SOLUTION: The rectangular floor of a shed has a length 4 feet longer than its width, and its area is 140 square feet. Let x be the width of the floor. (a) Write a quadratic equation whose

Algebra ->  Rectangles -> SOLUTION: The rectangular floor of a shed has a length 4 feet longer than its width, and its area is 140 square feet. Let x be the width of the floor. (a) Write a quadratic equation whose      Log On


   

Question 626824: The rectangular floor of a
shed has a length 4 feet longer than its width, and its
area is 140 square feet. Let x be the width of the floor.
(a) Write a quadratic equation whose solution gives
the width of the floor.
(b) Solve this equation.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the width of the floor (in feet).
The length is 4 feet longer than the width, so it is
x%2B4 (in feet).
(a) The area in square feet can be calculated by multiplying length (in feet) times width (in feet), and it equals 140 square feet, so
%28x%2B4%29%2Ax=140
That equation can be transformed into a more traditional form:
%28x%2B4%29%2Ax=140 --> x%5E2%2B4x=140 --> highlight%28x%5E2%2B4x-140=0%29

(b) The equation can be solved by factoring, or by completing the square, or by using the quadratic formula. I'll show you all 3 algebra ways of solving a quadratic equation, plus the way a fifth grader could solve the problem.

Someone who does not know algebra would not be able to write the equation, but could look for two numbers that multiply to give 140 and could be the width and length in feet. Possible products are:
1%2A140=140
2%2A70=140
4%2A35=140
5%2A28=140
7%2A20=140
10%2A14=140
The last one of those products represents the width (10 feet) times the length (14 feet, 4 more than the width) of the rectangle.

FACTORING (when it works, the easiest way):
The person who knows algebra (and became good at factoring polynomials) will do something similar, looking for integers that multiply to give the -140 in the equation, and that will add up to the 4 in the equation
x%5E2%2B4x-140=0
The integers, -10 and 14, are the numbers added to x in the factors of the quadratic polynomial:
x%5E2%2B4x-140=%28x-10%29%28x%2B14%29
The factored equation is
%28x-10%29%28x%2B14%29=0 with solutions that make x-10=0 or x%2B14=0:
highlight%28x=10%29 and x=-14
The only solution to the problem is
x-10=0 --> highlight%28x=10%29 (The width is 10 feet),
because the width in feet cannot be a negative number.
While x=-14 is a solution to x%2B14=0 and to the quadratic equation, it is not a solution to part (b) of the problem.

COMPLETING THE SQUARE:
x%5E2%2B4x-140=0 --> x%5E2%2B4x=140
Then, adding 4 to both sides of the equal sign,
we get x%5E2%2B4x%2B4=%28x%2B2%29%5E2 on the left side.
x%5E2%2B4x=140 --> x%5E2%2B4x%2B4=144 --> %28x%2B2%29%5E2=144
As 144=12%5E2 or sqrt%28144%29=12,
%28x%2B2%29%5E2=144 --> x%2B2=12 or x%2B2=-12
x%2B2=12 --> x%2B2-2=12-2--> highlight%28x=10%29
x%2B2=-12 --> x=-14, which is not a solution to the problem, because the width cannot be -14 feet.

USING THE QUADRATIC FORMULA:
The solutions to an equation of the form ax%5E2%2Bbx%2Bc=0 are given by
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In the case of equation x%5E2%2B4x-140=0,
a=1, b=4, and c=-140
So x+=+%28-4+%2B-+sqrt%28+4%5E2-4%2A1%2A%28-140%29+%29%29%2F%282%2A1%29+
x+=+%28-4+%2B-+sqrt%2816%2B560%29%29%2F2
x+=+%28-4+%2B-+sqrt%28576%29%29%2F2
x+=+%28-4+%2B-+24%29%2F2
x+=+%28-4%2B24%29%2F2 --> x=20%2F2 --> highlight%28x=10%29
x+=+%28-4-24%29%2F2 --> x=-28%2F2 --> x=-14, which is not a solution to the problem, because the width cannot be -14 feet.