SOLUTION: Page 132 (problem 95): A length of wire 16 inches is to be cut into two pieces, and then each piece will be bent to form a square. Find the length of the two pieces if the sum of t

Question 183085This question is from textbook Precalculus
: Page 132 (problem 95): A length of wire 16 inches is to be cut into two pieces, and then each piece will be bent to form a square. Find the length of the two pieces if the sum of the areas of the two squares is 10 square inches.
(x/4)^2 + ((16-x)/4)^2 = 10
I've gotten this far but can't figure out the next steps to solve it.
Thank you, Lynn Scott This question is from textbook Precalculus

Found 2 solutions by scott8148, jim_thompson5910:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
expanding __ (x^2)/16 + (256-32x+x^2)/16 = 10

multiplying by 16 and rearranging __ 2x^2-32x+256 = 160

subtracting 160 and dividing by 2 __ x^2-16x+48 = 0

factoring __ (x-12)(x-4) = 0

x equals 12 or 4 (which sum to 16)

one square is 9in^2 and the other is 1in^2
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You're on the right track, you just need to solve the equation:

Start with the given equation.

Square to get

Square to get

Combine the fractions.

Multiply both sides by 16.

Multiply

FOIL

Subtract 160 from both sides.

Combine like terms.

Notice we have a quadratic equation in the form of where , , and

Let's use the quadratic formula to solve for x

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the answers are or

This means that the second length of the square is either

or

Note: either way, the two side lengths are 12 and 4

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

Answer:

So the length of the two pieces are 12 and 4 inches.
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