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Two numbers differ by 4 and their product is 192. Find the numbers.
Given:
Two numbers differ by 4 and their product is 192.
To do:
We have to find the numbers.
Solution:
Let the two numbers be $x$ and $x+4$.
According to the question,
$x(x+4)=192$
$x^2+4x=192$
$x^2+4x-192=0$
Solving for $x$ by factorization method, we get,
$x^2+16x-12x-192=0$
$x(x+16)-12(x+16)=0$
$(x+16)(x-12)=0$
$x+16=0$ or $x-12=0$
$x=-16$ or $x=12$
If $x=-16$, then $x+4=-16+4=-12$
If $x=12$, then $x+4=12+4=16$
The required numbers are $-16$ and $-12$ or $12$ and $16$.
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