(2+sqrt{5})^{2}=?+4 sqrt{5} | vedclass

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Question

(B) To solve the expression $(2+\sqrt{5})^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2} + b^{2} + 2ab$.Here,$a = 2$ and $b = \sqrt{5}$.$(2+\sq

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) To solve the expression $(2+\sqrt{5})^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2} + b^{2} + 2ab$.Here,$a = 2$ and $b = \sqrt{5}$.$(2+\sqrt{5})^{2} = (2)^{2} + (\sqrt{5})^{2} + 2(2)(\sqrt{5})$.$= 4 + 5 + 4\sqrt{5}$.$= 9 + 4\sqrt{5}$.Comparing this with the given equation $? + 4\sqrt{5}$,we get $9 + 4\sqrt{5} = ? + 4\sqrt{5}$.Therefore,$? = 9$.

$(2+\sqrt{5})^{2}=?+4 \sqrt{5}$

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