SOLUTION: "Word Problems Involving Products and Factoring"
Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus 8 is the sum of their sq
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Question 251997: "Word Problems Involving Products and Factoring"
Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus 8 is the sum of their squares. can you solve this pls=( tyty so much
Answer by MRperkins(300) (Show Source): You can put this solution on YOUR website!
Two consecutive positive numbers can be represented by n and n+1
the product of the sum and difference of the two numbers:
sum= (n+n+1) or 2n+1
difference=n-(n+1)(distribute the negative)=n-n-1 =-1 Make sure to use parenthesis here or the answer will not come out right.
so the product of the sum and difference is -1(2n+1) or -2n-1
"plus 8" to this and you get -2n-1+8 or -2n+7
we are told this is equal to the sum of their squares.
their squares will be n^2 and (n+1)^2
so we have -2n+7={n^2)+(n+1)^2
(n+1)^2 is the same as (n+1)(n+1)
Foil to multiply and you get n^2+n+n+1 or n^2+2n+1
-2n+7=n^2+n^2+2n+1 or -2n+7=2n^2+2n+1
get everything on one side and equal to zero.
0=2n^2+4n-6
factor out a 2 and you get 0=2(n^2+2n-3)
divide both sides by 2
0=n^2+2n-3
factors of -3 that add and give you positive 2 are 3 and -1, so it factors to (x+3)(x-1)
x+3=0 or x-1=0
x=-3 or x=1
look at our question again and we see that we want two consecutive POSTIVE numbers, so x cannot be -3. so x=1. If x=1, then our 2 consecutive positive numbers are 1 and 2.
Check your answer:
(sum)(difference)+8=1^2+2^2
(3)(-1)+8=1+4
5=5
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