Question

Find the value of
$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..} } } }$
​

Answer 1

Answer:

3,-2

Step-by-step explanation:

$\rm \: Let \: x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..} } } } \\ \rm x = \sqrt{6 + x} \\ \rm \: Squaring \: on \: both \: sides \\ \rm \: {x}^{2} = 6 + x \\ \rm \dashrightarrow \: {x }^{2} - x - 6 = 0 \\ \rm \dashrightarrow {x}^{2} - 3x + 2x - 6 = 0 \\ \rm \dashrightarrow \: x(x - 3) + 2(x - 3) = 0 \\ \rm \dashrightarrow \:(x - 3)( x + 2) = 0 \\ \\ \bold{If \: x - 3 = 0} \\ \huge \red{ \rm \: x = 3} \\ \\ \bold{If \: x + 2 = 0} \\ \huge \green{ \rm \: x = - 2}$

[Both the value of x is correct]

Answer 2

Solution :

We have to find the value of √(6+√(6+√(6+ .. )).

Let this be equal to x.

> x = √(6+√(6+√(6+ .. ))

> x = √(6+ x)

Squaring this

x² = 6 + x

> x² - x - 6 = 0

> x² - 3x + 2x - 6 = 0

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

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

> x = -2 and x = 3

But the value of x can't be negative .

Answer - The value of the given expression is 3

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Additional Information :

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\ ^{a} log \ a= 1\\\\\circ \ ^{a}log \ 1 = 0 \\\\\circ \ ^{a ^{n}} log \ b^{m}= \dfrac{m}{n} \times\:^{a}log \ b \\\\\circ \ ^{a^{m}} \ log \ b^{m} = \ ^{a}log \ b \\\\\circ \ ^{a}log \ b = \dfrac{1}{^{b}log \ a} \\\\\circ \ ^{a}log \ b = \dfrac{^{m}log \ b}{^{m} log \ a} \\\\\circ \ a^{^{a} logb} = b \\\\\circ \ ^{a}log \ b + ^{a}log \ c = \ ^{a}log(bc) \\\\\circ \ ^{a}log \ b -\: ^{a}log \ c = \ ^{a}log \left( \dfrac{b}{c} \right) \\\circ \ ^{a}log \ b \:\cdot\: ^{a}log \ c = \ ^{a}log \ c \\\\\circ \ ^{a}log \left( \dfrac{b}{c} \right) = \ ^{a}log \left(\dfrac{c}{b}\right)\end{minipage}}}$

$\boxed{\boxed{\begin{minipage}{4.2cm}\circ\sf\:\ln 1=0\\\circ\sf\:\ln e=1\\\circ\sf\:\ln x=y\leftrightarrow e^y=x\\\circ\sf\:e^{\ln x}=x,\:x>0\\\circ\sf\:\ln(e^x)=x,\:x\in\mathbb{\large R}\\\circ\sf\:\ln(xy)=\ln x+\ln y\\\circ\sf\:\ln(x/y)=\ln x-\ln y\\\circ\sf\:\ln(x^r)=r\ln x\\\circ\sf\:\ln x=log_e\:x\end{minipage}}}$

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