SOLUTION: If the area of this rectangle is 105cm2 then the length is ____cm, the width is ____ cm, and the perimeter is ____cm. The width of the rectangle is (x) and the length is (2x+1). Th

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: If the area of this rectangle is 105cm2 then the length is ____cm, the width is ____ cm, and the perimeter is ____cm. The width of the rectangle is (x) and the length is (2x+1). Th      Log On


   

Question 68680: If the area of this rectangle is 105cm2 then the length is ____cm, the width is ____ cm, and the perimeter is ____cm. The width of the rectangle is (x) and the length is (2x+1). This is all the information the problem gives me. The problem was given to me on a study guide along with the answer key. The length is 15cm, and the width is 7 but I dont understand how they came about this answer. Can you please help me??
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with the formula for the area of a rectangle: A+=+L%2AW which is given as 105 sq.cm., you can substitute x for W and (2x+1) for L to get:
%282x%2B1%29%28x%29+=+105 Simplify and solve for x.
2x%5E2%2Bx+=+105 Subtract 105 from both sides of the equation.
2x%5E2%2Bx-105+=+0 Factor this quadratic equation.
%282x%2B15%29%28x-7%29+=+0 Apply the zero product principle.
2x%2B15+=+0 and/orx-7+=+0 from which you get:
2x%2B15+=+0 Subtract 15 from both sides.
2x+=+-15} Divide both sides by 2.
x+=+-%287.5%29 Discard this negative solution as the width must be a positive value.
x-7+=+0
x+=+7
The width is 7 cm.
The length is 2x+1 = 2(7)+1 = 15 cm.
The perimeter is:
P+=+2L+%2B+2W
P+=+2%2815%29%2B2%287%29
P+=+30%2B14
P+=+44 cm.