document.write( "Question 903553: A square vegetable garden is to be tilled and then enclosed with a fence. If the fence costs $4.00 per foot and the cost of preparing the soil is $1.00 per ft2, determine the size of the garden that can be enclosed for $561.00 \n" ); document.write( "

Algebra.Com's Answer #548135 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
perimeter is 4s area is s^2
\n" ); document.write( "4s*4+1*s^2=561
\n" ); document.write( "16s+s^2=561
\n" ); document.write( "rearrange
\n" ); document.write( "s^2+16s-561=0
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"s%5E2%2B16s-561\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"16\"$ , and the last term is $\"-561\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-561\"$ to get $\"%281%29%28-561%29=-561\"$ .

Now the question is: what two whole numbers multiply to $\"-561\"$ (the previous product) and add to the second coefficient $\"16\"$ ?

To find these two numbers, we need to list all of the factors of $\"-561\"$ (the previous product).

Factors of $\"-561\"$ :

1,3,11,17,33,51,187,561

-1,-3,-11,-17,-33,-51,-187,-561

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

These factors pair up and multiply to $\"-561\"$ .

1*(-561) = -561
3*(-187) = -561
11*(-51) = -561
17*(-33) = -561
(-1)*(561) = -561
(-3)*(187) = -561
(-11)*(51) = -561
(-17)*(33) = -561

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"16\"$ :

\n" ); document.write( "

First Number	Second Number	Sum
1	-561	1+(-561)=-560
3	-187	3+(-187)=-184
11	-51	11+(-51)=-40
17	-33	17+(-33)=-16
-1	561	-1+561=560
-3	187	-3+187=184
-11	51	-11+51=40
-17	33	-17+33=16

From the table, we can see that the two numbers $\"-17\"$ and $\"33\"$ add to $\"16\"$ (the middle coefficient).

So the two numbers $\"-17\"$ and $\"33\"$ both multiply to $\"-561\"$ and add to $\"16\"$

Now replace the middle term $\"16s\"$ with $\"-17s%2B33s\"$ . Remember, $\"-17\"$ and $\"33\"$ add to $\"16\"$ . So this shows us that $\"-17s%2B33s=16s\"$ .

$\"s%5E2%2Bhighlight%28-17s%2B33s%29-561\"$ Replace the second term $\"16s\"$ with $\"-17s%2B33s\"$ .

$\"%28s%5E2-17s%29%2B%2833s-561%29\"$ Group the terms into two pairs.

$\"s%28s-17%29%2B%2833s-561%29\"$ Factor out the GCF $\"s\"$ from the first group.

$\"s%28s-17%29%2B33%28s-17%29\"$ Factor out $\"33\"$ 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.

$\"%28s%2B33%29%28s-17%29\"$ Combine like terms. Or factor out the common term $\"s-17\"$

===============================================================

Answer:

So $\"s%5E2%2B16%2As-561\"$ factors to $\"%28s%2B33%29%28s-17%29\"$ .

In other words, $\"s%5E2%2B16%2As-561=%28s%2B33%29%28s-17%29\"$ .

Note: you can check the answer by expanding $\"%28s%2B33%29%28s-17%29\"$ to get $\"s%5E2%2B16%2As-561\"$ or by graphing the original expression and the answer (the two graphs should be identical).

\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics

Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert $\"1s%5E2%2B16s%2B-561=0\"$ to standard form by dividing both sides by 1:
\n" ); document.write( "We have: $\"1s%5E2%2B16s%2B-561=0\"$ . \n" ); document.write( "What we want to do now is to change this equation to a complete square $\"%28s%2Bsomenumber%29%5E2+%2B+othernumber\"$ . How can we find out values of somenumber and othernumber that would make it work?
\n" ); document.write( "Look at $\"%28s%2Bsomenumber%29%5E2\"$ : $\"%28s%2Bsomenumber%29%5E2+=+s%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2\"$ . Since the coefficient in our equation $\"1s%5E2%2Bhighlight_red%28+16%29+%2A+s%2B-561=0\"$ that goes in front of s is 16, we know that 16=2*somenumber, or $\"somenumber+=+16%2F2\"$ . So, we know that our equation can be rewritten as $\"%28s%2B16%2F2%29%5E2+%2B+othernumber\"$ , and we do not yet know the other number.
\n" ); document.write( "We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that $\"%28s%2B16%2F2%29%5E2+%2B+othernumber\"$ is equivalent to our original equation $\"1s%5E2%2B16s%2Bhighlight_green%28+-561+%29=0\"$ .
\n" ); document.write( "
\n" ); document.write( "

\n" ); document.write( " The highlighted red part must be equal to -561 (highlighted green part).
\n" ); document.write( "
\n" ); document.write( " $\"16%5E2%2F4+%2B+othernumber+=+-561\"$ , or $\"othernumber+=+-561-16%5E2%2F4+=+-625\"$ .
\n" ); document.write( "So, the equation converts to $\"%28s%2B16%2F2%29%5E2+%2B+-625+=+0\"$ , or $\"%28s%2B16%2F2%29%5E2+=+625\"$ .
\n" ); document.write( "
\n" ); document.write( " Our equation converted to a square $\"%28s%2B16%2F2%29%5E2\"$ , equated to a number (625).
\n" ); document.write( "
\n" ); document.write( " Since the right part 625 is greater than zero, there are two solutions:
\n" ); document.write( "
\n" ); document.write( " $\"system%28+%28s%2B16%2F2%29+=+%2Bsqrt%28+625+%29%2C+%28s%2B16%2F2%29+=+-sqrt%28+625+%29+%29\"$
\n" ); document.write( " , or
\n" ); document.write( "
\n" ); document.write( " $\"system%28+%28s%2B16%2F2%29+=+25%2C+%28s%2B16%2F2%29+=+-25+%29\"$
\n" ); document.write( " $\"system%28+s%2B16%2F2+=+25%2C+s%2B16%2F2+=+-25+%29\"$
\n" ); document.write( " $\"system%28+s+=+25-16%2F2%2C+s+=+-25-16%2F2+%29\"$
\n" ); document.write( "
\n" ); document.write( " $\"system%28+s+=+17%2C+s+=+-33+%29\"$
\n" ); document.write( "Answer: s=17, -33.\n" ); document.write( "

\n" ); document.write( "
\n" ); document.write( "discard negative answers
\n" ); document.write( "s=17
\n" ); document.write( " \n" ); document.write( "

\n" );