(2 sqrt{392}-21)+(sqrt{8}-7)^{2}=(?)^{2} | vedclass

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Question

(D) Given equation: $(2 \sqrt{392}-21)+(\sqrt{8}-7)^{2}=(?)^{2}$Step $1$: Simplify $\sqrt{392}$. Since $392 = 196 	imes 2$,$\sqrt{392} = 14 \sqrt{2}$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Given equation: $(2 \sqrt{392}-21)+(\sqrt{8}-7)^{2}=(?)^{2}$Step $1$: Simplify $\sqrt{392}$. Since $392 = 196 	imes 2$,$\sqrt{392} = 14 \sqrt{2}$.Step $2$: Substitute this into the equation: $(2 	imes 14 \sqrt{2}-21)+(\sqrt{8}-7)^{2}=(?)^{2}$Step $3$: Expand the squared term using $(a-b)^{2} = a^{2}-2ab+b^{2}$. Here $a = \sqrt{8}$ and $b = 7$.$(28 \sqrt{2}-21) + ((\sqrt{8})^{2} - 2 	imes \sqrt{8} 	imes 7 + 7^{2}) = (?)^{2}$Step $4$: Simplify the terms:$28 \sqrt{2} - 21 + 8 - 28 \sqrt{2} + 49 = (?)^{2}$Step $5$: Combine like terms. The $28 \sqrt{2}$ and $-28 \sqrt{2}$ cancel out:$-21 + 8 + 49 = (?)^{2}$$36 = (?)^{2}$Step $6$: Solve for $?$: $? = \sqrt{36} = 6$.

$(2 \sqrt{392}-21)+(\sqrt{8}-7)^{2}=(?)^{2}$

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