 1 = {x^2} \cr & \Leftrightarrow y = {x^2} - 1 \cr} $$

Solution:

 $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,1|\overline 2 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.414 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,24|\,\,1\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,96 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,281\,|\,\,\,\,\,\,\,\,\,400 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,281 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 2824\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11900 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11296 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt 2 = 1.414 \cr} $$

Solution:

 Given expression, $$ = \sqrt {\frac{1}{{100}}} + \sqrt {\frac{{81}}{{100}}} + \sqrt {\frac{{121}}{{100}}} + $$      $$\sqrt {\frac{9}{{10000}}} $$ $$\eqalign{ & = \frac{1}{{10}} + \frac{9}{{10}} + \frac{{11}}{{10}} + \frac{3}{{100}} \cr & = 0.1 + 0.9 + 1.1 + 0.03 \cr & = 2.13 \cr} $$

Solution:

 $$\eqalign{ & a + b + c = 27\sqrt {29} \cr & \Rightarrow 2c + \frac{3}{2}c + c = 27\sqrt {29} \cr & \Rightarrow \frac{9}{2}c = 27\sqrt {29} \cr & \Rightarrow c = 6\sqrt {29} \cr & \therefore \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \cr & = \sqrt {{{\left( {27\sqrt {29} } \right)}^2} - 2\left( {2c \times \frac{3}{2}c + \frac{3}{2}c \times c + c \times 2c} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2\left( {3{c^2} + \frac{3}{2}{c^2} + 2{c^2}} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2 \times \frac{{13}}{2}{c^2}} \cr & = \sqrt {\left( {729 \times 29} \right) - 13{{\left( {6\sqrt {29} } \right)}^2}} \cr & = \sqrt {29\left( {729 - 468} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$

Solution:

 $$\eqalign{ & \Leftrightarrow \frac{{\sqrt x }}{{\sqrt {441} }} = 0.02 \cr & \Leftrightarrow \frac{{\sqrt x }}{{21}} = 0.02 \cr & \Leftrightarrow \sqrt x = 0.02 \times 21 \cr & \Leftrightarrow \sqrt x = 0.42 \cr & \Leftrightarrow x = {\left( {0.42} \right)^2} \cr & \Leftrightarrow x = 0.1764{\text{ }} \cr} $$

Solution:

 $$\eqalign{ & {\left( {.000216} \right)^{\frac{1}{3}}} = {\left( {\frac{{216}}{{{{10}^6}}}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{{6 \times 6 \times 6}}{{{{10}^2} \times {{10}^2} \times {{10}^2}}}} \right)^{\frac{1}{3}}} \cr & = \frac{6}{{{{10}^2}}} \cr & = \frac{6}{{100}} \cr & = 0.06 \cr} $$

Solution:

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

Solution:

 Given expressing, $$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$     $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$ $$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$     $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$ $$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$   $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$   $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$   $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$   $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$   $$...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$ $$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$     $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$   $$ + ...... + $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$ $$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$     $$ + \sqrt 4 - \sqrt 3 $$   $$ + ...... + $$   $$\sqrt {121} - \sqrt {120} $$ $$ = - 1 + \sqrt {121} $$ $$ = - 1 + 11$$ $$ = 10$$