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Hint: The numbers can be determined by assuming them as two variables, then forming two equations using the information from the problem statements and then solving the two equations to find the values of the variables.

To determine the numbers, we will form equations using the statements from the question and solve them.

First, we will start by taking the two numbers as $x$ and $y$, such that $x>y$ .

Now, we will form an equation using the information from the question.

The first statement says that difference of the squares of two numbers is $45.$

So, ${{x}^{2}}-{{y}^{2}}=45...........\left( i \right)$

The second statement says that the square of the smaller number is $4$ times the larger number.

So, ${{y}^{2}}=4x...........\left( ii \right)$

Now, we will substitute the value of ${{y}^{2}}$ from equation $\left( ii \right)$ in equation $\left( i \right)$ .

On substituting value of ${{y}^{2}}$ from equation $\left( ii \right)$ in equation $\left( i \right)$, we get;

${{x}^{2}}-4x=45$

So, ${{x}^{2}}-4x-45=0.......(iii)$

Clearly, we can see equation$(iii)$ is a quadratic in $x$.

Now, we will use middle term splitting to solve the quadratic.

On splitting the middle term into two numbers such that their sum $=-4x$ and their product $=-45{{x}^{2}}$, we get

${{x}^{2}}-9x+5x-45=0$

Taking $x$ common from first two terms and $5$ common from last two terms we get

$x\left( x-9 \right)+5\left( x-9 \right)=0$

$\Rightarrow \left( x-9 \right)\left( x+5 \right)=0$

So , $\text{ }x=9\text{ or }x=-5$

Now, the second statement square of the smaller number is $4$times the larger number and we also know that the larger number cannot be negative as the square of any number is always positive. So, the smaller number cannot be negative.

So,$\text{ }x\ne -5$

Hence, $x=9$

Now, substituting $x=9$ in equation $\left( ii \right)$,

$\begin{align}

& {{y}^{2}}=4\times 9 \\

& \Rightarrow {{y}^{2}}=36 \\

& \Rightarrow y=6 \\

\end{align}$

Hence, $9$ and $6$ are the two numbers.

Note: While writing the equation, remember which number is larger if the first equation is made taking $x>y$, then the second equation must be made taking $x>y$. Otherwise the solution of the equation will give wrong values or imaginary values.

To determine the numbers, we will form equations using the statements from the question and solve them.

First, we will start by taking the two numbers as $x$ and $y$, such that $x>y$ .

Now, we will form an equation using the information from the question.

The first statement says that difference of the squares of two numbers is $45.$

So, ${{x}^{2}}-{{y}^{2}}=45...........\left( i \right)$

The second statement says that the square of the smaller number is $4$ times the larger number.

So, ${{y}^{2}}=4x...........\left( ii \right)$

Now, we will substitute the value of ${{y}^{2}}$ from equation $\left( ii \right)$ in equation $\left( i \right)$ .

On substituting value of ${{y}^{2}}$ from equation $\left( ii \right)$ in equation $\left( i \right)$, we get;

${{x}^{2}}-4x=45$

So, ${{x}^{2}}-4x-45=0.......(iii)$

Clearly, we can see equation$(iii)$ is a quadratic in $x$.

Now, we will use middle term splitting to solve the quadratic.

On splitting the middle term into two numbers such that their sum $=-4x$ and their product $=-45{{x}^{2}}$, we get

${{x}^{2}}-9x+5x-45=0$

Taking $x$ common from first two terms and $5$ common from last two terms we get

$x\left( x-9 \right)+5\left( x-9 \right)=0$

$\Rightarrow \left( x-9 \right)\left( x+5 \right)=0$

So , $\text{ }x=9\text{ or }x=-5$

Now, the second statement square of the smaller number is $4$times the larger number and we also know that the larger number cannot be negative as the square of any number is always positive. So, the smaller number cannot be negative.

So,$\text{ }x\ne -5$

Hence, $x=9$

Now, substituting $x=9$ in equation $\left( ii \right)$,

$\begin{align}

& {{y}^{2}}=4\times 9 \\

& \Rightarrow {{y}^{2}}=36 \\

& \Rightarrow y=6 \\

\end{align}$

Hence, $9$ and $6$ are the two numbers.

Note: While writing the equation, remember which number is larger if the first equation is made taking $x>y$, then the second equation must be made taking $x>y$. Otherwise the solution of the equation will give wrong values or imaginary values.

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