Question

two consecutive positive numbers are such that the sum of theirs squares 113. find the two numbers

Answer 1

as the numbers are consecutive ..
their difference is 1 ....
let the two numbers be X and y ...
x-y =1
X =y+1
.....

as their sum of squares is 113...
X square +y square =113
substitute X =y+1 in above equation
y+1 square +y square =113


after solving ulll get the answer........




hope it helps you

Answer 2

Let the two consecutive positive numbers be X and X+1...

So,
${x }^{2} + {(x + 1)}^{2} = 113 \\ {x}^{2} + {x}^{2} + 1 + 2x = 113 \\ 2 {x}^{2} + 2x - 112 = 0 \\ {x}^{2} + x - 56 = 0 \\ {x}^{2} + 8x - 7x - 56 = 0 \\ x(x + 8) - 7(x + 8) = 0 \\ (x - 7)(x + 8) = 0 \\ \\ x = 7 \: or \: x = - 8$