First slide

Theory of equations

Question

If $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . to infinity}}}, then x =$

Moderate

A

$\frac{1 + \sqrt{5}}{2}$
B

$\frac{1 - \sqrt{5}}{2}$
C

$\frac{1 \pm \sqrt{5}}{2}$
D

None of these

Solution

$x = \sqrt{1 + \sqrt{1 + \sqrt{1 + . . .}}} to \infty$
We have $x = \sqrt{1 + x}$
$\Rightarrow x^{2} = 1 + x \Rightarrow x^{2} - x - 1 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$
As $x > 0, we get x = \frac{1 + \sqrt{5}}{2}$

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