Question-208681

Question Number 208681 by Noorzai last updated on 20/Jun/24

Answered by Rasheed.Sindhi last updated on 21/Jun/24

\sqrt{7 + \sqrt{48}} = \sqrt{7 + 4 \sqrt{3}} = a + b \sqrt{3} (> 0) (l e t)

W h e r e a, b \in Z

7 + 4 \sqrt{3} = a^{2} + 3 b^{2} + 2 a b \sqrt{3}

a^{2} + 3 b^{2} = 7 & 2 a b = 4 \Rightarrow b = \frac{2}{a}

a^{2} + 3 {(\frac{2}{a})}^{2} = 7

a^{2} + \frac{12}{a^{2}} = 7

a^{4} - 7 a^{2} + 12 = 0

(a^{2} - 3) (a^{2} - 4) = 0

a^{2} = 4 \Rightarrow a = \pm 2

b = \frac{2}{\pm 2} = \pm 1

a + b \sqrt{3} = 2 + \sqrt{3}, - 2 - \sqrt{3} (< 0)

∴ a + b \sqrt{3} = 2 + \sqrt{3}

Answered by a.lgnaoui last updated on 20/Jun/24

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

= \sqrt{(4 + 3) + 4 \sqrt{3}}

= \sqrt{{(2 + \sqrt{3})}^{2}}

= 2 + \sqrt{3}