The length of a rectangle is 7 less than twice of it's width if the area of the rectangle is 15 square meters find the value of x
Let the width of the rectangle be x meters.
Therefore, the length of the rectangle is 2x - 7 meters.
The area of the rectangle is given by the formula:
Area = length x width
15 = (2x - 7) x x
15 = 2x^2 - 7x
Rearrange the equation into standard form:
2x^2 - 7x - 15 = 0
Now we need to solve for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = -7 and c = -15
Plugging in the values:
x = (7 ± √((-7)^2 - 4*2*(-15))) / 2*2
x = (7 ± √(49 + 120)) / 4
x = (7 ± √169) / 4
x = (7 ± 13) / 4
Therefore, we have two possible values for x:
x = (7 + 13) / 4 = 20 / 4 = 5
x = (7 - 13) / 4 = -6 / 4 = -1.5
Since the width can't be negative, x = 5 meters.