Practice MCQ Questions and Answer on Square Root and Cube Root

Solution:

 $$\eqalign{ & {\text{We have to find}} \cr & = \sqrt {9{x^2} + 25 - 30x} \cr} $$  $$ = \sqrt {{{\left( {3x} \right)}^2} - 2.3x.5 + {{\left( { - 5} \right)}^2}} $$   $$\,\,\,\,\,\,\,\left\{ {\because {a^2} - 2ab + {b^2} = {{\left( {a + b} \right)}^2}} \right\}$$ $$\eqalign{ & = \sqrt {{{\left( {3x - 5} \right)}^2}} \cr & = 3x - 5 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \times \frac{{\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)}} \cr & = \frac{{3\sqrt 3 - 3\sqrt 2 + 3\sqrt 2 - 2\sqrt 3 }}{{\left( {3 - 2} \right)}} \cr & = \sqrt 3 \cr & = 1.732 \cr} $$

Solution:

 $$\eqalign{ & \,\,\,\,\,\,\,1|\overline 1 \,.\,\overline {56} \,\,\overline {25} \,(1.25 \cr & \,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,22|\,\,\,\,\,\,56 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 245\,|\,\,\,\,\,\,\,12\,25 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,12\,25 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\text{x} \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt {1.5625} = 1.25 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr & \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr & \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$

Solution:

 $$\eqalign{ & = \sqrt {\frac{3}{{11}}} \cr & = \sqrt {\frac{{3 \times 11}}{{11 \times 11}}} \cr & = \frac{{\sqrt {33} }}{{11}} \cr & = \frac{{5.745}}{{11}} \cr & = 0.5223 \cr} $$

Solution:

 $$\eqalign{ & = \sqrt {\left( {{{272}^2} - {{128}^2}} \right)} \cr & = \sqrt {\left( {272 + 128} \right)\left( {272 - 128} \right)} \cr & = \sqrt {400 \times 144} \cr & = \sqrt {57600} \cr & = 240 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \because x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr & \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$ $$\therefore {x^2} + \frac{1}{{{x^2}}}$$ $$ = \left( {17 + 12\sqrt 2 } \right)$$   $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$   $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$ $$\eqalign{ & = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr & = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr & = 17 + 17 \cr & = 34 \cr} $$

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:

 $$\eqalign{ & {\text{Let,}} \cr & {\text{ }}99 \times 21 - \root 3 \of x = 1968 \cr & {\text{Then,}} \cr & \Leftrightarrow 2079 - \root 3 \of x = 1968 \cr & \Leftrightarrow \root 3 \of x = 2079 - 1968 \cr & \Leftrightarrow \root 3 \of x = 111 \cr & \Leftrightarrow x = {\left( {111} \right)^3} \cr & \Leftrightarrow x = 1367631 \cr} $$