Sqrt - -1-sqrt - -1-sqrt - -1-infty - - - -1-1omega - omega - 2 -

Question

Toppr Admin · Accepted Answer

Let x-x221A-x2212-1-x2212-x221A-x2212-1-x2212-x221A-x2212-1-x2212-x221E-Then x-x221A-x2212-1-x2212-x-x21D2-x2-x2212-1-x2212-x-x21D2-x2-x-1-0-x2234-x-x2212-1-xB1-x221A-1-x2212-4-1-12-1-x2212-1-xB1-x221A-x2212-32-x2212-1-xB1-x221A-32-x3C9- or -x3C9-2

$\sqrt{- 1 - \sqrt{- 1 - \sqrt{- 1 - . . . \infty}}} =$
$1$
$- 1$
$ω$
$ω^{2}$

Let $x = \sqrt{- 1 - \sqrt{- 1 - \sqrt{- 1 - . . . \infty}}}$

Then $x = \sqrt{- 1 - x}$

$\Rightarrow x^{2} = - 1 - x \Rightarrow x^{2} + x + 1 = 0$

$∴ x = \frac{- 1 \pm \sqrt{1 - 4.1.1}}{2.1} = \frac{- 1 \pm \sqrt{- 3}}{2}$

$= \frac{- 1 \pm \sqrt{3}}{2} = ω$ or $ω^{2}$

√−1−√−1−√−1−...∞=1−1ωω2

Let x=√−1−√−1−√−1−...∞Then x=√−1−x⇒x2=−1−x⇒x2+x+1=0∴x=−1±√1−4.1.12.1=−1±√−32=−1±√32=ω or ω2

$\sqrt{- 1 - \sqrt{- 1 - \sqrt{- 1 - . . . \infty}}} =$
$1$
$- 1$
$ω$
$ω^{2}$

Let $x = \sqrt{- 1 - \sqrt{- 1 - \sqrt{- 1 - . . . \infty}}}$

Then $x = \sqrt{- 1 - x}$

$\Rightarrow x^{2} = - 1 - x \Rightarrow x^{2} + x + 1 = 0$

$∴ x = \frac{- 1 \pm \sqrt{1 - 4.1.1}}{2.1} = \frac{- 1 \pm \sqrt{- 3}}{2}$

$= \frac{- 1 \pm \sqrt{3}}{2} = ω$ or $ω^{2}$