sqrt{7+2 sqrt{5}} = dots | vedclass

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Question

(A) To simplify the expression $\sqrt{7+2 \sqrt{5}}$,we look for two numbers $a$ and $b$ such that $(a+b)^2 = a^2 + b^2 + 2ab = 7 + 2\sqrt{5}$.Compari

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does not exist as a binomial surd

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(A) To simplify the expression $\sqrt{7+2 \sqrt{5}}$,we look for two numbers $a$ and $b$ such that $(a+b)^2 = a^2 + b^2 + 2ab = 7 + 2\sqrt{5}$.Comparing the terms,we have $a^2 + b^2 = 7$ and $2ab = 2\sqrt{5}$,which implies $ab = \sqrt{5}$.If we assume $a = \sqrt{5}$ and $b = 1$,then $a^2 + b^2 = (\sqrt{5})^2 + (1)^2 = 5 + 1 = 6$.Since $6 
eq 7$,the expression cannot be simplified into the form $\sqrt{a} + \sqrt{b}$ where $a$ and $b$ are rational numbers.Thus,$\sqrt{7+2 \sqrt{5}}$ does not exist as a binomial surd of the form $\sqrt{a} + \sqrt{b}$.

$\sqrt{7+2 \sqrt{5}} = \dots$

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