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2 Creating Quadratic Equations to Fi... 15 Error Analysis Can You Find the Error? A student was solving Example D and made a mistake. Identify the error made by the student. Example D: A rectangular lot is 20 yards longer than it is wide and its area is 2400 square yards. Find the dimensions of the lot. As a Visual: As an Equation: Solution to the Problem: W + 20 w(w+20) = 2400 w+20w = 2400 w+20w - 2400 = 0 (w +60)w - 40) = 0 The rectangle is 40 yards by 60 yards. 3 W+60 = 0w-40=0 w = 60 and w = 40 The student's mistake was... The dimensions of the lot are...

2 Creating Quadratic Equations to Fi 15 Error Analysis Can You Find the Error A student was solving Example D and made a mistake Identify the error made by the class=

Respuesta :

Given data:

The expression for length of rectangle is l=w+20.

The area of rectangle A=2400 square-yards.

The expression for the area of area of rectangle is,

[tex]\begin{gathered} A=l\times w \\ 2400=(w+20_{})w \\ w^2+20w-2400=0 \\ w^2+60w-40w-2400=0 \\ w(w+60)-40(w+60)=0 \\ (w-40)(w+60)=0 \\ w=\text{ 40 yards, or w= -60 yards} \end{gathered}[/tex]

As the side of the rectangle can't be negative, so width is 40 yards.

The length of rectangle is,

l=(40+20) yards

=60 yards.

Thus, the width of rectangle is 40 yards and length is 60 yards, the mistake made by the student he took width 60 yards but it is not possible.

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