Let the consecutive positive integers be x and x + 1. Either x + 14 = 0 or x − 13 = 0, i.e., x _{ }= −14 or x _{ }= 13 Since the integers are positive, x can only be 13. ∴ x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14. Posted by Ishan Goyal(student)on 1/11/12 This conversation is already closed by Expert

Let the consecutive positive integers be x and x + 1. Either x + 14 = 0 or x − 13 = 0, i.e., x _{ }= −14 or x _{ }= 13 Since the integers are positive, x can only be 13. ∴ x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14. Posted by Ishan Goyal(student)on 1/11/12 This conversation is already closed by Expert

Let the consecutive positive integers be x and x + 1. Either x + 14 = 0 or x − 13 = 0, i.e., x _{ }= −14 or x _{ }= 13 Since the integers are positive, x can only be 13. ∴ x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14. Posted by Ishan Goyal(student)on 1/11/12 This conversation is already closed by Expert

Let The Numbers Be x And x+1 Squares x^{2} And (x+1)^{2} Sum = 365 Hence x^{2}+(x+1)^{2} = 365 x^{ 2 }+x^{2}+1+2x = 365 2x^{2}+2x+1 = 365 2x^{2}+2x-364 = 0 x^{2}+x-182 = 0 Use Quadratic Formula -b+Square Root (b^{2}-4ac) / 2a And -b-Square Root (b^{2}-4ac) / 2a -1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2 -1+27/2 And -1-27/2 13 And -14 As Positive Integers So 13 Other Integer = 13+1 = 14 Posted by Prakhar Bindal(student)on 1/11/12

Let The Numbers Be x And x+1 Squares x^{2} And (x+1)^{2} Sum = 365 Hence x^{2}+(x+1)^{2} = 365 x^{ 2 }+x^{2}+1+2x = 365 2x^{2}+2x+1 = 365 2x^{2}+2x-364 = 0 x^{2}+x-182 = 0 Use Quadratic Formula -b+Square Root (b^{2}-4ac) / 2a And -b-Square Root (b^{2}-4ac) / 2a -1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2 -1+27/2 And -1-27/2 13 And -14 As Positive Integers So 13 Other Integer = 13+1 = 14 Posted by Prakhar Bindal(student)on 1/11/12

Let The Numbers Be x And x+1 Squares x^{2} And (x+1)^{2} Sum = 365 Hence x^{2}+(x+1)^{2} = 365 x^{ 2 }+x^{2}+1+2x = 365 2x^{2}+2x+1 = 365 2x^{2}+2x-364 = 0 x^{2}+x-182 = 0 Use Quadratic Formula -b+Square Root (b^{2}-4ac) / 2a And -b-Square Root (b^{2}-4ac) / 2a -1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2 -1+27/2 And -1-27/2 13 And -14 As Positive Integers So 13 Other Integer = 13+1 = 14 Posted by Prakhar Bindal(student)on 1/11/12