Question

Find two consecutive positive interes sum of whose square is 365

Answer 1

Answer:

X=13

y=14

Step-by-step explanation:

two consecutive positive integers are (X) and (X+1)

(X)^2+(X+1)^2= 365

(a+b)^2= x^2+2ab+b^2

x^2+x^2+2x+(1)^2=365

2x^2+2x+1-365=0

2x^2+2x-364=0

2(x^2+x-364)=0

x^2+x-364=0

x^2+(14-13)x-182=0

x^2+14x-13x-182=0

x(X+14)-13(X+14)=0

(x+14)(x-13)=0

X+14=0

X=-14(invalid)

x-13=0

X=13.

x=13

X+1=13+1

=14.