Just need a little boost to solve this question..

" The Parks department is planning a new flower bed outside city hall. It will be rectangular with dimensions by 9 m by 6m. The flower bed will be surrounded by a path of constant width with the same area as the flower bed. Calculate the perimeter of the path."

Diagram attatched.

A) Set up a quadratic Equation To model the problem

2. Originally Posted by letsgofam
Just need a little boost to solve this question..

" The Parks department is planning a new flower bed outside city hall. It will be rectangular with dimensions by 9 m by 6m. The flower bed will be surrounded by a path of constant width with the same area as the flower bed. Calculate the perimeter of the path."

Diagram attatched.

A) Set up a quadratic Equation To model the problem
(2x+6)(2x+9) = 2(54)

solve for x, then determine the path's perimeter

3. Originally Posted by skeeter
(2x+6)(2x+9) = 2(54)

solve for x, then determine the path's perimeter
Alright
So
(2x+6)(2x+9) = 2(54)
I simplify that
Get

4x^2 + 30x +54 = 108
4x^2 + 30x -54=0

Apply the quadratic formula, solve for x ?

4. Okay, I got -3 and -4.5 for the the values of X, but dimensions can't be negative numbers , what's wrong?

5. Originally Posted by letsgofam
Okay, I got -3 and -4.5 for the the values of X, but dimensions can't be negative numbers , what's wrong?

6. Originally Posted by harish21
How?

2x + 6 = 0
2x = =-6
x= -3

2x+9 = 0
2x = -9
x = -4.5

??

7. Originally Posted by letsgofam
How?

2x + 6 = 0
2x = =-6
x= -3

2x+9 = 0
2x = -9
x = -4.5

??
You are assuming that (2x + 6)(2x+9)=0 This is NOT true in your case..

Look at what you have done in post number 3 of this thread.. you obtained a quadratic equation:4x^2 + 30x -54=0

this is where you need to find your x from.

An unknown x in a qudratic equation $ax^2+bx+c$ (where a,b,c,are known) can be calculated as:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

now compare the equation you obtained 4x^2 + 30x -54=0 with the above formula and find x.

8. Great, I solved the equation.

Got 9M for the widths, and 12 for the lengths.
Thanks alot guys!