Math word problem about rectangle

Question

The length of a rectangular garden is 25 feet longer than its width. If the area of the garden is 186 square feet, what are the dimensions of the garden?

A) First, write a quadratic equation in the from aw2+bw+c=0 you can solve to answer the question given. Let w be the width of the garden.

Answer: The equation is: 

B) Factor your equation from part [a] into the form (w+A)(w−B)=0 and enter the full factored equation below.

Answer: The factored equation is: 

Don't forget to include the right hand side of the equation and to use the correct variable.

C)Use your factored equation to find the length and width of the garden.

Length =
31
feet.

Width =
6
feet.

Answer 1

Hello, Katelyn,

   Apparently, they have given you the length(31 feet) and the width (6 feet) already. 

Note:  It is helpful to underline key words and numbers in the story problem.

 

1.  Then start by drawing a picture of the garden.      _____________

                                                                            [                     ] w    Note:  Area of rectangle= L* W

                                                                              ---------------- -                                                 

                                                                             L=(w + 25)                  

 

Area = 186 sq. ft.

 

 

                                                                                                                

Substitute the length and width of the rectangle into the area formula(A=L*W).        186 = (w+25) (w)    Multiply it out.

 

 

w² +25w=186      Now place quadratic in standard from: ax² +bx +C=0 by subtracting 186 from both sides of equation.

 

w² +25W-186=0                 Answer                         Now they want you to factor it.

 

 

(w+31)(w-6)=0      Ask yourself what you need to multiply together to get -186.  Also, when you add these two, previous factors together, you get 25w, the middle term of the quadratic.

 

(w+31)=0   or  w-6=0  Solve the equations. The width can't be negative.  Therefore, only one answer is usable here.

 

Finally, check your work by substituting the positive answer back into the original equation and into the drawing to see if it works.

 

If I have helped you, please give me a thumbs up.

                                                      Susan C.