# If two consecutive positive integers have the property that one integer times twice the other equals 612 what is the sum of these two integers?

and the two consecutive positive integers are 17 and 18

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Let ( x ) be the first positive integer. Then, the next consecutive positive integer is ( x + 1 ). According to the given condition, we have the equation ( x(2(x + 1)) = 612 ).

Expanding and simplifying:

( x(2x + 2) = 612 )

( 2x^2 + 2x = 612 )

Divide both sides by 2:

( x^2 + x = 306 )

Rearrange the equation:

( x^2 + x - 306 = 0 )

Now, we can use the quadratic formula to solve for ( x ):

( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )

Where ( a = 1 ), ( b = 1 ), and ( c = -306 ).

( x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-306)}}{2(1)} )

( x = \frac{-1 \pm \sqrt{1 + 1224}}{2} )

( x = \frac{-1 \pm \sqrt{1225}}{2} )

( x = \frac{-1 \pm 35}{2} )

Since ( x ) must be positive, we take the positive root:

( x = \frac{-1 + 35}{2} = \frac{34}{2} = 17 )

So, the first positive integer is 17, and the next consecutive positive integer is ( 17 + 1 = 18 ).

Therefore, the sum of these two integers is ( 17 + 18 = 35 ).

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