Practice MCQ Questions and Answer on Square Root and Cube Root

Solution:

 $$\eqalign{ & \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr & = 5 + 2\sqrt 6 \cr & \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr & = 5 - 2\sqrt 6 \cr & \therefore {\text{ }}{a^2} + {b^2} \cr & = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr & = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr & = 2\left( {25 + 24} \right) \cr & = 2 \times 49 \cr & = 98 \cr} $$

Solution:

Answer & Solution Answer: Option E Solution: $$\eqalign{ & \sqrt {\sqrt {17956} + \sqrt {24025} } \cr & = \sqrt {134 + 155} \cr & = \sqrt {289} \cr & = 17 \cr} $$

Solution:

 $$\eqalign{ & \Leftrightarrow \sqrt y = 4x \cr & \Leftrightarrow y = {\left( {4x} \right)^2} \cr & \Leftrightarrow y = 16{x^2} \cr & \Leftrightarrow \frac{{{x^2}}}{y} = \frac{1}{{16}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & = \root 4 \of {{{\left( {625} \right)}^3}} \cr & = {\left( {625} \right)^{\frac{3}{4}}} \cr & = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr & = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr & = {5^3} \cr & = 125 \cr} $$

Solution:

 $$\eqalign{ & \Leftrightarrow \sqrt {{3^n}} = 729 \cr & \Leftrightarrow \sqrt {{3^n}} = {3^6} \cr & \Leftrightarrow {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2} \cr & \Leftrightarrow {3^n} = {3^{12}} \cr & \Leftrightarrow n = 12 \cr} $$

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:

 $$\eqalign{ & {\text{Let, }} \cr & {\left( {15} \right)^2} + {\left( {18} \right)^2} - 20 = \sqrt x \cr & {\text{Then, }} \cr & \sqrt x = 225 + 324 - 20 \cr & \Leftrightarrow x = {\left( {529} \right)^2} \cr & \Leftrightarrow x = 279841 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \,\,\,\,\,\,\,\,8|\overline {68} \,\,\overline {06} \,\,\overline {21} \,(824 \cr & \,\,\,\,\,\,\,\,\,\,\,\,|\,\,64 \cr & \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,162|\,\,\,\,4\,06 \cr & \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,3\,24 \cr & \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 1644|\,\,\,\,\,\,\,\,\,\,82\,21 \cr & \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,65\,76 \cr & \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,45 \cr & \therefore {\text{Number to be added}} \cr & = {\left( {825} \right)^2} - 680621 \cr & = 680625 - 680621 \cr & = 4 \cr} $$

Solution:

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