Question

The sum of the square of two positive integer is 208. if the square of the larger is 18 times the smaller number. Find the number?

Answer 1

let x and be the number such that x>y then ,

according to the statement ,

 x^2 + y^2 = 208   -(1)

and x^2 = 18y  - (2)

ie. x = (18y)^(1/2)

replacing in equation (1) 
ie. 18y + y^2 - 208 = 0

factorizing we can write 

y^2 + 26y - 8y -208 = 0

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

then y cannot be equal to -26 as square of x is always positive  from (2)

ie y = 8 and x = 12 

:P

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption,

Smaller number be p

Also,

Square of larger number = 18p

Square of smaller number = p²

Sum of squares of integers = 208

Situation,

⇒ p² + 18p = 208

⇒ p² + 18p - 208 = 0

Factorize we get,

⇒ p² + 26p - 8p - 208 = 0

⇒ p(p + 26) - 8(p + 26) = 0

⇒ (p + 26)(p - 8) = 0

⇒ p + 26 = 0

⇒ p = -25 (Not possible)

Or,

⇒ p - 8 = 0

⇒ p = 8

Square of larger number :-

= 18p

= 18 × 8

= 144

Larger number :-

= √144

= 12

Therefore,

Numbers are 8 and 12