Question

Find two consecutive positive integers sum of whose square is 421

Answer 1

let one integer be x
second integer=x+1

A.T.Q.

${x}^{2} + {(x + 1)}^{2} = 421 \\ {x}^{2} + {x}^{2} + 1 = 421 \\ 2 {x}^{2} = 421 - 1 = 420 \\ {x}^{2} = \frac{420}{2} = 210 \\ {x}^{2} = 210 \\ x = \sqrt{210} = 14.4913767$
hope tis will help u

first integer=14.4913767
second integer=14.4913767+1=15.4913767

Answer 2

let the 1st no. be x
other no. is (x+1)
ATQ
(x)^2 + (x +1)^2 = 421
x^2 + x^2 + 1 + 2x = 421
2x^2 + 2x +1 -421 = 0
2x^2 + 2x - 420 = 0
x^2 + x - 210 = 0
x^2 + 15x - 14x - 210 = 0
x( x + 15) -14(x+15) = 0
(x-14)(x+15)=0
x = 14 x= -15
x can not be equal to -15 because x is a positive integer
So x = 14
So the no.s are 14 and 15