Math, asked by Vickypanjiyar, 10 months ago

solve this question​

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Answers

Answered by umiko28
3

Step-by-step explanation:

y=-2(not possible)

y use

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Answered by Anonymous
3

Answer:

 \bf\ your \: answer \\ \\ \bf\\: y =  - 2(not \: possible) \: and \:y = 3

Step-by-step explanation:

\bf\ \:  \sqrt{ 6 + \sqrt{6 +  \sqrt{6 + ................ .........\infty } } }  \\  \\  \bf\ \: let  \:  \:  \:  \sqrt{6 +  \sqrt{6 +  \sqrt{6} } }  = y  \\  \\  \sf\  \therefore \sqrt{6 + y}  = y \\  \\  \sf\  \implies: 6 + y =  {y}^{2} \\  \\  \sf\  \implies: {y}^{2}  - y - 6 = 0 \\  \\  \sf\  \implies: {y}^{2}   - (3y - 2y) - 6 = 0 \\  \\ \sf\  \implies: {y}^{2} - 3y + 2y - 6 = 0 \\  \\ \sf\  \implies:y(y - 3) + 2(y - 3) = 0 \\  \\   \sf\boxed{   \sf\implies:(y - 3)(y + 2)  = 0} \\  \\  \\  \sf\ y - 3 = 0 \\  \\  \sf\  \implies:y  = 3 \\  \\  \sf\boxed{  \implies: \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ................ +  \infty } } }  = 3 }\\  \\   \bf\huge\red{or} \\  \\  \sf\  y  + 2 = 0 \\  \\ \sf\  \implies:y =  - 2 \\  \\   \sf\boxed{  \implies: \sqrt{6 +  \sqrt{6 +  \sqrt{6 + .................. +  \infty } }  =  - 2(not \: possible)}}

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