Question

$find \: \: the \: value \: of \sqrt{6 + \sqrt{6 \tim \di + \sqrt{6 + ......} } }$
solve this question please​

Answer 1

Let the value of $\tt \sqrt{6+\sqrt{6+\sqrt{6+....}}}= x$

Now solving the Question

$\tt \sqrt{6+x}= x \ \ \ \ [\because \sqrt{6+\sqrt{6+\sqrt{6+....}}}= x] \\\\ \tt Squaring \ \ Both \ \ sides \\\\ \Rightarrow \tt (\sqrt{6+x})^2=x^2 \\\\ \tt \Rightarrow 6 + x = x^2 \\\\ \tt \Rightarrow x^2 -x-6 = 0 \\\\ \tt \Rightarrow x^2+2x-3x-6=0 \\\\ \tt \Rightarrow x(x+2)-3(x+2) =0 \\\\ \tt \Rightarrow (x+2)(x-3) = 0$

Equating the factors with Zero

$\tt x +2 =0 \\\\ \tt \Rightarrow x = -2$

and

$\tt x -3 =0 \\\\ \tt \Rightarrow x = 3$

So, the answer is 3,

-2 is not the answer because we cannot take a negative number inside a root.

Points To Remember While Solving this type of Question:

• If the answer is in negative, Ignore the negative value as square root of negative number don't fall in Real Numbers
• You can use Quadratic method/Formula also to solve the Quadratic Equation.

Answer 2

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QUESTION:

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

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ANSWER:

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6.....} } } } = 3$

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GIVEN:

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

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TO FIND:

$The \: \: value \: \: of \: \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6.....} } } }$

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EXPLANATION:

$Let \: \: x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6.....} } } }$

As the term has no end we can consider that $x = \sqrt{6 + x }$

Squaring on both sides of the equation

${x}^{2} = {\bigg( \sqrt{6 + x} \bigg)}^{2}$

x² = 6 + x

x² - x - 6 = 0

By adding and subtracting 2x

Since adding and subtracting the same term will have no change and will not affect the original value.

x² - 3x + 2x - 6 = 0

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

Take x - 3 as common

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

x = 3 (or) -2

$HENCE \: \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6.....} } } } = 3$

NOTE : HERE WE SHOULD NOT TAKE x = - 2 BECAUSE AS THERE IS A SQUARE ROOT HENCE - 2 IS NEGLECTED.

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