Math, asked by frooti24, 1 year ago

please solve it now it's my homework

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Answered by Mankuthemonkey01
6
Let

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


Now square both the sides

 =  > ( \sqrt{6 +  \sqrt{6 +  \sqrt{6....} } } ) {}^{2}  = x {}^{2}  \\  \\  =  > 6 +  \sqrt{6 +  \sqrt{6 +  \sqrt{6...} } }  =  {x}^{2}


But we have assumed that

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

Hence,
6 +  \sqrt{6 +  \sqrt{6 +  \sqrt{6.....} } }  =  {x}^{2}   \\  \\ can \: be \: written \: as \:  \\  \\ 6 + x =  {x}^{2}

Now solve the quadratic equation

 {x}^{2}  = 6 + x \\  \\  =  >  {x}^{2}  - x - 6 = 0 \\  \\  =  >  {x}^{2}  - 3x + 2x - 6 = 0 \\ \\   =  > x(x - 3) + 2(x - 3) = 0 \\  \\  =  > (x + 2)(x - 3) = 0

Now,

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

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

=> x + 2 = 0

=> x = -2


Similarly,

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

=> x - 3 = 0

=> x = 3

So x = -2, 3

We generally take the positive greater value.

=> x = 3


Answer :- 3
Answered by Anonymous
3
ĀNSWĒR ⬆⬆


THANKS ✌☺


#DILDAAR NAVI ♥
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