The length of a square doubled and the width of the square increased by 5 to form a rectangle. If the area of the rectangle is 48 cm2. What is the side length of the original square?
Help with this please!!
if the square has side s, then
(2s)(s+5) = 48
Old square ..... x by x
new square .... 2x by x+5
2x(x+5) = 48
expand, re-arrange, then solve this quadratic
To solve this problem, we can set up the following equation:
Area of rectangle = length * width
We are given that the area of the rectangle is 48 cm². Let's assume the side length of the original square is "x" cm.
So, the length of the rectangle will be 2x cm (since the length of the square is doubled), and the width will be (x + 5) cm (since the width of the square is increased by 5).
Now we can substitute these values into the equation:
48 cm² = (2x cm) * (x + 5 cm)
To simplify further, we can distribute the 2x through the parentheses:
48 cm² = 2x² + 10x cm
Now, let's rearrange the equation to get it in standard quadratic form:
2x² + 10x cm - 48 cm² = 0
To solve this quadratic equation for x, we can either factor, complete the square, or use the quadratic formula. Let's use factoring:
2x² + 10x cm - 48 cm² = 0
Divide the equation by 2 to simplify:
x² + 5x - 24 cm² = 0
Now we need to find two numbers that multiply to -24 and add up to 5. The numbers are +8 and -3.
So, the factored equation becomes:
(x + 8 cm)(x - 3 cm) = 0
Setting each factor equal to zero gives us:
x + 8 = 0 or x - 3 = 0
Solving for x, we get:
x = -8 cm or x = 3 cm
We can ignore the negative value, as lengths cannot be negative.
Therefore, the side length of the original square is 3 cm.