Solution:-
Let us assume the two numbers be A and B, B being the bigger number.
As per the condition given in the question,
![\[2{{B}^{2}}~=\text{ }3{{A}^{2}}~+\text{ }29\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5e0d63e6fe799d6e2b004f35b90b5da_l3.png "Rendered by QuickLaTeX.com")
Now, consider the equation (i),
Then, substitute the value of B is equation (ii),
![\[2{{\left( 25-A \right)}^{2}}~=\text{ }3{{A}^{2}}~+\text{ }29\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e80ce902841f6a9e6aee11374eb44462_l3.png "Rendered by QuickLaTeX.com")
We know that,
![\[{{\left( a-b \right)}^{2}}~=\text{ }{{a}^{2}}~-2ab\text{ }+\text{ }{{b}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-25ac56a216f7b29b167aa383a28e7c9e_l3.png "Rendered by QuickLaTeX.com")
![\[2({{25}^{2}}-\left( 2\text{ }\times \text{ }25\text{ }\times \text{ }A \right)\text{ }+\text{ }{{A}^{2}})\text{ }=\text{ }3{{A}^{2}}~+\text{ }29\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b82540f4de5dbe22faf86a4fa4ff7b6a_l3.png "Rendered by QuickLaTeX.com")
![\[1250\text{ }+\text{ }2{{A}^{2}}-100A\text{ }=\text{ }3{{A}^{2}}~+\text{ }29\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-84a3c91f019f57349a205a7a005c4d5e_l3.png "Rendered by QuickLaTeX.com")
![\[{{A}^{2}}~+\text{ }100A-1221\text{ }=\text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7717d058e13752fa86289cefc8028602_l3.png "Rendered by QuickLaTeX.com")
![\[{{A}^{2}}~-11A\text{ }+\text{ }111A-1221\text{ }=\text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5916eb3af3132b25ab9fe92229554c5_l3.png "Rendered by QuickLaTeX.com")
Take out common in each terms,
Equate both to zero,
Therefore,
So,
