Surds And Indices - Study Mode

[#66] If $$x = 5 + 2sqrt 6 { ext{,}}$$ xa0xa0 then $$sqrt x - frac{1}{{sqrt x }}$$ xa0 = is?
Correct Answer

(A) $${ ext{2}}sqrt 2 $$

Explanation

Solution: $$eqalign{
& {left( {sqrt x - frac{1}{{sqrt x }}}
ight)^2} cr
& = x + frac{1}{x} - 2 cr
& = left( {5 + 2sqrt 6 }
ight) + frac{1}{{left( {5 + 2sqrt 6 }
ight)}} - 2 cr
& = left( {5 + 2sqrt 6 }
ight) + frac{1}{{left( {5 + 2sqrt 6 }
ight)}} imes frac{{left( {5 - 2sqrt 6 }
ight)}}{{left( {5 - 2sqrt 6 }
ight)}} - 2 cr
& = left( {5 + 2sqrt 6 }
ight) + left( {5 - 2sqrt 6 }
ight) - 2 cr
& = 10 - 2 cr
& = 8 cr
& herefore left( {sqrt x - frac{1}{{sqrt x }}}
ight) = sqrt 8 = 2sqrt 2 cr} $$

[#67] $$left( {4 + sqrt 7 }
ight),$$ xa0 expressed as a perfect square, is equal to = ?
Correct Answer

(C) $$left{ {frac{1}{2}{{left( {sqrt 7 + 1} ight)}^2}} ight}$$

Explanation

Solution: $$eqalign{
& left( {4 + sqrt 7 }
ight) cr
& = frac{7}{2} + frac{1}{2} + 2 imes frac{{sqrt 7 }}{{sqrt 2 }} imes frac{1}{{sqrt 2 }} cr
& = {left( {frac{{sqrt 7 }}{{sqrt 2 }}}
ight)^2} + {left( {frac{1}{{sqrt 2 }}}
ight)^2} + 2 imes frac{{sqrt 7 }}{{sqrt 2 }} imes frac{1}{{sqrt 2 }} cr
& = {left( {frac{{sqrt 7 }}{{sqrt 2 }} + frac{1}{{sqrt 2 }}}
ight)^2} cr
& = frac{1}{2}{left( {sqrt 7 + 1}
ight)^2} cr} $$

[#68] If 3 x+y = 81 and 81 x-y = 3, then the value of $$frac{x}{y}$$ is = ?
Correct Answer

(D) $$frac{{17}}{{15}}$$

Explanation

Solution: $$eqalign{
& { ext{According to question,}} cr
& Rightarrow {{ ext{3}}^{x + y}}{ ext{ = 81}},{ ext{and}},{ ext{8}}{{ ext{1}}^{x - y}}{ ext{ = 3}} cr
& Rightarrow {{ ext{3}}^{x + y}}{ ext{ = (3}}{{ ext{)}}^4},{ ext{and}},{left( 3
ight)^{4(}}^{x - y)}{ ext{ = }}{{ ext{3}}^1} cr
& Rightarrow x + y = 4,{ ext{and}},x - y = frac{1}{4} cr
& x + y = 4......{ ext{(i)}} cr
& { ext{ }}x - y = frac{1}{4}.....(ii) cr
& { ext{Solve the equation of (i) and (ii)}} cr
& x = frac{{17}}{8}, cr
& y = frac{{15}}{8}, cr
& Rightarrow frac{x}{y} = frac{{17}}{{15}} cr} $$

[#69] $$left( {frac{{1 + sqrt 2 }}{{sqrt 5 + sqrt 3 }} + frac{{1 - sqrt 2 }}{{sqrt 5 - sqrt 3 }}}
ight)$$ xa0 xa0 simplifies to = ?
Correct Answer

(C) $$sqrt 5 - sqrt 6 $$

Explanation

Solution: $$frac{{1 + sqrt 2 }}{{sqrt 5 + sqrt 3 }} + frac{{1 - sqrt 2 }}{{sqrt 5 - sqrt 3 }}$$ $$ Rightarrow frac{{left( {1 + sqrt 2 }
ight)left( {sqrt 5 - sqrt 3 }
ight) + left( {1 - sqrt 2 }
ight)left( {sqrt 5 + sqrt 3 }
ight)}}{{left( {sqrt 5 + sqrt 3 }
ight)left( {sqrt 5 - sqrt 3 }
ight)}}$$ $$ Rightarrow frac{{sqrt 5 - sqrt 3 + sqrt {10} - sqrt 6 + sqrt 5 + sqrt 3 - sqrt {10} - sqrt 6 }}{{5 - 3}}$$ $$eqalign{
& Rightarrow frac{{2sqrt 5 - 2sqrt 6 }}{2} cr
& Rightarrow frac{{2left( {sqrt 5 - sqrt 6 }
ight)}}{2} cr
& Rightarrow sqrt 5 - sqrt 6 cr} $$

[#70] The value of $${left( {sqrt 8 }
ight)^{frac{1}{3}}}$$xa0 is = ?
Correct Answer

(C) $$sqrt 2 $$

Explanation

Solution: $$eqalign{
& {left( {sqrt 8 }
ight)^{frac{1}{3}}} cr
& = {left( {{8^{frac{1}{2}}}}
ight)^{frac{1}{3}}} cr
& = {8^{left( {frac{1}{2} imes frac{1}{3}}
ight)}} cr
& = {8^{frac{1}{6}}} cr
& = {left( {{2^3}}
ight)^{frac{1}{6}}} cr
& = {2^{left( {3 imes frac{1}{6}}
ight)}} cr
& = {2^{frac{1}{2}}} cr
& = sqrt 2 cr} $$