Question

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James wants to build a rectangular field with an area of 135 square meters, and he wants the length of the field to be 6 meters more than its width. James created a quadratic equation to model the situation and graphed the equation to determine the width of the field.





Use the equation and the graph to select the true statement about the solution(s) for the equation.

A.
The equation has only one solution, and it is a reasonable width.
B.
The equation has two solutions, and both are reasonable widths.
C.
The equation has two solutions, but one must be discarded because it is not a reasonable width.
D.
The equation has only one solution, but it must be discarded because it is not a reasonable width.

Answer 1

Let L and W be the length and width

LW = 135

L = W + 6

(W+6)W = 135

W² + 6W - 135 = 0

There's our quadratic equation.  I don't see the graph, but it factors as

(W - 9)(W + 15) = 0

so the graph is of a CUP (concave up) parabola that goes below the x axis (ok, the W axis here), intersecting at W=-15 and W=9.

We have a negative and a positive solution.  A negative width isn't reasonable, so the answer is

Answer: C

Answer 2

The correct option is C, The equation has two solutions, but one must be discarded because it is not a reasonable width.

What is a rectangle?

A parallelogram in which adjacent sides are perpendicular to each other is called a rectangle. A rectangle is always a parallelogram and a quadrilateral but the reverse statement may or may not be true.

Given the length of the rectangle is 6 more than the length. Therefore, if the width of the rectangle is x units, then the length of the rectangle will be (x+6).

The area of the rectangle will be,

x(x+6) = 135

x² + 6x - 135 = 0

x² + 15x - 9x -135=0

(x+15)(x-9)=0

x = -15, 9

Since -15 can not be the width of the rectangle, the width of the rectangle will be 9.

Hence, the correct option is C, The equation has two solutions, but one must be discarded because it is not a reasonable width.

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