Question is in description

Question

Please explain answer with work/explanation. Very very confused. 

If the area of a rectangle is 126 square inches and the width of the rectangle is 7 inches shorter than the length, what is the length, in inches, of the rectangle (actual question) 

Answer 1

Area of a rectangle = width * height

 

Assuming, that length of the rectangle = x inches

width of the rectangle would be = x - 7 inches

 

Plug these in the area formula:

126 = (x - 7)*x

126 = x² - 7x

x² - 7x -126 = 0

 

Now, we apply quadratic formula to solve for x:

 

x = [-b ±√ (b² - 4ac)] / 2a

x = [7 ±√{(-7)² - 4*1*(-126)}] / 2

x = [7 ±√{49 + 504}] / 2

x = [7 ±√553] / 2

x = [7 ±23.52] / 2
x = 15.26 and -8.26

 

Since length can not be negative, we take the positive value.

 

Therefore, length of the rectangle = 15.26 inches

Answer 2

If one side's length is x, then the other is x+7.  The area of the rectangle is found by multiplying the two sides or

   x(x+7)= 126.  Multiply that out gives:  x² +  7x -126 =0.

 

Use the quadratic equation -b +/- Sqr(b²-4ac)/2a gives -7 +/- sqr(49+504)/2

 

Negative answers are impossible, so

 

(-7+23.5)/2 = 16.5/2 = 8.25 in

longer side is +7 or 15.25 inches

 

Check your answer:

 

8.25 x 15.25 = 125.8, different from 126 because of rounding off numbers.

 

Answer 3

Let x = length

 

Then, width = x - 7

 

So, x(x-7) = 126

 

x²- 7x - 126 = 0

 

Using the quadratic formula, x = [7 ± √(49 + 504)]/2

 

                                                      = [7 ± √553]/2

 

                                                      ≈ 15.26, -8.26

 

Since length can't be negative, x = length ≈ 15.26"