[tex] \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \infty } } } } [/tex]
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[tex] \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \infty } } } } [/tex]
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Verified answer
√{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.}
hance, the answer is x = 3
Verified answer
The ans is 3. because square roots won't be negative.:-)hope it helps u.