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

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(B) Let the missing value be $x$. The equation is $(2 \sqrt{392} - 21) + (\sqrt{8} - 7)^{2} = x^{2}$.First,simplify $\sqrt{392}$: $\sqrt{392} = \sqrt{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let the missing value be $x$. The equation is $(2 \sqrt{392} - 21) + (\sqrt{8} - 7)^{2} = x^{2}$.First,simplify $\sqrt{392}$: $\sqrt{392} = \sqrt{196 	imes 2} = 14 \sqrt{2}$.So,$2 \sqrt{392} = 2 	imes 14 \sqrt{2} = 28 \sqrt{2}$.Next,expand $(\sqrt{8} - 7)^{2}$ using $(a - b)^{2} = a^{2} - 2ab + b^{2}$:$(\sqrt{8} - 7)^{2} = (\sqrt{8})^{2} - 2 	imes \sqrt{8} 	imes 7 + 7^{2} = 8 - 14 \sqrt{8} + 49 = 57 - 14 \sqrt{8}$.Since $\sqrt{8} = 2 \sqrt{2}$,then $14 \sqrt{8} = 14 	imes 2 \sqrt{2} = 28 \sqrt{2}$.Substituting these back into the equation:$(28 \sqrt{2} - 21) + (57 - 28 \sqrt{2}) = x^{2}$.$28 \sqrt{2} - 21 + 57 - 28 \sqrt{2} = x^{2}$.$57 - 21 = x^{2}$.$36 = x^{2}$.$x = 6$.

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

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