Question

the sum of the square of two consecutive natural numbers is113 find the numbers​

Answer 1

Let the 1st number be x & 2nd number be (x + 1).

$\large {x}^{2} + {(x + 1)}^{2} = 113 \\ = > {x}^{2} + {x}^{2} + 2x + 1 = 113 \\ = > \: \: \: \: \: \: \: \: \: \: \: {2x}^{2} + 2x = 113 - 1 \\ = > 2 {x}^{2} + 2x - 112 = 0 \\ = > 2( {x}^{2} + x - 56) = 0 \\ = > \: \: \: \: \: \: {x}^{2} + x - 56 = \frac{0}{2} \\ \\ = > {x}^{2} + x - 56 = 0 \\ = > {x}^{2} + 8x - 7x - 56 = 0 \\ = > x(x + 8) - 7(x + 8) = 0 \\ = > (x + 8)(x - 7) = 0 \\ \\ \therefore \: \: \: \: \: x = \red7 \: \: \: or \: \: \: ( - 8)$

Thus,

$\sf {1}^{st} \: \: \: number = \red8 \\ \sf {2}^{nd} \: \: number = \red9$