Respuesta :
Answer:
3 metres
Step-by-step explanation:
Let the width of the pathway be x
Area: (9+2x)(4+2x)
36 + 8x + 18x + 4x²
4x² + 26x + 36
Increased area is 150:
4x² + 26x + 36 = 150
4x² + 26x - 114 = 0
2x² + 13x - 57 = 0
2x² + 19x - 6x - 57 = 0
x(2x + 19) - 3(2x + 19) = 0
(x - 3)(2x + 19) = 0
x = 3, x = -19/2
The width of the considered pathway after increasing the total area to 150 square meters is 3 meters.
How to find the area and the perimeter of a rectangle?
For a rectangle with length and width L and W units, we get:
- Area of the rectangle = [tex]L \times W \: \rm unit^2[/tex]
- Perimeter of the rectangle = [tex]2(L + W) \: \rm unit[/tex]
How to find the roots of a quadratic equation?
Suppose that the given quadratic equation is
[tex]ax^2 + bx + c = 0[/tex]
Then its roots are given as:
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
For this case, we're given that:
- The dimensions of the rectangular garden is 9 m by 4 m (assume its rectangular shaped)
- Pathway is constructed around the garden
- The area of the total garden + pathway = 150 sq. meters.
- Pathway is of same width on all sides of the garden.
Consider the diagram given below.
The pathway is rectangular.
Let its width be x.
One dimension of the pathway = width on either side parallel to that garden's one dimension + that dimension of garden = x + x + 9 = 2x + 9 meters (as width on all the sides is 'x' m)
Other dimension of the pathway = width on either side parallel to that garden's other dimension + that dimension of garden = x + x + 4 = 2x + 4 meters
Thus, the dimensions of the rectangular pathway are 2x + 9 m and 2x + 4 m.
The area of the rectangle of dimension 2x +9 and 2x + 4 is covering area of the pathway and area of the garden.
This summed area is 150 sq. meters.
This area is [tex](2x+9)(2x+4)[/tex] sq. meters.
Thus, we get:
[tex](2x + 9)(2x +4) = 150\\4x^2 + 26x + 36 = 150\\4x^2 + 26x -114 = 0\\[/tex]
Using the formula for solving quadratic equations, we get:
[tex]x = \dfrac{-26 \pm \sqrt{26^2 - 4(4)(-114)}}{2(4)} = \dfrac{-26 \pm \sqrt{2500}}{8}\\\\x = \dfrac{-26 \pm 50}{8}\\\\x = 3, -76/8[/tex]
As x represents width (a measurement of how long a thing) cannot be negative, therefore, we take only x = 3 (in meters)
This is the width of the pathway.
Thus, the width of the considered pathway after increasing the total area to 150 square meters is 3 meters.
Learn more about finding the solutions of a quadratic equation here:
https://brainly.com/question/3358603
#SPJ2