Square Root And Cube Root - Study Mode

[#106] If $$a = frac{{sqrt 3 }}{2}{ ext{,}}$$ xa0 then $$sqrt {1 + a} + sqrt {1 - a} = ?$$
Correct Answer

(D) $$sqrt 3 $$

Explanation

Solution: $$eqalign{
& a = frac{{sqrt 3 }}{2}{ ext{ (given)}} cr
& herefore sqrt {1 + a} + sqrt {1 - a} cr
& = sqrt {1 + frac{{sqrt 3 }}{2}} + sqrt {1 - frac{{sqrt 3 }}{2}} cr
& = sqrt {frac{{2 + sqrt 3 }}{2}} + sqrt {frac{{2 - sqrt 3 }}{2}} cr
& = sqrt {frac{{2left( {2 + sqrt 3 }
ight)}}{4}} + sqrt {frac{{2left( {2 - sqrt 3 }
ight)}}{4}} cr
& = sqrt {frac{{4 + 2sqrt 3 }}{4}} + sqrt {frac{{4 - 2sqrt 3 }}{4}} cr} $$ $$ = sqrt {frac{{3 + 1 + 2 imes sqrt 3 imes 1}}{2}} + $$ xa0 xa0 $$sqrt {frac{{3 + 1 - 2 imes sqrt 3 imes 1}}{2}} $$ xa0 xa0 [x08ecause left{ x08egin{gathered}
{left( {sqrt 3 }
ight)^2} + {left( 1
ight)^2} - 2.sqrt 3 .1 = {left( {sqrt 3 - 1}
ight)^2} hfill \
{left( {sqrt 3 }
ight)^2} + {left( 1
ight)^2} + 2.sqrt 3 .1 = {left( {sqrt 3 + 1}
ight)^2} hfill \
{a^2} + {b^2} - 2ab = {left( {a - b}
ight)^2} hfill \
{a^2} + {b^2} - 2ab = {left( {a + b}
ight)^2} hfill \
end{gathered}
ight}] $$eqalign{
& = sqrt {frac{{{{left( {sqrt 3 + 1}
ight)}^2}}}{2}} + sqrt {frac{{{{left( {sqrt 3 - 1}
ight)}^2}}}{2}} cr
& = frac{{sqrt 3 + 1 + sqrt 3 - 1}}{2} cr
& = frac{{2sqrt 3 }}{2} cr
& = sqrt 3 cr} $$

[#107] What is $$frac{{5 + sqrt {10} }}{{5sqrt 5 - 2sqrt {20} - sqrt {32} + sqrt {50} }}$$ xa0 xa0xa0 equal to ?
Correct Answer

(D) $$sqrt 5 $$

Explanation

Solution: $$eqalign{
& { ext{Given,}} cr
& frac{{5 + sqrt {10} }}{{5sqrt 5 - 2sqrt {20} - sqrt {32} + sqrt {50} }}{ ext{ }} cr
& = frac{{5 + sqrt {10} }}{{5sqrt 5 - 2 imes 2sqrt 5 - 2 imes 2sqrt 2 + 5sqrt 2 }} cr
& = frac{{5 + sqrt {10} }}{{5sqrt 5 - 4sqrt 5 - 4sqrt 2 + 5sqrt 2 }} cr
& = frac{{5 + sqrt {10} }}{{sqrt 5 + sqrt 2 }} cr
& = frac{{sqrt 5 left( {sqrt 5 + sqrt 2 }
ight)}}{{sqrt 5 + sqrt 2 }} cr
& = sqrt 5 cr} $$

[#108] The square root of $$frac{{{{left( {0.75}
ight)}^3}}}{{1 - 0.75}}$$ $${ ext{ + }}$$$$left[ {0.75 + {{left( {0.75}
ight)}^2} + 1}
ight]$$ xa0xa0 is = ?
Correct Answer

(B) 2

Explanation

Solution: $$frac{{{{left( {0.75}
ight)}^3}}}{{1 - 0.75}}$$ $${ ext{ + }}$$$$left[ {0.75 + {{left( {0.75}
ight)}^2} + 1}
ight]$$ $$ = frac{{{{left( {0.75}
ight)}^2} imes 0.75}}{{0.25}}$$ xa0 $${ ext{ + }}$$ $$left[ {0.75 + 0.5625 + 1}
ight]$$ $$ = 0.5625 imes 3 ,, + $$ xa0 $$left[ {0.75 + 0.5625 + 1}
ight]$$ $$eqalign{
& = 1.6875 + 2.3125 cr
& = 4 cr} $$ Square root of 4 = 2

[#109] The cube root of .000216 is:
Correct Answer

(B) .06

Explanation

Solution: $$eqalign{
& {left( {.000216}
ight)^{frac{1}{3}}} = {left( {frac{{216}}{{{{10}^6}}}}
ight)^{frac{1}{3}}} cr
& = {left( {frac{{6 imes 6 imes 6}}{{{{10}^2} imes {{10}^2} imes {{10}^2}}}}
ight)^{frac{1}{3}}} cr
& = frac{6}{{{{10}^2}}} cr
& = frac{6}{{100}} cr
& = 0.06 cr} $$

[#110] What should come in place of both x in the equation $$frac{x}{{sqrt {128} }} = frac{{sqrt {162} }}{x}$$
Correct Answer

(A) 12

Explanation

Solution: $$eqalign{
& { ext{Let}},frac{x}{{sqrt {128} }} = frac{{sqrt {162} }}{x} cr
& { ext{Then}},{x^2} = sqrt {128 imes 162} cr
& = sqrt {64 imes 2 imes 18 imes 9} cr
& = sqrt {{8^2} imes {6^2} imes {3^2}} cr
& = 8 imes 6 imes 3 cr
& = 144 cr
& herefore x = sqrt {144} = 12 cr} $$