The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number.
The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.
Let the greater number be $x$ and the smaller number be $y$.
According to the question:
$x^{2}-y^{2}=45 \quad \ldots($ i)
$y^{2}=4 x \quad \ldots$ (ii)
From (i) and (ii), we get:
$x^{2}-4 x=45$
$\Rightarrow x^{2}-4 x-45=0$
$\Rightarrow x^{2}-(9-5) x-45=0$
$\Rightarrow x^{2}-9 x+5 x-45=0$
$\Rightarrow x(x-9)+5(x-9)=0$
$\Rightarrow(x-9)(x+5)=0$
$\Rightarrow x-9=0$ or $x+5=0$
$\Rightarrow x=9$ or $x=-5$
$\Rightarrow x=9 \quad(\because x$ is a natural number $)$
Putting the value of $x$ in equation (ii), we get:
$y^{2}=4 \times 9$
$\Rightarrow y^{2}=36$
$\Rightarrow y=6$
Hence, the two numbers are 9 and 6 .
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