Question

please ans it BRAINLIST COMFORMEDMark

Answer 1

Answer:

Step-by-step explanation:

let x=root6+root6........infinitt

by squaring

x2=6+root6+root6.....

x2=6+x

x2-x-6=0

x2-3x+2x-6=0

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

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

x=3,x=-2

So x=3

Hope it helps u mark as brainliest please

Answer 2

Heya

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$let \: the \: sum \: of \: this \: series \: be \: = x \\ \\ i.e \\ \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 \sqrt{ + \sqrt{6 + ...infinity \: } } } } } } } = x \\ \\ \\ \\ squaring \: both \: sides \: we \: get \\ \\ \\ x {}^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ...infinity} } } } \\ \\ \\ \\ x {}^{2} = 6 + x \\ \\ \\ becoz \: \: x \: = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ..infinit} } } } \\ \\ \\ \\ \\ x {}^{2} - x - 6 = 0 \\ \\ \\ \\ x {}^{2} - 3x +2x - 6 = 0 \\ \\ \\ x(x - 3) + 2(x - 3) = 0 \\ \\ \\ \\ (x - 3) = 0 \\ \\ \\ or \: \: \: \: (x + 2) = 0 \\ \\ \\ \\ \\ x = 3 \: \: \: \: or \: \: \: x \: = - 2 \\ \\ \\ \\ \\ x = - 2 \: will \: be \: rejected \: becoz \: \: sum \: of \: numbers \: in \:square \: even \: root \: is \: always \: even \: \\ \\ \\ \\ so \: x \: = 3$