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Two numbers differ by 3 and their product is 504. Find the numbers.
Given:
Two numbers differ by 3 and their product is 504.
To do:
We have to find the numbers.
Solution:
Let the two numbers be $x$ and $x+3$.
According to the question,
$x(x+3)=504$
$x^2+3x=504$
$x^2+3x-504=0$
Solving for $x$ by factorization method, we get,
$x^2+24x-21x-504=0$
$x(x+24)-21(x+24)=0$
$(x+24)(x-21)=0$
$x+24=0$ or $x-21=0$
$x=-24$ or $x=21$
If $x=-24$, then $x+3=-24+3=-21$
If $x=21$, then $x+3=21+3=24$
The required numbers are $-24$ and $-21$ or $21$ and $24$.
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