Question

please solve question no.2

Answer 1

$(i) \: \: x + {x}^{2} = 30 \\ {x}^{2} + x - 30 = 0$
$(ii) \: \: y(y - 3) = 42 \\ {y}^{2} - 3y - 42 = 0$
$(iii) \: \: x + \frac{1}{x} = \frac{37}{6} \\ \frac{ {x}^{2} + 1}{x} = \frac{37}{6} \\ 6 {x}^{2} - 37x + 6 = 0$
$(iv) \: \: let \: digit \: at \: unit \: place \: be \: x \\ and \: digit \: at \: tens \: place \: be \: y \\ given \: \: y = {x}^{2} + 5 - - - (1)\\ also \: \: 10y + x = 61 - - - (2) \\ substituting \: \: y \: in \: \: (1) \: to \: (2) \\ 10( {x}^{2} + 5) + x = 61 \\ 10 {x}^{2} + x - 11 = 0$
$(v) \: \: length \: is \: \: x \\ breadth \: is \: \: x - 3\\ area \: x( x - 3) = 70 \\ {x}^{2} - 3x - 70 = 0$