The sum of squares of two numbers is equal to 5 times their sum. The sum of the reciprocals of the squares of these two numbers is $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers. Find the numbers.
Answer
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Hint: We have been given two numbers and two conditions on those numbers. We assume those two numbers as variables. We represent those given conditions in the algebraic form of two variables. So, we got two equations of two variables. Solving those two equations we get the numbers.
Complete step-by-step answer:
Let’s assume that the numbers are x and y.
First condition is that the sum of squares of two numbers is equal to 5 times their sum.
So, the sum of squares of two numbers will be ${{x}^{2}}+{{y}^{2}}$ which is equal to 5 times their sum.
So, the equation becomes ${{x}^{2}}+{{y}^{2}}=5\left( x+y \right)$ …(i)
From the equation we get $\dfrac{{{x}^{2}}+{{y}^{2}}}{5\left( x+y \right)}=1$ …(ii)
Second condition is the sum of the reciprocals of the squares of these two numbers is $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers.
The squares of these numbers are ${{x}^{2}}$ and ${{y}^{2}}$. Reciprocals of the numbers are $\dfrac{1}{{{x}^{2}}}$ and $\dfrac{1}{{{y}^{2}}}$.
So, sum of the reciprocals of the squares of these two numbers $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}$ which is equal to $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers.
So, $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers is $\dfrac{5}{18}\left( \dfrac{1}{x}+\dfrac{1}{y} \right)$.
So, the equation becomes $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}=\dfrac{5}{18}\left( \dfrac{1}{x}+\dfrac{1}{y} \right)$ …(iii)
From the equation we get
\[\begin{align}
& \dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}=\dfrac{5}{18}\left( \dfrac{1}{x}+\dfrac{1}{y} \right) \\
& \Rightarrow \dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}{{y}^{2}}}=\dfrac{5}{18}\times \dfrac{x+y}{xy}=\dfrac{5\left( x+y \right)}{18xy} \\
& \Rightarrow \dfrac{{{x}^{2}}+{{y}^{2}}}{5\left( x+y \right)}=\dfrac{{{x}^{2}}{{y}^{2}}}{18xy}=\dfrac{xy}{18} \\
\end{align}\]
We replace the left-hand side part from equation (ii).
Replacing we get \[\dfrac{{{x}^{2}}+{{y}^{2}}}{5\left( x+y \right)}=\dfrac{xy}{18}\Rightarrow \dfrac{xy}{18}=1\Rightarrow xy=18\].
So, the multiplication of two numbers is 1 which means they are reciprocal numbers to each other.
We add the value \[xy=18\Rightarrow 2xy=36\] in equation (i).
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=5\left( x+y \right) \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}+2xy=5\left( x+y \right)+36 \\
\end{align}\].
We get a quadratic equation of $\left( x+y \right)$ which we solve to get value.
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}+2xy=5\left( x+y \right)+36 \\
& \Rightarrow {{\left( x+y \right)}^{2}}=5\left( x+y \right)+36 \\
& \Rightarrow {{\left( x+y \right)}^{2}}-5\left( x+y \right)-36=0 \\
& \Rightarrow {{\left( x+y \right)}^{2}}-9\left( x+y \right)+4\left( x+y \right)-36=0 \\
& \Rightarrow \left( x+y-9 \right)\left( x+y+4 \right)=0 \\
\end{align}\]
Upon solving we get either $\left( x+y \right)=9$ or $\left( x+y \right)=-4$.
From the equation \[xy=18\] we can tell that the multiplication of those two numbers is positive which means they are both positive or both negative.
Let’s assume they are both negative then their sum $\left( x+y \right)$ will be negative. So, $\left( x+y \right)=-4$ is the only choice but it has no solution as \[xy=18\]. Maximum value of the multiplication can be 4.
So, another choice is the numbers are positive and the conditions are $\left( x+y \right)=9$ and \[xy=18\].
The only possible choice is 3 and 6. There are no other number combinations which can satisfy those two conditions.
So, the numbers are 3 and 6.
Note: We can prove the last part also but it’s totally unnecessary.
As \[xy=18\] the divisors combo of 18 are $1\times 18$, $2\times 9$, $3\times 6$. Now, addition of 1 and 18 gives 19 which doesn’t satisfy. Similarly, addition of 2 and 9 gives 11 which doesn’t satisfy. The remaining choice is 3 and 6 which satisfy.
