MCQ
If $x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \infty}}}$ and $x$ is a natural number, then
- A$x^2+x-2=0$
- B$x^2+2 x+2=0$
- ✓$x^2-x-2=0$
- D$x^2-x+2=0$
✓
Answer
Correct option: C.
$x^2-x-2=0$
(c) : We have, $x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \infty}}}$
$
\Rightarrow x=\sqrt{2+x} \quad[\because x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \infty}}}]
$
On squaring both sides, we get
$
x^2=2+x \Rightarrow x^2-x-2=0
$
$
\Rightarrow x=\sqrt{2+x} \quad[\because x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \infty}}}]
$
On squaring both sides, we get
$
x^2=2+x \Rightarrow x^2-x-2=0
$
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