Question

the sum of the square of 2 positive integers is 208 .if the square of the integers is 8 times smaller .find the number

Answer 1

let the numbers be x,y (x>y) here x2=18yaccording to the problemx2+y2=208....................(1) 18y+y2=208y2+18y-208=0 y2+26y-8y-208=0y(y+26)-8(y+26)=0 (y+26)(y-8)=0 as the numbers are +ve y-8=0 y=8 substitute y in (1) x2+(8)2=208 x2+64=208 x2=208-64x2=144 x=12 the nos. are 12,8

Answer 2

Hey mate!!

let the numbers be x,y (x>y) here

x² = 18y

according to the problem

x² +y² = 208.................(1)

18y+y2=208

y2+18y-208=0

 y2+26y-8y-208=0

y(y+26)-8(y+26)=0 

(y+26)(y-8)=0

 as the numbers are +ve

 y-8=0 =>  y=8 

substitute y in (1) x2+(8)2=208

 x2+64=208 

x2=208-64

x2=144 

x = √144

x=12 

Hence, The required nymbers are 12,8.

Thankyou!