what is the answer to: There are two consecutive positive integers such that the square of the larger number is 73 more than 8 times the smaller number. and how do you solve
what is the answer to: There are two consecutive positive integers such that the square of the larger number is 73 more than 8 times the smaller number. and how do you solve.
Let the first number be=N, then we have,
N+1=the second number, but we have,
(N+1)^2=8N+73, therefore the two numbers are
N=12 and 12+1=13
Solve for N:
(N+1)^2 = 8 N+73
Subtract 8 N+73 from both sides:
-73-8 N+(N+1)^2 = 0
Expand out terms of the left hand side:
N^2-6 N-72 = 0
The left hand side factors into a product with two terms:
(N-12) (N+6) = 0
Split into two equations:
N-12 = 0 or N+6 = 0
Add 12 to both sides:
N = 12 or N+6 = 0
Subtract 6 from both sides:
Answer: | N = 12 or N = -6 we discard this
