The sum of the squares of two positive integers is 208. If the square of the
larger number is 18 times the smaller number, find the numbers
Answers
✪ x = 12 , y = 8 ✪
☯ Let the two numbers be x,y , and here let the great no. be x . So , by the given information ,
♦ x² + y² = 208 ...eq(1)
♦ x² = 18y
So , now substituting x² = 18y in eq(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
⇒ y = - 26 or 8
Here , given that no.'s are positive , so here y = 8 . Now , finding x
→ x² = 18y
→ x = √(18y)
→ x = √(9×2)(y)
→ x = 3√(2y)
Now , substituting y = 8
=> x = 3√(2×8)
=> x = 3√16
=> x = 3√4²
=> x = 3 × 4
=> x = 12
Hence , x = 12 & y = 8 .
- P & Q are two positive integers.
- P is greater than Q.
- The sum of the squares of two positive integers is 208.
- The square of the larger number is 18 times to the smaller number.
- The numbers.
↝ Putting the value of P² in equation (1), we get
[NOTE ➛ It is given that both numbers are positive integers.]
↝ Putting the value of Q in the equation (2), we get
[NOTE ➛ It is given that number is a positive integer.]
The two positive integers are 12 & 8.