SOLUTION: 1) Two consecutive integers are added. The square of their sum is 361. What are the integers ? 2) The product of two consecutive even numbers is 288. What are the numbers ?

Click here to see ALL problems on Numbers Word Problems

Question 249047: 1) Two consecutive integers are added. The square of their sum is 361. What are the integers ?
2) The product of two consecutive even numbers is 288. What are the numbers ?
Found 2 solutions by checkley77, actuary:
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
LET X & X+1 ARE THE 2 INTEGERS.
(X+X+1)^2=361
(2X+1)^2=361
4X^2+4X+1=361
4X^2+4X+1-361=0
4(X^2+X-90)=0
4(X+10)(X-9)=0
X+10=0
X=-10 ANS & -9
X-9=0
X=9 ANS & 10
PROOF:
(9+10)^2=361
19^2=361
361=361

Answer by actuary(112) (Show Source):
You can put this solution on YOUR website!
PROBLEM 1
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.
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
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
and -9.
Problem 2
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,
k=-9 and k=8. So, the first pair of integers is n=2*k = -18 and the second integer is -16.
The second pair of integers is 2*8=16 and 18.
It is important to recognize the pairs of solutions for each question.
Good luck with your study of math.
Larry