The expression (2 + sqrt{2})^4 has a value ly | vedclass

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(B) We expand $(2 + \sqrt{2})^4$ using the binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.$(2 + \sqrt{2})^4 = \binom{4}{0} 2^

Vedclass · Accepted Answer

$135$ and $136$

Vedclass · Answer

(B) We expand $(2 + \sqrt{2})^4$ using the binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.$(2 + \sqrt{2})^4 = \binom{4}{0} 2^4 + \binom{4}{1} 2^3 (\sqrt{2}) + \binom{4}{2} 2^2 (\sqrt{2})^2 + \binom{4}{3} 2^1 (\sqrt{2})^3 + \binom{4}{4} (\sqrt{2})^4$.$= 1(16) + 4(8)(\sqrt{2}) + 6(4)(2) + 4(2)(2\sqrt{2}) + 1(4)$.$= 16 + 32\sqrt{2} + 48 + 16\sqrt{2} + 4$.$= 68 + 48\sqrt{2}$.Given $\sqrt{2} \approx 1.414$,we have $48 	imes 1.414 = 67.872$.So,$68 + 67.872 = 135.872$.This value lies between $135$ and $136$.

The expression $(2 + \sqrt{2})^4$ has a value lying between

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