If x-1-1-1- to infinity - then x-

The correct option is C x = √(1+√(1+√(1+....to nbsp;∞))) We have x = √(1+x) ⇒ x^2 = 1 + x nbsp;⇒ x^2 x 1 =0 ⇒ x = 1± √(1+4)/2 = nbsp;1± √(5)/2 As x ...

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Byju's Answer

Standard IX

Mathematics

Geometrical Method to Find Square Root of a Number

If x=√1+√1+√1...

Question

If x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . t o i n f i n i t y}}}$ , then x =

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Solution

The correct option is A

x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . t o \infty}}}$

We have x = $\sqrt{1 + x}$

⇒ $x^{2} = 1 + x$ ⇒ $x^{2} - x - 1 = 0$

⇒ $x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

As x > 0, we get $x = \frac{1 + \sqrt{5}}{2}$

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If x = √1+√1+√1+....to infinity, then x =

The correct option is A x = √1+√1+√1+....to ∞ We have x = √1+x ⇒ x2=1+x ⇒ x2−x−1=0 ⇒ x=1±√1+42=1±√52 As x > 0, we get x=1+√52

If x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . t o i n f i n i t y}}}$ , then x =

The correct option is A

x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . t o \infty}}}$

We have x = $\sqrt{1 + x}$

⇒ $x^{2} = 1 + x$ ⇒ $x^{2} - x - 1 = 0$

⇒ $x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

As x > 0, we get $x = \frac{1 + \sqrt{5}}{2}$