x-1-2-find-x-1-x-2-1-

Question Number 157220 by MathSh last updated on 21/Oct/21

x = \frac{1}{2}

find \sqrt{x + 1 + \sqrt{x^{2} + 1}} = ?

Answered by Rasheed.Sindhi last updated on 21/Oct/21

x = \frac{1}{2}; \sqrt{x + 1 + \sqrt{x^{2} + 1}} = ?

\sqrt{x + 1 + \sqrt{x^{2} + 1}}

\sqrt{\frac{1}{2} + 1 + \sqrt{{(\frac{1}{2})}^{2} + 1}}

\sqrt{\frac{3}{2} + \sqrt{\frac{1 + 4}{4}}}

\sqrt{\frac{3}{2} + \frac{\sqrt{5}}{2}} = \sqrt{\frac{3 + \sqrt{5}}{2}} = \frac{\sqrt{3 + \sqrt{5}}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

= \frac{\sqrt{6 + 2 \sqrt{5}}}{2}

▸ \sqrt{6 + 2 \sqrt{5}} = p + q \sqrt{5} (S a y)

{(\sqrt{6 + 2 \sqrt{5}} = p + q \sqrt{5})}^{2}; p, q \in Q

= 6 + 2 \sqrt{5} = p^{2} + 2 p q \sqrt{5} + 5 q^{2}

p^{2} + 5 q^{2} = 6 \land 2 p q = 2

{(\frac{1}{q})}^{2} + 5 q^{2} = 6

\frac{1}{q^{2}} + 5 q^{2} = 6 \Rightarrow 5 q^{4} - 6 q^{2} + 1 = 0

(q - 1) (q + 1) (\underset{q \notin Q}{\underset{⏟}{5 q^{2} - 1}}) = 0

q = \pm 1 \Rightarrow p = \frac{1}{q} = \frac{1}{\pm 1} = \pm 1

\sqrt{6 + 2 \sqrt{5}} = p + q \sqrt{5} = \pm 1 \pm \sqrt{5}

= 1 + \sqrt{5}, - 1 - \sqrt{5}

x = \frac{\sqrt{6 + 2 \sqrt{5}}}{2} = \frac{1 + \sqrt{5}}{2}, \frac{- 1 - \sqrt{5}}{2}

Commented by MathSh last updated on 21/Oct/21

Thank you so much dear Ser

Commented by Tawa11 last updated on 21/Oct/21

Great sir

Commented by Rasheed.Sindhi last updated on 22/Oct/21

T h a n k s m i s s f o r e n c o u r a g i n g u s!