The number of pairs of natural numbers the difference of whose squares is 45 will be ?

Answer:- 1

Explanation:-

Solution:

Let the numbers be x and y
According to question

\begin{aligned} (x > y) \\ x^{2} - y^{2} = 45 \\ (x + y) (x - y) = 45 \\ Make factor of 45 \\ 15 \times 3 \\ 9 \times 5 (3 pairs) \\ 45 \times 1 \end{aligned}

$\eqalign{ & \left( {x > y} \right) \cr & {x^2} - {y^2} = 45 \cr & \left( {x + y} \right)\left( {x - y} \right) = 45 \cr & {\text{Make factor of 45}} \cr & {\text{15}} \times 3 \cr & \,\,9 \times 5\,\,\,\,\,\,\,\,\,\,(3\,{\text{pairs}}) \cr & 45 \times 1 \cr}$
These pairs gives the value of x and y which satisfy the given condition.

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These pairs gives the value of x and y which satisfy the given condition. ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

Let the numbers be x and y
According to question