Question

There are three consecutive positive integers such that the sum of the squares of first and the product of other two is 154​

Answer 1

Answer:

$let \: the \: three \: consecutive \: positive \: integers \: </p><p></p><p>be \:( x - 1) \\ and \: x \: and \: (x + 1) \\ \\ according \: to \: the \: question \\ = > {(x - 1)}^{2} + x(x + 1) = 154 \\ = > {x}^{2} - 2x + 1 + {x}^{2} + x = 154 \\ = > 2 {x}^{2} - x - 153 = 0 \\ = > 2 {x}^{2} - 18x + 17x - 153 = 0 \\ = > 2x(x - 9) + 17(x - 9) = 0 \\ = > (x - 9)(2x + 17) = 0 \\ = > x = 9 \: \\ \: \: \: \: \: \: \: \: or \: \\ x = - \frac{17}{2}which \: is \: not \: possible \: since \: x \\ should \: be \: a \: positive \: integer. \\ \\ threrefore \: the \: three \: consecutive \: positive \: \\ integers \: are \: 8 \: and \: 9 \: and \: 10.$