Question

The sum of the squares of the two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the number.

Answer 1

Hint: In this question we have the square of the two positive integers and the other one is the square of the larger number, so we have two equations. By solving those equations we can get the value of two positive integers.

Complete step-by-step answer:
Let x and y be two integer
Sum of the squares of two positive integer, is 208
\[{x^2} + {y^2} = 208 \cdot \cdot \cdot \left( i \right)\]
Also given that the square of the larger number is 18 times the smaller
${x^2} = 18y \cdot \cdot \cdot \left( {ii} \right)$
Substituting the value of ${x^2}$ from equation $\left( {ii} \right)$ in equation $\left( i \right)$ we have
$
  18y + {y^2} = 208 \\
  {y^2} + 18y - 208 = 0 \\
  {y^2} + 26y - 8y - 208 = 0 \\
  \left( {y - 8} \right)\left( {y + 26} \right) = 0 \\
  y = 8, - 26 \\
 $
Since we use positive integer $y = 8$
$
  {x^2} = 18y = 18 \times 8 = 144 \\
  x = \sqrt {144} \\
  x = 12 \\
 $
Thus, two positive integers are 12 and 8.

Note: Since we can get the combine equation by using both the equation then we can use the formula of finding x which is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and we can get the values of the integers.