Question

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. Plz guys help me plz tommorow is my exam.!!

Answer 1

${x}^{2} + {y}^{2} = 208$
let X be larger number and y be smaller one
${x}^{2} = 18 \times y$
$18y + {y}^{2} = 208$
${y}^{2} + 18y - 208 = 0$
$y(y - 8) + 26(y - 8) = 0 \\ y = 8$
${x}^{2} = 18 \times 8 \\ {x}^{2} = 14 \\ x = 12$

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