alculator AllowedAlgebra I, Subpart 3The length of a garden is 6 feet more than the width. The area of the garden is 40 square feet.What is the length, in feet, of the garden?Enter your answer in the space provided.
Q. alculator AllowedAlgebra I, Subpart 3The length of a garden is 6 feet more than the width. The area of the garden is 40 square feet.What is the length, in feet, of the garden?Enter your answer in the space provided.
Define Variables: Let's call the width of the garden w (in feet). Then the length is w+6 feet.
Calculate Area: The area of a rectangle is given by the formula length×width. So, the area of the garden is (w+6)×w.
Set Up Equation: We know the area is 40 square feet, so we set up the equation (w+6)×w=40.
Expand and Simplify: Now we need to solve for w. Expanding the equation gives us w2+6w=40.
Factor the Equation: To solve the quadratic equation, we need to set it to zero. So, w2+6w−40=0.
Set Factors to Zero: We can factor this equation to find the values of w. Factoring gives us (w+10)(w−4)=0.
Solve for Width: Setting each factor equal to zero gives us w+10=0 or w−4=0.
Find Length: Solving for w, we get w=−10 or w=4. Since width can't be negative, we take w=4 feet.
Find Length: Solving for w, we get w=−10 or w=4. Since width can't be negative, we take w=4 feet.Now we find the length by adding 6 to the width. So, the length is 4+6=10 feet.
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