I don't know where to go about creating a polynomial equations for the following problem: A piece of wire 52 in. long is cut into two pieces, and then each piece is formed into a square. If the sum of the areas of the two squares is 97 in.^2, how long are the pieces of wire? My attempted starting point is 97 = (1/4(x))^2 + ((1/4)(52-x))^2, but my numbers don't seem to work. Help is appreciated!
Your equation is perfectly correct.
What did you get for solution?
Once I move numbers around to achieve a trinomial, I have (1/8)x^2 - (13/2)x + 72, I'm just not sure how I would factor that.
Expand
97 = (x/4)^2 + ((52-x)/4)^2
to get
x^2/8-13x/2+72=0
multiply by 8 throughout to get
x^2-52x+576=0
factor and solve for x.
You should get two positive integers which add up to 52.
To solve this problem, you need to set up a correct equation for the sum of the areas of the two squares. Let's walk through the process step by step:
1. Start by letting x represent the length of one of the pieces of wire.
2. Since the total length of the wire is 52 inches and you are cutting it into two pieces, the length of the other piece would be (52 - x) inches.
3. Next, determine the side length of each square. Recall that the perimeter of a square is equal to four times the length of its side. So, for each square, the side length would be (1/4)x and (1/4)(52 - x) inches, respectively.
4. The area of a square is calculated by squaring the length of its side. Therefore, the first square's area would be ((1/4)x)^2 and the second square's area would be ((1/4)(52 - x))^2.
5. Finally, you need to set up the equation using the information given in the problem. The sum of the areas of the two squares is 97 square inches, so you can write the equation as follows:
((1/4)x)^2 + ((1/4)(52 - x))^2 = 97
Now, you can solve this equation to find the value of x, which represents the length of one of the wire pieces.