Question

Find the square root of the surd $\sf 7 + \sqrt{48}$

Answer 1

Question :-

Find the square root of the surd $\sf 7 + \sqrt{48}$

Answer :-

According to the question, we have to find the value of $\sf \sqrt{7 + \sqrt{48} }$,

By prime factorisation,

48 = 2 × 2 × 2 × 2 × 3

48 = 2⁴ × 3

$\sf \sqrt{48} = \sqrt{2^{2} \times 2^{2} \times 3}$

Let,

$\implies \sf \sqrt{48} = 2\sqrt{12}$

$\sf \sqrt{7 + \sqrt{48} } = \sqrt{x} + \sqrt{y}$

$\implies \sf \sqrt{7 + 2\sqrt{12}} = \sqrt{x} + \sqrt{y}$

Squaring both sides,

$\implies \sf (\sqrt{7 + 2\sqrt{12}} )^{2} = (\sqrt{x} + \sqrt{y})^{2}$

Applying (a+b)² = a² + 2ab + b² identity on the RHS (Right hand side of the equation)

$\implies \sf 7 + 2\sqrt{12} = (\sqrt{x})^{2} + 2(\sqrt{x})(\sqrt{y}) + (\sqrt{y})^{2}$

$\implies \sf 7 + 2\sqrt{12} = x + 2(\sqrt{xy}) + y$

Rearranging,

$\implies \sf 7 + 2\sqrt{12} = x + y + 2(\sqrt{xy})$

We can conclude that,

$\implies \sf x + y = 7\\\implies \sf 2\sqrt{12} = 2\sqrt{xy}$

By the above equations,

$\implies \sf x + y = 7\\ \implies \sf xy = 12$

Two numbers which when added results in 7 and multiplied results in 12 is 4 and 3.

Hence,

x can be either 4 or 3, and y can either be 4 or 3.

$\implies \sf \sqrt{ \sf 7 + \sqrt{48}} = \sqrt{4} + \sqrt{3}$

$\implies \sf \sqrt{\sf 7 + \sqrt{48}} = 2 + \sqrt{3}$

Identities :-

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a - b)(a + b)
• (x + a)(x + b) = x² + x(a + b) + ab
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)