Help with algebraic number problem?

Question

A rectangular swimming pool is 12m long by 6m wide. It is surrounded by a pavement of uniform width, the area of the pavement being 7/8 of the are of the pool.

a) If the pavements x meters wide, show that the area of the pavement is 4xˆ2 + 36 mˆ2

b) Hence, show that 4xˆ2+36x-63=0

c) How wide is the pavement?

Happy for help + answers for any of the questions!!!

Answer 1

Area of the rectangular swimming pool: blue area



a = 12 * 6



a = 72 m²





Area of the pavements: yellow area



p = top + left + bottom + right



p = [(x + 12 + x) * x] + [6 * x] + [(x + 12 + x) * x] + [6 * x]



p = [(2x + 12) * x] + 6x + [(2x + 12) * x] + 6x



p = [2x² + 12x] + 6x + [2x² + 12x] + 6x



p = 2x² + 12x + 6x + 2x² + 12x + 6x



p = 4x² + 36x (in m² of course) ← answer a)







The area of the pavement being 7/8 of the area of the swimming pool



p = (7/8) * a



4x² + 36x = (7/8) * 72



4x² + 36x = (7 * 72)/8



4x² + 36x = (7 * 8 * 9)/8



4x² + 36x = 63



4x² + 36x - 63 = 0 ← answer b)









4x² + 36x - 63 = 0



4.[x² + 9x - (63/4)] = 0



x² + 9x - (63/4) = 0



x² + 9x + (9/2)² - (9/2)² - (63/4) = 0



x² + 9x + (9/2)² - (81/4) - (63/4) = 0



x² + 9x + (9/2)² - (144/4) = 0



x² + 9x + (9/2)² = 144/4



x² + 9x + (9/2)² = 36



[x + (9/2)]² = (± 6)²



x + (9/2) = ± 6



x = ± 6 - (9/2) → but x is a distance, so x must be a positive value



x = + 6 - (9/2)



x = (12/2) - (9/2)



x = 3/2 m ← answer c)

Answer 2

Did you make a diagram ? Why not ??



Area of pool: A = 12 * 6 = 72 m^2



Area of walkway: (7/8) * (72) = 63 m^2



a) if the pavement is x meters wide, then the dimensions of the pool AND walkway are:

(12 + 2x)m by (6 + 2x)m

Look at that diagram that you made...each side of the pool is extended by "x"



Area of (pool + pavement) - Area of pool = Area of pavement

(12 + 2x) * (6 + 2x) = 72 + 36x + 4x^2 - 72 = (4x^2 + 36x) m^2



b) Hence, Area of pavement is:

4x^2 + 36x = 63

4x^2 + 36x - 63 = 0



c) solving for x: solve the quadratic eqauation in (b)...can you do that ? Can you TRY ? Why not ??



What did you get ?



[4x^2 - 6x + 42x - 63 = 2x(2x - 3) + 21(2x - 3) = (2x - 3)(2x + 21) = 0

2x + 21 = 0, x = - 21/2 [extraneous]

or

x = 3/2 m

the pavement is 1 and 1/2 meters wide

check !!]

Answer 3

area of pool = 12×6 = 72 m²

area of pavement = 7/8 of 72 = 63 m²

total area = 72+63 = 135 m²



dimensions of combined area (pool+pavement) = (2x+12) by (2x+6)

(2x+12)(2x+6) = 135

4x²+36x+72 = 135

Subtract area of pool to get area of pavement:

4x²+36x = 63



By the quadratic formula,

x = [-36 ± √(36² – 4·4(-63))] / [2·4]

   = [-36 ± √2304] / 8

   = [-36 ± 48] /8

   = -10.5, 1.5

-10.5 is an extraneous solution. x = 1.5 meters

Answer 4

If the pool os 12 by 6, then the surround (of width x) must be 2 strips of 12 by x, 2 strips of 6 by x and for corner squares of x^2.

Area = 2(12x) + 2(6x) + 4x^2

= 4x^2 + 24x + 12x

= 4x^2 + 36x

And if the area is 7/8 of the area of the pool (12 by 6 is 72 m^2) . . .

4x^2 + 36x = 72 * 7/8

4x^2 + 36x = 63

4x^2 + 36x - 63 = 0

Solve by completing the square

4x^2 + 36x - 63 = 0

Take loose number over

4x^2 + 36x = 63

Divide by multiplier of squared term (4)

x^2 + 9x = 15.75

Take half of multiplier of x term, square it and add to both sides (9 becomes 4.5, becomes 20.25)

x^2 + 9x + 20.25 = 15.75 + 20.25

(x + 4.5)^2 = 36

x + 4.5 = +/-6

x = 1.5 or -10.5

x = 1.5m

Answer 5

Area of pool = 12(6) = 72..................[1]

Area of pavement = (7/8)(72) = 63......[2]



(a) If the pavement is x meters wide, dimensions of rectangle formed by pavement are:

Length = Length of pool + 2x = 12+2x

Width = Width of pool + 2x = 6 + 2x



The area of the rectangle formed by the pavement is therefore

(12+2x)(6+2x) = 72 + 36x + 4x²........[3]



The area of the pavement is just [3] - [1]

72 + 36x + 4x² - 72 =

4x² + 36x.....................ANS

(You have a typo in the answer you provided: it's not 4x²+36)

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

(b) We know from [2] that the area of the pavement is 63.

We also know from the answer to (a) that the area of the pavement is 4x² + 36x. Therefore, we can set them equal to each other:

4x² + 36x = 63

or,

4x² + 36x - 63 = 0...............ANS

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

(c) Given 4x² + 36x - 63 = 0, we can use the quadratic formula to solve for x to find out how wide the pavement actually is.



4x² + 36x - 63 = 0



......-36±√(36² - 4(4)(-63))

x = ------------------------------

.................2(4)



......-36±√(1296 + 1008)

x = -----------------------------

.................8



......-36±√2304

x = ---------------

...........8



......-36±48

x = ----------

...........8



......12...........-84

x = ---- or x = -----

.......8..............8



We can ignore the negative solution because you can't have a pavement of negative width. Therefore, the width of the pavement is

x = 12/8

x = 1.5m...............ANS

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