Question

Quadratic equation pls solve it fast​

Answer 1

Question The sum of the poditive squares of two positive integers is 208. if the square of the larger number is 18 times the smaller number, find the numbers .

Solution

Let x and y be two positive integer .

as given in the question - sum of square of two number is 208 , we have :

${x}^{2} + {y}^{2} = 208 - - - > (1)$

Also given that square of the larger integer number is 18 times smaller number , we have:

${x}^{2} = 18y - - - - > (2)$

Substituting the value of x^2 from equation 2 in equation 1 . we have,

$18y + {y}^{2} = 208 \\ {y}^{2} + 18y = 208 \\ {y}^{2} + 18y - 208 = 0 \\ {y}^{2} + 26y - 8y - 208 = 0 \\ y(y + 26) - 8(y + 26) = 0 \\ (y - 8)(y + 26) = 0 \\ (y - 8) = 0 \\ or \\ (y + 26) = 0 \\ y = 8 \\ y = - 26 \\ since \: the \: integers \: are \: positve \: answr \: woud \: be : \\ y = 8 \\ {x}^{2} + {y}^{2} = 208 \\ {x}^{2} + {8 }^{2} = 208 \\ {x}^{2} + 64 = 208 \\ {x}^{2} = 208 - 64 \\ x = \sqrt{144} \\ x = 12$

hence ,  two positive numbers are 8 and 12

Answer 2

Answer:

the numbers are 8 ,12 or 8,- 12

refer to the above attachment for the solution.

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