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$\sqrt{7+4\sqrt{3}}$ =

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Simplify radical expressions involving fractions

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Q.

\sqrt{7+4\sqrt{3}}

=

Apply Simplification Principle: Apply the principle of simplifying a nested radical, if possible, by finding a perfect square that can be added and subtracted inside the radical to create a binomial square.
Identify Potential Binomial Square: Identify the expression inside the radical as a potential binomial square. We are looking for an expression of the form $(a + b\sqrt{c})^2$ that equals $7 + 4\sqrt{3}$ . To find $a$ and $b$ , we need to solve the system of equations derived from the binomial square: $\newline$ $a^2 + 2ab\sqrt{c} + b^2c = 7 + 4\sqrt{3}$
Set Up System of Equations: Set up the system of equations by comparing coefficients: $\newline$ From $a^2 + b^2c = 7$ and $2ab\sqrt{c} = 4\sqrt{3}$ , we can deduce that $b^2c = 3$ and $2ab = 4$ .
Solve for $a$ and $b$ : Solve for $a$ and $b$ from the equations $b^2c = 3$ and $2ab = 4$ . Since $c = 3$ , we have $b^2 = 1$ and thus $b = 1$ (since $b$ must be positive for the square root to be real). Then, from $2ab = 4$ , we get $b$ $1$ .
Check Values in Original Equation: Check if the values of $a$ and $b$ satisfy the original equation $(a + b\sqrt{c})^2 = 7 + 4\sqrt{3}$ . Substitute $a = 2$ and $b = 1$ into the binomial square to get $(2 + \sqrt{3})^2$ .
Expand Binomial Square: Expand the binomial square $(2 + \sqrt{3})^2$ to verify that it equals $7 + 4\sqrt{3}$ . The expansion is $(2 + \sqrt{3})^2 = 2^2 + 2\cdot2\cdot\sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3$ .
Combine Terms for Verification: Combine the terms from the expansion to see if they match the original expression: $4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$ . This confirms that $(2 + \sqrt{3})^2$ is indeed equal to $7 + 4\sqrt{3}$ .
Conclude Simplified Form: Conclude that the simplified form of $\sqrt{7+4\sqrt{3}}$ is $2 + \sqrt{3}$ , since we have shown that $7 + 4\sqrt{3}$ is a perfect square, namely $(2 + \sqrt{3})^2$ .

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