For an equation to be viable, the result of the equation must be positive (i.e. greater than 1).
The 2 templates, the farmer created have two solutions, but only one of the solutions are viable.
For the first template, we have:
[tex]x \to[/tex] length of the shorter leg of the triangle
[tex]z \to[/tex] length of the longer leg of the triangle
From the question, we have:
[tex]z = 4 + 6x[/tex]
The area (y) of the first template is:
[tex]y = \frac{1}{2}xz[/tex] --- area of a triangle
So, we have:
[tex]y = \frac{1}{2}x(4 + 6x)[/tex]
Open bracket
[tex]y = 2x + 3x^2[/tex]
For the second template, we have:
[tex]Width = 5 + x[/tex]
[tex]Length = 3[/tex]
So, the area (y) of the second template is:
[tex]y = Length \times Width[/tex] --- area of a rectangle
This gives:
[tex]y = 3 \times (5 + x)[/tex]
[tex]y = 3(5 + x)[/tex]
[tex]y = 15 + 3x[/tex]
So, the expression for both areas are:
[tex]y = 2x + 3x^2[/tex] --- template 1
[tex]y = 15 + 3x[/tex] --- template 2
Both areas must be equal. This is represented as:
[tex]2x + 3x^2 = 15 + 3x[/tex]
Collect like terms
[tex]3x^2 +2x - 3x - 15 = 0[/tex]
[tex]3x^2 - x - 15 = 0[/tex]
Solve for x using quadratic formula, we have:
[tex]x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]
Where:
[tex]a = 3; b =-1; c=-15[/tex]
So, we have:
[tex]x = \frac{-(-1) \± \sqrt{(-1)^2 - 4\times 3 \times -15}}{2 \times 3}[/tex]
[tex]x = \frac{1 \± \sqrt{181}}{6}[/tex]
[tex]x = \frac{1 \± 13.5}{6}[/tex]
Split
[tex]x = \frac{1 + 13.5}{6} \ or \ x = \frac{1 + 13.5}{6}[/tex]
[tex]x = \frac{-12.5}{6} \ or \ x = \frac{14.5}{6}[/tex]
[tex]x = -2.08 \ or \ x = 2.42[/tex]
We can see that x has 2 solutions, but only 1 of the solutions is viable because the other is negative.
Hence, option (a) is correct
Read more about viable and non-viable solutions at:
https://brainly.com/question/10558256
