[tex]\sf{\sqrt{1 + \sqrt{1+ \sqrt{1 + ...\infty}}}}[/tex] Where, x > 0
To solve this problem, let that :
[tex]\implies \sf{\alpha = \sqrt{1 + \sqrt{1 + \sqrt{1 + ...\infty}}}}[/tex]
By squaring the equation we get :
[tex]\implies \sf{{\alpha}^{2} = {\bigg(\sqrt{1 + \sqrt{1 + \sqrt{1 +...\infty}}}\bigg)}^{2}}[/tex]
[tex]\implies \sf{{\alpha}^{2} = 1 + {\sqrt{1 + \sqrt{1 + \sqrt{1 + ...\infty}}}}}[/tex]
Now since [tex]\sf{\sqrt{1 + \sqrt{1 + \sqrt{1 + ...\infty}}} = alpha}[/tex], so we can substitute it for that, we get :
[tex]\implies \sf{{\alpha}^{2} = 1 + \alpha}[/tex]
By shuffling a bit, we get :
[tex]\implies \sf{{\alpha}^{2} - \alpha - 1 = 0}[/tex]
Now using the quadratic formula, i.e,
[tex]\boxed{\sf{\alpha = \dfrac{-b \pm \sqrt{{b}^{2} - 4ac}}{2a}}}[/tex]
Which gives :
[tex]\implies \sf{\alpha = \dfrac{-1 \pm \sqrt{{1}^{2} - 41}}{2}}[/tex]
[tex]\implies \sf{\alpha = \dfrac{-1 \pm \sqrt{-3}}{2}}[/tex]
[tex]\implies \sf{\alpha = \dfrac{-1 \pm \sqrt{3}i}{2}}[/tex]
Where i is the iota, which stands for √-1.
So we got the result, now you can apply the same method for other parts except (i), try it once :p
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Answers & Comments
Solution :
[tex]\sf{\sqrt{1 + \sqrt{1+ \sqrt{1 + ...\infty}}}}[/tex] Where, x > 0
To solve this problem, let that :
[tex]\implies \sf{\alpha = \sqrt{1 + \sqrt{1 + \sqrt{1 + ...\infty}}}}[/tex]
By squaring the equation we get :
[tex]\implies \sf{{\alpha}^{2} = {\bigg(\sqrt{1 + \sqrt{1 + \sqrt{1 +...\infty}}}\bigg)}^{2}}[/tex]
[tex]\implies \sf{{\alpha}^{2} = 1 + {\sqrt{1 + \sqrt{1 + \sqrt{1 + ...\infty}}}}}[/tex]
Now since [tex]\sf{\sqrt{1 + \sqrt{1 + \sqrt{1 + ...\infty}}} = alpha}[/tex], so we can substitute it for that, we get :
[tex]\implies \sf{{\alpha}^{2} = 1 + \alpha}[/tex]
By shuffling a bit, we get :
[tex]\implies \sf{{\alpha}^{2} - \alpha - 1 = 0}[/tex]
Now using the quadratic formula, i.e,
[tex]\boxed{\sf{\alpha = \dfrac{-b \pm \sqrt{{b}^{2} - 4ac}}{2a}}}[/tex]
Which gives :
[tex]\implies \sf{\alpha = \dfrac{-1 \pm \sqrt{{1}^{2} - 41}}{2}}[/tex]
[tex]\implies \sf{\alpha = \dfrac{-1 \pm \sqrt{-3}}{2}}[/tex]
[tex]\implies \sf{\alpha = \dfrac{-1 \pm \sqrt{3}i}{2}}[/tex]
Where i is the iota, which stands for √-1.
So we got the result, now you can apply the same method for other parts except (i), try it once :p