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the product of 2 positive intergers is 90 find the 2 intergers if the larger is 3 more than twice the smaller
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x= the smaller 2x+3= the larger \[x(2x+3)=90\]
\[2x^2+3x-90=0\]
\[(2x+15)(x-6)=0\]
\[2x+15=0\] \[x-6=0\]
Discard 2x+15=0 because -15/2 is not positive and is not an integer.
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\[x=6\]
\[2x+3=15\]
So the integers and 6 and 15
it was cut off the work of the intergers
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