Question 211832
The product of two consecutive odd integers is 1 less than four times their sum.
Find the two integers.
 

Step 1.  Let n be one odd integer.


Step 2.  Let n+2 be the next consecutive odd integer.


Step 3.  Translate problem statement into an equation such than n(n+2)=4(n+n+2)-1 since product is 1 less than four times their sum


where n(n+2) is the product of the two consecutive integers and (n+n+2) is the sum of the two integers


{{{n(n+2)=4(n+n+2)-1}}}


{{{n^2+2n=4(2n+2)-1}}}


{{{n^2+2n=8n+8-1}}}


{{{n^2+2n=8n+7}}}


Step 4.  Simplify to obtain a quadratic equation.  Subtract 8n+7 from both sides of equation to get the right side of equation to be zero.


{{{n^2+2n-8n-7=8n+7-8n-7}}}


{{{n^2-6n-7=0}}}


Step 5.  Now we have a quadratic equation.  So use quadratic formula given as 


{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where


a=1, b=-6 and  c=-7


The following steps solves the quadratic equation


*[invoke quadratic "n", 1,  -6,  -7]


Step 6.  ANSWER:  Based on the above steps, choose 7 since it is positive.  The integers are then 7 and 9.


Step 7.  Verify the answer using n(n+2)=4(n+n+2)+1 in Step 3.


{{{(7)(9)=4(7+9)-1}}}


{{{63=4(16)-1}}} 


{{{63=63}}}  So it checks out.


I hope the above steps were helpful. 

 

And good luck in your studies!


Respectfully,
Dr J


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