document.write( "Question 282422: Solve using the five-step problem-solving process. Show all steps necessary to arrive at your solution. \r
\n" ); document.write( "\n" ); document.write( "The product of two consecutive positive integers is 272. Find the integers. \n" ); document.write( "

Algebra.Com's Answer #205058 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
x*(x+1)=272
\n" ); document.write( "x^2+x-272=0
\n" ); document.write( "16 and 17\r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"x%5E2%2Bx-272\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"1\"$ , and the last term is $\"-272\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-272\"$ to get $\"%281%29%28-272%29=-272\"$ .

Now the question is: what two whole numbers multiply to $\"-272\"$ (the previous product) and add to the second coefficient $\"1\"$ ?

To find these two numbers, we need to list all of the factors of $\"-272\"$ (the previous product).

Factors of $\"-272\"$ :

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

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

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to $\"-272\"$ .

1*(-272) = -272
2*(-136) = -272
4*(-68) = -272
8*(-34) = -272
16*(-17) = -272
(-1)*(272) = -272
(-2)*(136) = -272
(-4)*(68) = -272
(-8)*(34) = -272
(-16)*(17) = -272

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"1\"$ :

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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

From the table, we can see that the two numbers $\"-16\"$ and $\"17\"$ add to $\"1\"$ (the middle coefficient).

So the two numbers $\"-16\"$ and $\"17\"$ both multiply to $\"-272\"$ and add to $\"1\"$

Now replace the middle term $\"1x\"$ with $\"-16x%2B17x\"$ . Remember, $\"-16\"$ and $\"17\"$ add to $\"1\"$ . So this shows us that $\"-16x%2B17x=1x\"$ .

$\"x%5E2%2Bhighlight%28-16x%2B17x%29-272\"$ Replace the second term $\"1x\"$ with $\"-16x%2B17x\"$ .

$\"%28x%5E2-16x%29%2B%2817x-272%29\"$ Group the terms into two pairs.

$\"x%28x-16%29%2B%2817x-272%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%28x-16%29%2B17%28x-16%29\"$ Factor out $\"17\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

$\"%28x%2B17%29%28x-16%29\"$ Combine like terms. Or factor out the common term $\"x-16\"$

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Answer:

So $\"x%5E2%2Bx-272\"$ factors to $\"%28x%2B17%29%28x-16%29\"$ .

In other words, $\"x%5E2%2Bx-272=%28x%2B17%29%28x-16%29\"$ .

Note: you can check the answer by expanding $\"%28x%2B17%29%28x-16%29\"$ to get $\"x%5E2%2Bx-272\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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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:
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\n" ); document.write( " $\"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"$
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\n" ); document.write( " For these solutions to exist, the discriminant $\"b%5E2-4ac\"$ should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant $\"b%5E2-4ac\"$ : $\"b%5E2-4ac=%281%29%5E2-4%2A1%2A-272=1089\"$ .
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\n" ); document.write( " Discriminant d=1089 is greater than zero. That means that there are two solutions: $\"+x%5B12%5D+=+%28-1%2B-sqrt%28+1089+%29%29%2F2%5Ca\"$ .
\n" ); document.write( "
\n" ); document.write( " $\"x%5B1%5D+=+%28-%281%29%2Bsqrt%28+1089+%29%29%2F2%5C1+=+16\"$
\n" ); document.write( " $\"x%5B2%5D+=+%28-%281%29-sqrt%28+1089+%29%29%2F2%5C1+=+-17\"$
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\n" ); document.write( " Quadratic expression $\"1x%5E2%2B1x%2B-272\"$ can be factored:
\n" ); document.write( " $\"1x%5E2%2B1x%2B-272+=+1%28x-16%29%2A%28x--17%29\"$
\n" ); document.write( " Again, the answer is: 16, -17.\n" ); document.write( "Here's your graph:
\n" ); document.write( " $\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-272+%29\"$

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