Question

%20%5Cfrac%7B%20%5Csqrt%7B7%20-%20%20%5Csqrt%7B24%7D%20%7D%20%7D%7B2%7D%20" id="TexFormula1" title=" \frac{ \sqrt{7 + \sqrt{24} } } 2{ + } \frac{ \sqrt{7 - \sqrt{24} } }{2} " alt=" \frac{ \sqrt{7 + \sqrt{24} } } 2{ + } \frac{ \sqrt{7 - \sqrt{24} } }{2} " align="absmiddle" class="latex-formula">
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Answer 1

Answer:

√6

Step-by-step explanation:

We have to find the value of the expression $\frac{\sqrt{7+\sqrt{24} } }{2} +\frac{\sqrt{7-\sqrt{24} } }{2}$.

Let, $A = \frac{\sqrt{7+\sqrt{24} } }{2} +\frac{\sqrt{7-\sqrt{24} } }{2}$

⇒ $A^{2} = \frac{7+\sqrt{24} }{4} +\frac{7-\sqrt{24} }{4}+ 2\times \frac{\sqrt{7+\sqrt{24} } }{2} \times \frac{\sqrt{7-\sqrt{24} } }{2}$ {Squaring both sides. Since (a + b)² = a² + b² + 2ab}

⇒ $A^{2} = \frac{7+\sqrt{24}+7-\sqrt{24} }{4} + 2 \times \frac{\sqrt{49-24} }{4}$ {Since (a + b)(a - b) = a² - b²}

⇒ $A^{2}= \frac{7}{2} +2 \times \frac{5}{4}$

⇒ $A^{2} = \frac{7}{2} +\frac{5}{2} = 6$

⇒ A = √6 (Answer) {Neglecting the negative root as the original expression is positive}