sqrt[4]{17 + 12sqrt{2}} = | vedclass

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Question

(A) We are given the expression $\sqrt[4]{17 + 12\sqrt{2}}$.First,consider the square root of the expression: $\sqrt{17 + 12\sqrt{2}}$.We can write $1

Vedclass · Accepted Answer

$\sqrt{2} + 1$

Vedclass · Answer

(A) We are given the expression $\sqrt[4]{17 + 12\sqrt{2}}$.First,consider the square root of the expression: $\sqrt{17 + 12\sqrt{2}}$.We can write $17 + 12\sqrt{2}$ as $3^2 + (2\sqrt{2})^2 + 2 	imes 3 	imes 2\sqrt{2}$.This is of the form $a^2 + b^2 + 2ab = (a+b)^2$,where $a=3$ and $b=2\sqrt{2}$.So,$\sqrt{17 + 12\sqrt{2}} = \sqrt{(3 + 2\sqrt{2})^2} = 3 + 2\sqrt{2}$.Now,we need to find $\sqrt{3 + 2\sqrt{2}}$.We can write $3 + 2\sqrt{2}$ as $(\sqrt{2})^2 + 1^2 + 2 	imes \sqrt{2} 	imes 1 = (\sqrt{2} + 1)^2$.Therefore,$\sqrt{3 + 2\sqrt{2}} = \sqrt{(\sqrt{2} + 1)^2} = \sqrt{2} + 1$.

Column-$I$	Column-$II$
$(A)$ Bryophytes	$(i)$ Equisetum
$(B)$ Gymnosperms	$(ii)$ Onion
$(C)$ Angiosperms	$(iii)$ Anthoceros
$(D)$ Pteridophytes	$(iv)$ Thuja

$\sqrt[4]{17 + 12\sqrt{2}} = $

Similar Questions

Match the following columns:

Column-$I$ Column-$II$

$(A)$ Bryophytes $(i)$ Equisetum

$(B)$ Gymnosperms $(ii)$ Onion

$(C)$ Angiosperms $(iii)$ Anthoceros

$(D)$ Pteridophytes $(iv)$ Thuja

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$A$ bucket containing water is revolved in a vertical circle of radius $r$. To prevent the water from falling down, the minimum frequency of revolution required is $[g = \text{acceleration due to gravity}]$.

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