(5+(5+(5+......)^1/2)^1/2)^1/2
Answers
Answered by
3
Ha! Finally I got Ur answer think it helps u if it helps plz mark my answer brainliest
Attachments:
![](https://hi-static.z-dn.net/files/d8f/8438f711c7e45fcb373f81263b9e76cd.png)
Answered by
3
question is -----> Evaluate : ![\sqrt{5+\sqrt{5+\sqrt{5+......}}} \sqrt{5+\sqrt{5+\sqrt{5+......}}}](https://tex.z-dn.net/?f=%5Csqrt%7B5%2B%5Csqrt%7B5%2B%5Csqrt%7B5%2B......%7D%7D%7D)
![x = \sqrt{5+\sqrt{5+\sqrt{5+......}}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{x} x = \sqrt{5+\sqrt{5+\sqrt{5+......}}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{x}](https://tex.z-dn.net/?f=+x+%3D+%5Csqrt%7B5%2B%5Csqrt%7B5%2B%5Csqrt%7B5%2B......%7D%7D%7D+%5C%5C+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5Cunderbrace%7Bx%7D)
or,![x=\sqrt{5+x} x=\sqrt{5+x}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B5%2Bx%7D)
squaring both sides,
x² = 5 + x
x² - x - 5 = 0
use quadratic formula to find value of x.
x =![\frac{-(-1)\pm\sqrt{(-1)^2-4(-5)(1)}}{2(1)} \frac{-(-1)\pm\sqrt{(-1)^2-4(-5)(1)}}{2(1)}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28-1%29%5Cpm%5Csqrt%7B%28-1%29%5E2-4%28-5%29%281%29%7D%7D%7B2%281%29%7D)
x=![\frac{1\pm\sqrt{21}}{2} \frac{1\pm\sqrt{21}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%5Cpm%5Csqrt%7B21%7D%7D%7B2%7D)
but here x ≥ 0 [ because every square root function is positive function. ]
so, x ≠![\frac{1-\sqrt{21}}{2} \frac{1-\sqrt{21}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1-%5Csqrt%7B21%7D%7D%7B2%7D)
hence, x =
or,
squaring both sides,
x² = 5 + x
x² - x - 5 = 0
use quadratic formula to find value of x.
x =
x=
but here x ≥ 0 [ because every square root function is positive function. ]
so, x ≠
hence, x =
Similar questions