The area of the front cover of a daily journal is 273cm^2. , and the length is 8cm greater than the width. What are the dimensions of the cover?
Let x = width, then the length = x +8
x(x+8) = 273
x^2 + 8x - 273 = 0
Factor and solve for x.
To find the dimensions of the cover, let's represent the width as 'x' cm.
According to the given problem, the length is 8 cm greater than the width. So, the length can be represented as 'x + 8' cm.
Now, we have the dimensions of the cover. The area of a rectangle is calculated by multiplying its length and width.
So, we can set up the equation as follows:
Area = Length * Width
273 cm^2 = (x + 8) cm * x cm
Expanding the equation:
273 cm^2 = x^2 + 8x
Rearranging the equation to make it quadratic:
x^2 + 8x - 273 = 0
To solve this quadratic equation, we can either factorize it or use the quadratic formula. Since the quadratic does not easily factorize, let's use the quadratic formula.
The quadratic formula is given as:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation x^2 + 8x - 273 = 0, the values of a, b, and c are:
a = 1, b = 8, c = -273
Now, let's substitute these values into the quadratic formula:
x = (-8 ± √(8^2 - 4*1*(-273))) / (2*1)
Simplifying the equation:
x = (-8 ± √(64 + 1092)) / 2
x = (-8 ± √1156) / 2
x = (-8 ± 34) / 2
We get two possible values for 'x':
x = (-8 + 34) / 2 = 26 / 2 = 13
x = (-8 - 34) / 2 = -42 / 2 = -21
Since dimensions cannot be negative, we discard the negative value. Therefore, the width (x) of the cover is 13 cm.
Now, let's find the length by using the previously determined relationship:
Length = Width + 8
Length = 13 + 8 = 21 cm
Hence, the dimensions of the cover are width = 13 cm and length = 21 cm.