Question

Please answer this question
find the value of x.

Answer 1

Hola there

Given:
$x = \sqrt{6 + \sqrt {6 + \sqrt {6 + ...... \infty } } }$
$x = \sqrt{6 + x}$
${x}^{2} = 6 + x$
${x }^{2} - x - 6 = 0$
${x}^{2} - 3x + 2x - 6$
$x(x - 3) + 2(x - 3) = 0$
$(x - 3)(x + 2) = 0$
$x = - 2 \: and \: 3$
Hope this helps...:)

Answer 2

Hi friend, Harish here,

Here is your answer.

$x = \sqrt{6+ \sqrt{6+ \sqrt{6+ \sqrt{6 +............\infty} } } } = \sqrt{6+x}$

⇒  $x = \sqrt{6+x}$

⇒ Infinity is not defined, So, it will be continuing till the end, So the rest of the part is also x. 

That's  y we got the above equation..

Now square on both sides.

Then,

⇒ $x^{2} = 6+x$

⇒ $x^{2} -6-x=0$

⇒ $x^{2} +2x -3x -6 =0$

⇒ $x(x+2)-3(x+2) = 0$

⇒ $(x-3)(x+2)=0$

From the equation we get values of x as 3 & -2.

As x cannot have a -ve value,

Therefore x can be 3 .
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Hope my answer is helpful to you.