SOLUTION: Solving a word promblem using a quadratic equation with rational roots The length of a rectangle is 1 yd less than three times ite width and the area of the rectangle is 24yd^2.

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Question 91488: Solving a word promblem using a quadratic equation with rational roots
The length of a rectangle is 1 yd less than three times ite width and the area of the rectangle is 24yd^2. Find the diminsions.
Found 2 solutions by ptaylor, Earlsdon:
Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
Area of a rectangle=Length times Width or A=L*W
Let W=width of the rectangle
Then Length (L)=3W-1
And Area= 24 yd^2 =W*(3W-1)=3W^2-W So our equation to solve is:
3W^2-W=24 subtract 24 from both sides
3W^2-W-24=24-24 simplifying, we get:
3W^2-W-24=0 quadratic in standard form. We'll solve using the quadratic formula

and
discount negative value for W

yds--------------------------------width
yds ---------------------------length
CK
A=24=3*8
24=24

Hope this helps---ptaylor

Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website!
Let W = the width of the rectangle and L = the length.
The problem description tells you that the length, L, is (=) 1 yd less than three times its width (L = 3W-1).
You also know that the area, A = 24 sq.yds.
Starting with the formula for the area of a rectangle: A = L*W
Making the appropriate substitutions into the formula, you get:
Simplifying this:
Subtract 24 from both sides.
You can solve this quadratic equation by factoring.
Apply the zero products principle:
or , so...
Subtract 8 from both sides.
Divide both sides by 3.
Discard this solution a negative width is not meaningful.
Add 3 to both sides.
The width is 3 yards.

The length is 8 yards.
Check:

sq.yds.