MCQ
The value of $\sqrt{6+\sqrt{6+\sqrt{6+}}} ...$ is :
- A$4$
- ✓$3$
- C$-2$
- D$3.5$
✓
Answer
Correct option: B.
$3$
Let $\text{x}=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$ \Rightarrow x^2=6+x $
$ \Rightarrow x^2-x-6=0 $
$ \Rightarrow x^2-3 x+2 x-6=0 $
$\Rightarrow x(x - 3) + 2(x - 3) = 0$
$\Rightarrow (x - 3)(x + 2) = 0$
Either $x - 3 = 0,$ then $x = 3$
Or $x + 2 = 0,$ then $x = -2$
Now if $x = 3,$ then
$3=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$=\sqrt{6+ 3}=\sqrt{9}$
$=3$
If $x = -2,$ then
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$\Rightarrow-2=\sqrt{6-2}$
$\Rightarrow-2=\sqrt{4}$
$\Rightarrow-2\neq2$
Which is not possible $x = 3$ is correct.
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$ \Rightarrow x^2=6+x $
$ \Rightarrow x^2-x-6=0 $
$ \Rightarrow x^2-3 x+2 x-6=0 $
$\Rightarrow x(x - 3) + 2(x - 3) = 0$
$\Rightarrow (x - 3)(x + 2) = 0$
Either $x - 3 = 0,$ then $x = 3$
Or $x + 2 = 0,$ then $x = -2$
Now if $x = 3,$ then
$3=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$=\sqrt{6+ 3}=\sqrt{9}$
$=3$
If $x = -2,$ then
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$\Rightarrow-2=\sqrt{6-2}$
$\Rightarrow-2=\sqrt{4}$
$\Rightarrow-2\neq2$
Which is not possible $x = 3$ is correct.
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