Question

$\sqrt{6 + \sqrt{6 + \sqrt{6 + ... = \infty } } }$
is equal to
​

Answer 1

$\huge{\underline{\mathfrak{\green{Answer}}}}$

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

$\boxed{\bold{\red{Assume\:that:}}}$

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

$\boxed{\bold{\red{Then:}}}$

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

$\boxed{\red{\bold{Squaring \:both\: sides:}}}$

$6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } } = x^2$

$6 + x = x^2$

$x^2 - x - 6 = 0$

$x^2 - 3x + 2x - 6 = 0$

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

$\boxed{\bold{\red{By\:Grouping \:Method:}}}$

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

$x = 3 \; or\; x = -2$

$Root \:can \:never \:be \:negative\: ,\: so \:x = 3$

Hence , $\sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } } = 3$

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Answer 2

Step-by-step explanation:

√{6 + x} = x

6 + x = x²

x² - x - 6 = 0

x² - 3x + 2x - 6 = 0

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

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

x = - 2, 3

because x ≠ -2 { square root can't be negative.}