document.write( "Question 249047: 1) Two consecutive integers are added. The square of their sum is 361. What are the integers ?\r
\n" ); document.write( "\n" ); document.write( "2) The product of two consecutive even numbers is 288. What are the numbers ?
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Algebra.Com's Answer #181458 by actuary(112)\"\" \"About 
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PROBLEM 1\r
\n" ); document.write( "\n" ); document.write( "Let n = the first integer, so n+1 is the next integer. Now translate the word problem into an algebraic expresion. So, (n+(n+1))^2=361. Take the square root of both sides of the equation, so, n+(n+1)=19 and n+(n+1)=-19.\r
\n" ); document.write( "\n" ); document.write( "Case 1 yields the following linear equation 2n+1=19. Solving this linear equation results in the value n = (19-1)/2=9. The pair is 9 and 10 \r
\n" ); document.write( "\n" ); document.write( "Case 2 yields the folowing linear equation n+(n+1)=-19. Solving this linear equation results in the value for n = (-19-1)/2=-10. The pair is then -10
\n" ); document.write( "and -9.\r
\n" ); document.write( "\n" ); document.write( "Problem 2 \r
\n" ); document.write( "\n" ); document.write( "Let n = the first even integer. So n = 2k for some k. The next even integer is then n+2 = 2k+2. Therefore, the words of the problem can be translated into the following algebraic equation 2k*(2k+2)=288. Expanding this equation we have 4k^2+4k-288=0. Dividing both sides by 4 produces k^2+k-72=0. This equation can be solved for k by factoring. k^2+k-72=(k+9)(k-8)=0. Therefore,
\n" ); document.write( "k=-9 and k=8. So, the first pair of integers is n=2*k = -18 and the second integer is -16.\r
\n" ); document.write( "\n" ); document.write( "The second pair of integers is 2*8=16 and 18.\r
\n" ); document.write( "\n" ); document.write( "It is important to recognize the pairs of solutions for each question.\r
\n" ); document.write( "\n" ); document.write( "Good luck with your study of math.\r
\n" ); document.write( "\n" ); document.write( "Larry
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