If-x-4a-2-b-4b-2-1-where-b-gt-1-2-then-a-2-x-a-2-x-a-2-x-a-2-x-

Question Number 211761 by MATHEMATICSAM last updated on 20/Sep/24

If x = \frac{4 a^{2} b}{4 b^{2} + 1} where b > \frac{1}{2} then

\frac{\sqrt{a^{2} + x} + \sqrt{a^{2} - x}}{\sqrt{a^{2} + x} - \sqrt{a^{2} - x}} = ?

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

l e t \frac{p + q}{p - q} = \frac{\sqrt{a^{2} + x} + \sqrt{a^{2} - x}}{\sqrt{a^{2} + x} - \sqrt{a^{2} - x}}

\frac{p}{q} = \frac{\sqrt{a^{2} + x}}{\sqrt{a^{2} - x}}

\frac{p^{2}}{q^{2}} = \frac{a^{2} + x}{a^{2} - x}

\frac{p^{2} + q^{2}}{p^{2} - q^{2}} = \frac{a^{2}}{x} = \frac{a^{2}}{\frac{4 a^{2} b}{4 b^{2} + 1}}

= a^{2} \times \frac{4 b^{2} + 1}{4 a^{2} b}

= \frac{4 b^{2} + 1}{4 b}

\frac{p^{2}}{q^{2}} = \frac{4 b^{2} + 4 b + 1}{4 b^{2} - 4 b + 1} = \frac{{(2 b + 1)}^{2}}{{(2 b - 1)}^{2}}

\frac{p}{q} = \frac{2 b + 1}{2 b - 1} ∣ \frac{p}{q} = \frac{1 + 2 b}{1 - 2 b}

\frac{p + q}{p - q} = \frac{4 b}{2} = 2 b ∣ \frac{p + q}{p - q} = \frac{2}{4 b} = \frac{1}{2 b}

Answered by som(math1967) last updated on 21/Sep/24

\frac{\sqrt{a^{2} + x} + \sqrt{a^{2} - x}}{\sqrt{a^{z} + x} - \sqrt{a^{2} - x}}

= \frac{{(\sqrt{a^{2} + x} + \sqrt{a^{2} - x})}^{2}}{{(\sqrt{a^{2} + x})}^{2} - {(\sqrt{a^{2} - x})}^{2}}

= \frac{a^{2} + x + a^{2} - x + 2 \sqrt{a^{4} - x^{2}}}{a^{2} + x - a^{2} + x}

= \frac{2 (a^{2} + \sqrt{a^{4} - x^{2}}}{2 x}

= \frac{a^{2} + \sqrt{a^{4} - \frac{16 a^{4} b^{2}}{{(4 b^{2} + 1)}^{2}}}}{\frac{4 a^{2} b}{4 b^{2} + 1}}

= \frac{a^{2} + a^{2} \sqrt{\frac{{(4 b^{2} + 1)}^{2} - 4.4 b^{2} . 1}{{(4 b^{2} + 1)}^{2}}}}{\frac{4 a^{2} b}{(4 b^{2} + 1)}}

= \frac{a^{2} (1 + \frac{4 b^{2} - 1}{4 b^{2} + 1})}{\frac{4 a^{2} b}{4 b^{2} + 1}}

= \frac{a^{2} \times 8 b^{2}}{4 b^{2} + 1} \times \frac{4 b^{2} + 1}{4 a^{2} b} = 2 b

Commented by som(math1967) last updated on 21/Sep/24

y e s s i r,

Commented by MATHEMATICSAM last updated on 21/Sep/24

sir the answer should be 2 b