Question

The front walkway from the street to Pam's house has an area of 250ft^2. Its length is two less than four times its width. Find the length and width of the walkway. Round to the nearest tenth of a foot.

Answer 1

Answer:

Step-by-step explanation:

rectangle is 250ft2. Substituting into the formula for the area of a rectangle, A=length×width, we have

250250=(4w−2)(w)=4w2−2w

In the standard form aw2+bw+c=0, this is

4w2−2w−250=0

Substituting the coefficients a=4, b=−2, and c=−250 into the quadratic formula, we have

w=−b±b2−4ac‾‾‾‾‾‾‾‾√2a=−(−2)±(−2)2−4(4)(−250)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√2(4)=2±4,004‾‾‾‾‾√8

There are two solutions for w, which we can evaluate on a calculator.

w=≈2+4,004‾‾‾‾‾√88.2andw=≈2−4,004‾‾‾‾‾√8−7.7

The width of the rectangle must be positive, so w=8.2. The length is then given by

4w−2=4(8.2)−2=30.8

Thus, Pam's front walkway has a width of 8.2ft and a length of 30.8ft.

Answer 2

The length and width of the walkway is mathematically given as

L= 30.64 ft

W = 8.16

What is the length and width of the walkway?

Question Parameters:

The front walkway from the street to Pam's house has an area of 250ft^2

Generally, the equation for the  Area  is mathematically given as

a=LW

a= 250

Therefore

W(4W - 2) = 250

4W^2 - 2W- 250 = 0

Generally, the equation for the  quadratic equations is mathematically given as

W = [- b ± √b^2 - 4ac]/2a

Therefore

$W= [- (- 2) \pm \sqrt{-2^2 -4(4 * -250)}/2*4$

W= (2 ±63.277)/8

W = 8.16 or W = - 7.91

In conclusion, considering the positive figure, the lenght is

LW = 250

8.16L = 250

L = 250/8.16

L= 30.64 ft

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