I would like some help on this problem....
A rectangular photograph has an area of 40 square inches. If the width of the photograph is 3 inches more than its height, find the dimensions of the photograph.
well, 5*8 = 40
Not
Bot
To solve this problem, we need to set up and solve an algebraic equation based on the given information.
Let's assume the height of the photograph is "x" inches. Since the width is 3 inches more than the height, the width can be expressed as "x + 3" inches.
We know that the area of a rectangle is given by multiplying its length (width) by its width (height). Therefore, we can set up the following equation:
Area = Length x Width
40 = x(x + 3)
Now, we can solve this quadratic equation to find the value of x.
To solve the quadratic equation, we need to multiply out the terms:
40 = x^2 + 3x
Next, we'll subtract 40 from both sides to set the equation equal to zero:
x^2 + 3x - 40 = 0
Finally, we'll factor the quadratic equation:
(x + 8)(x - 5) = 0
Setting each factor equal to zero gives us two possible solutions:
x + 8 = 0 or x - 5 = 0
If x + 8 = 0, then x = -8, which is not a valid solution since the height of a rectangle cannot be negative.
If x - 5 = 0, then x = 5. This is a valid solution, and it represents the height of the rectangle.
Now that we have the height, we can find the width as x + 3:
Width = 5 + 3 = 8 inches
Therefore, the dimensions of the rectangle photograph are 5 inches by 8 inches.