Math, asked by sailekshmi, 1 year ago

pls solve this... as much fast as you can

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smartboyhemant: hiii
smartboyhemant: in which class you read

Answers

Answered by Anonymous
1

 \sqrt{6  + \sqrt{6  + \sqrt{6}..... } }  \\ let \: it \: be \: x \\  \sqrt{6 +  \sqrt{6 +  \sqrt{6.......} }  }  = x \\  {x}^{2}  = 6 +  \sqrt{6 +  \sqrt{6.....} }  \\ now \: we \: have \: considered \:  \\  \sqrt{6 +  \sqrt{6 +  \sqrt{6... = x} } }  \\ hence \:  {x}^{2}  = 6 + x \\  {x}^{2}  - x - 6 = 0 \\  {x }^{2}   - 3x + 2x - 6 = 0 \\ x(x - 3) + 2(x - 3) = 0 \\ x =  - 2 \: or \: x = 3 \\ so \: the \: value \: of \: x \: can \: not \: be  \\  negative \: because \: it \: is \: sum \: of  \\ \: positive \: roots \: hence \:  \\ x = 3

sailekshmi: But they have asked what is the value of p
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