1Cademy - Simplifying $$(2 + \sqrt{3})^2$$ and $$(4

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Simplifying $(2 + \sqrt{3})^2$ and $(4 - 2\sqrt{5})^2$

To expand a binomial containing square roots that is squared, apply the Binomial Squares Pattern.

ⓐ $(2 + \sqrt{3})^2$ : Use the pattern for the square of a sum: $(a + b)^2 = a^2 + 2ab + b^2$ , where $a = 2$ and $b = \sqrt{3}$ .

Square the first term: $2^2 = 4$ .
Double the product of the two terms: $2 \cdot 2 \cdot \sqrt{3} = 4\sqrt{3}$ .
Square the last term: $(\sqrt{3})^2 = 3$ . Combining these parts yields $4 + 4\sqrt{3} + 3$ . Adding the integer constants 4 and 3 gives the final simplified expression: $7 + 4\sqrt{3}$ .

ⓑ $(4 - 2\sqrt{5})^2$ : Use the pattern for the square of a difference: $(a - b)^2 = a^2 - 2ab + b^2$ , where $a = 4$ and $b = 2\sqrt{5}$ .

Square the first term: $4^2 = 16$ .
Double the product of the terms: $2 \cdot 4 \cdot 2\sqrt{5} = 16\sqrt{5}$ . Because it is a difference, this term is subtracted: $-16\sqrt{5}$ .
Square the last term: $(2\sqrt{5})^2$ . Square both the numerical coefficient and the radical: $2^2 \cdot (\sqrt{5})^2 = 4 \cdot 5 = 20$ . Putting it all together gives $16 - 16\sqrt{5} + 20$ . Combining the integers 16 and 20 results in: $36 - 16\sqrt{5}$ .

A critical detail when squaring binomials with radicals is remembering the middle $2ab$ term, rather than just squaring the first and last terms. Furthermore, when a term like $2\sqrt{5}$ is squared, both the coefficient and the radicand must be squared independently.

Updated 2026-05-01

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