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# 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:A. $14,10$ B. $12,6$ C. $12,8$ D. $16,12$

Last updated date: 04th Aug 2024
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Hint: Assume both the integers and form the equations of the two variables assumed according to the given in question and then solve both the equations to get both the integers.

The sum of the squares of two positive integers is $208$ . The square of the larger number is $18$ times the smaller number.
Assume that the larger integer be $x$ and the other or the smaller integer be $y$ .
As per the given condition, the sum of the squares of two positive integers is $208$ . This condition can be written in equation form as
${x^2} + {y^2} = 208\;\;\; \ldots \left( 1 \right)$ .
It is also given in the question that the square of the larger number is $18$ times the smaller number. This condition can be written in equation form as
${x^2} = 18y\;\;\;\; \ldots \left( 2 \right)$ .
Now substitute the value of ${x^2}$ from first equation to the second equation:
${x^2} + {y^2} = 208 \\ \Rightarrow {y^2} + 18y = 208 \\ \Rightarrow {y^2} + 18y - 208 = 0 \;$
Solve the quadratic equation:
${y^2} + 18y - 208 = 0 \\ \Rightarrow {y^2} + 26y - 8y - 208 = 0 \\ \Rightarrow y\left( {y + 26} \right) - 8\left( {y + 26} \right) = 0 \\ \Rightarrow \left( {y - 8} \right)\left( {y - 26} \right) = 0 \\ \Rightarrow y = 8, - 26 \;$
As the square of a number can’t be a negative value. So, neglect the value $y = - 26$ .
Now substitute the value of $y = 8$ in the second equation:
${x^2} = 18y \\ \Rightarrow {x^2} = 18\left( 8 \right) \\ \Rightarrow {x^2} = 144 \\ \Rightarrow x = \sqrt {144} \\ \Rightarrow x = 12 \;$
So, the two numbers are $12,8$ .
So, the correct answer is “Option C”.

Note: As the square of any number can’t be equal to a negative number. So, we need to ignore that value and then take it to find the value of $x$ and also need to take only the positive value of under root.