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The Second Solution
High-school algebra is more than five decades in my rearview mirror, but
the more things change, the more high-school algebra remains pretty
much the same. The books are a lot more interesting graphically and a
whole lot more expensive—but they still contain problems such as the one
in the next paragraph.
Susan’s garden has the shape of a rectangle. The area of the garden is 50
square yards, and the length of the garden exceeds the width by 5 yards.
What are the dimensions of the garden?
The setup for this problem is straightforward. If you let L and W denote
the dimensions of the garden, then you have the following equations.
LW 50 (area 50 square yards)
L 5 W (length exceeds width by 5 yards)Substituting the second equation into the first results in the quadratic
equation L(L5)50. Regrouping and factoring, L^2 5 L 50 0 (L10)
(L5). There are two solutions to this equation. One of these is L10;