Complete step-by-step answer:
Let’s assume that the numbers are x and y.
First condition is that the sum of squares of two numbers is equal to 5 times their sum.
So, the sum of squares of two numbers will be ${{x}^{2}}+{{y}^{2}}$ which is equal to 5 times their sum.
So, the equation becomes ${{x}^{2}}+{{y}^{2}}=5\left( x+y \right)$ …(i)
From the equation we get $\dfrac{{{x}^{2}}+{{y}^{2}}}{5\left( x+y \right)}=1$ …(ii)
Second condition is the sum of the reciprocals of the squares of these two numbers is $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers.
The squares of these numbers are ${{x}^{2}}$ and ${{y}^{2}}$. Reciprocals of the numbers are $\dfrac{1}{{{x}^{2}}}$ and $\dfrac{1}{{{y}^{2}}}$.
So, sum of the reciprocals of the squares of these two numbers $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}$ which is equal to $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers.
So, $\dfrac{5}{18}$ times sum of the reciprocals of the given numbers is $\dfrac{5}{18}\left( \dfrac{1}{x}+\dfrac{1}{y} \right)$.
So, the equation becomes $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}=\dfrac{5}{18}\left( \dfrac{1}{x}+\dfrac{1}{y} \right)$ …(iii)
From the equation we get
\[\begin{align}
& \dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}=\dfrac{5}{18}\left( \dfrac{1}{x}+\dfrac{1}{y} \right) \\
& \Rightarrow \dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}{{y}^{2}}}=\dfrac{5}{18}\times \dfrac{x+y}{xy}=\dfrac{5\left( x+y \right)}{18xy} \\
& \Rightarrow \dfrac{{{x}^{2}}+{{y}^{2}}}{5\left( x+y \right)}=\dfrac{{{x}^{2}}{{y}^{2}}}{18xy}=\dfrac{xy}{18} \\
\end{align}\]
We replace the left-hand side part from equation (ii).
Replacing we get \[\dfrac{{{x}^{2}}+{{y}^{2}}}{5\left( x+y \right)}=\dfrac{xy}{18}\Rightarrow \dfrac{xy}{18}=1\Rightarrow xy=18\].
So, the multiplication of two numbers is 1 which means they are reciprocal numbers to each other.
We add the value \[xy=18\Rightarrow 2xy=36\] in equation (i).
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=5\left( x+y \right) \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}+2xy=5\left( x+y \right)+36 \\
\end{align}\].
We get a quadratic equation of $\left( x+y \right)$ which we solve to get value.
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}+2xy=5\left( x+y \right)+36 \\
& \Rightarrow {{\left( x+y \right)}^{2}}=5\left( x+y \right)+36 \\
& \Rightarrow {{\left( x+y \right)}^{2}}-5\left( x+y \right)-36=0 \\
& \Rightarrow {{\left( x+y \right)}^{2}}-9\left( x+y \right)+4\left( x+y \right)-36=0 \\
& \Rightarrow \left( x+y-9 \right)\left( x+y+4 \right)=0 \\
\end{align}\]
Upon solving we get either $\left( x+y \right)=9$ or $\left( x+y \right)=-4$.
From the equation \[xy=18\] we can tell that the multiplication of those two numbers is positive which means they are both positive or both negative.
Let’s assume they are both negative then their sum $\left( x+y \right)$ will be negative. So, $\left( x+y \right)=-4$ is the only choice but it has no solution as \[xy=18\]. Maximum value of the multiplication can be 4.
So, another choice is the numbers are positive and the conditions are $\left( x+y \right)=9$ and \[xy=18\].
The only possible choice is 3 and 6. There are no other number combinations which can satisfy those two conditions.
So, the numbers are 3 and 6.
Note: We can prove the last part also but it’s totally unnecessary.
As \[xy=18\] the divisors combo of 18 are $1\times 18$, $2\times 9$, $3\times 6$. Now, addition of 1 and 18 gives 19 which doesn’t satisfy. Similarly, addition of 2 and 9 gives 11 which doesn’t satisfy. The remaining choice is 3 and 6 which satisfy.
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