SOLUTION: Solve using the five-step problem-solving process. Show all steps necessary to arrive at your solution. The product of two consecutive positive integers is 272. Find the intege

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Question 282422: Solve using the five-step problem-solving process. Show all steps necessary to arrive at your solution.
The product of two consecutive positive integers is 272. Find the integers.

Found 2 solutions by richwmiller, solver91311:
Answer by richwmiller(9135) About Me  (Show Source):
You can put this solution on YOUR website!
x*(x+1)=272
x^2+x-272=0
16 and 17
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)
In order to factor x%5E2%2Bx-272, first multiply the leading coefficient 1 and the last term -272 to get -272. Now we need to ask ourselves: What two numbers multiply to -272 and add to 1? Lets find out by listing all of the possible factors of -272


Factors:

1,2,4,8,16,17,34,68,136,272,

-1,-2,-4,-8,-16,-17,-34,-68,-136,-272, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -272.

(-1)*(272)=-272

(-2)*(136)=-272

(-4)*(68)=-272

(-8)*(34)=-272

(-16)*(17)=-272

Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1

||||||||||
First Number|Second Number|Sum
1|-272|1+(-272)=-271
2|-136|2+(-136)=-134
4|-68|4+(-68)=-64
8|-34|8+(-34)=-26
16|-17|16+(-17)=-1
-1|272|(-1)+272=271
-2|136|(-2)+136=134
-4|68|(-4)+68=64
-8|34|(-8)+34=26
-16|17|(-16)+17=1


We can see from the table that -16 and 17 add to 1. So the two numbers that multiply to -272 and add to 1 are: -16 and 17

So the original quadratic


x%5E2%2Bx-272


breaks down to this (just replace 1%2Ax with the two numbers that multiply to -272 and add to 1, which are: -16 and 17)


x%5E2%2Bhighlight%28-16x%2B17x%29-272 Replace 1%2Ax with -16x%2B17x

Group the first two terms together and the last two terms together like this:

%28x%5E2-16x%29%2B%2817x-272%29

Factor a 1x out of the first group and factor a 17 out of the second group.


1x%28x-16%29%2B17%28x-16%29


Now since we have a common term x-16 we can combine the two terms.


%28x%2B17%29%28x-16%29 Combine like terms.
==============================================================================

Answer:


So the quadratic x%5E2%2Bx-272 factors to %28x%2B17%29%28x-16%29




Notice how %28x%2B17%29%28x-16%29 foils back to our original problem x%5E2%2Bx-272. This verifies our answer.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-272+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-272=1089.

Discriminant d=1089 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+1089+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+1089+%29%29%2F2%5C1+=+16
x%5B2%5D+=+%28-%281%29-sqrt%28+1089+%29%29%2F2%5C1+=+-17

Quadratic expression 1x%5E2%2B1x%2B-272 can be factored:
1x%5E2%2B1x%2B-272+=+1%28x-16%29%2A%28x--17%29
Again, the answer is: 16, -17. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-272+%29

Answer by solver91311(16872) About Me  (Show Source):
You can put this solution on YOUR website!


Let represent the first of two consecutive positive integers. Then the next consecutive positive integer must be

The product of these two integers is then

This product is equal to 272, so:



Just put the equation in standard form, factor, and solve. Exclude the negative root because the question asks for positive integers.

John