Question

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Answer 1

Answer:

-2,3

Step-by-step explanation:

Let x=√6(+√6.....

Then,

x=√(6+x)

x²=6+x

x²-x-6=0

x²-3x+2x-6=0

x(x-3)+2(x-3)=0

(x+2)(x-3)=0

So, x= -2 , 3

Answer 2

$\red{ \underline {\underline{ \star \: {Given:-}}}}$

The Value of

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 ..... \: \: is } } } }$

$\red{ \underline {\underline{ \star \: {Solution:-}}}}$

${ \large \star}\: Let \: the \: \sqrt{6........} be \: \: \: x \:$

So,

Our equation will be :-

$x = \sqrt{6 + x}$

Squaring both side ,we get.

${x}^{2} = {(\sqrt{6 + x} )}^{2} \\ \\ \implies {x}^{2} = 6 + x. \\ \\ \implies {x}^{2} - x - 6. \\ \\ \implies {x}^{2} - 3x + 2x - 6. \\ \\ \implies x(x - 3) + 2(x - 3). \\ \\ \implies (x + 2)(x - 3)$

So,

You can say that Either.

$\large\tt{x = (- 2 \: \: \: and \: \: \: 3)}$

$\longrightarrow \red{\huge\underline {\boxed{x = 3.}}}$