Question

The sum of the squares of two positive integers is 208. If the square of the large number is 18 times the smaller. Find the numbers

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

Let x be the smaller number.

So, the square of larger number = 18x

A/C,

Square of smaller number + Square of larger number = 208

$\longrightarrow$$\sf\red{ {x}^{2} + 18x = 208}$

$\longrightarrow$$\sf\red{{x}^{2} + 18x - 208 = 0}$

$\longrightarrow$$\sf\red{{x}^{2} + 26x - 8x - 208 = 0}$

$\longrightarrow$$\sf\red{x(x + 26) - 8(x + 26) = 0}$

$\longrightarrow$$\sf\red{(x + 26)(x - 8) = 0}$

$\longrightarrow$$\sf\red{x = 8, x = - 26}$

As x is a positive integer, x ≠ -26 ,so x = 8

$\longrightarrow$ Smaller number = 8 and

Smaller number = 8 andLarger number = √18×8 = 12