the length of a rectangle is 6 cm more than its width. Find its length and width when its area is 40 cm
width --- x
length --- x+6
x(x+6) = 42
x^2 + 6x - 42 = 0
x^2 + 9x + 9 = 42+9
(x+3)^2 = 51
x+3 = √51
x = √51 - 3 = appr 4.14
width is 4.14 , length is 10.14
check:
4.14(10.14) = appr 41.9796 , close enough for my round-off
I assume you know how to solve a quadratic that does not factor nicely.
I used completing the square instead of the formula.
If the middle term is even, that is easier and faster.
To solve this problem, we can use the formula for the area of a rectangle, which is length multiplied by width. We are given that the area of the rectangle is 40 cm². Let's call the width of the rectangle "w" and the length "l."
Based on the given information, we can create the equation: l × w = 40.
We are also given that the length is 6 cm more than the width. So we have: l = w + 6.
Now we can substitute the value of l in terms of w into the area equation:
(w + 6) × w = 40.
Expanding this equation gives us: w² + 6w = 40.
Rearranging the equation to have zero on one side gives us: w² + 6w - 40 = 0.
Now we have a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula. In this case, factoring is the most efficient method.
The equation factors to: (w + 10)(w - 4) = 0.
Setting each factor equal to zero gives us two possible solutions: w + 10 = 0 or w - 4 = 0.
For w + 10 = 0, we get: w = -10. Since the width of a rectangle cannot be negative, we can discard this solution.
For w - 4 = 0, we get: w = 4.
So the width of the rectangle is 4 cm. To find the length, we can substitute this value back into our equation l = w + 6.
l = 4 + 6 = 10.
Therefore, the length of the rectangle is 10 cm and the width is 4 cm when its area is 40 cm².