Question

post test: relationships between functions
3
select the correct answer.
a carpenter is creating two new templates for his designs. one template will be in the shape of a right triangle, where the longer leg is 4 inches more than six times the shorter leg.
the second template will be in the shape of a rectangle, where the width is 5 inches more than the triangles shorter leg, and the length is 3 inches.
the carpenter needs the areas of the two templates to be the same. write a system of equations to represent this situation, where y is the area, and x is the length of the shorter leg of the triangle. which statement describes the number and viability of the systems solutions?
a. the system has only one solution, but it is not viable because it results in negative side lengths.
b. the system has two solutions, and both are viable because they result in positive side lengths.
c. the system has only one solution, and it is viable because it results in positive side lengths.
d. the system has two solutions, but only one is viable because the other results in negative side lengths.

Explanation:

Step1: Find area of right - triangle

The shorter leg of the right - triangle is $x$. The longer leg is $6x + 4$. The area of a right - triangle $y_1=\frac{1}{2}x(6x + 4)=3x^{2}+2x$.

Step2: Find area of rectangle

The width of the rectangle is $x + 5$ and the length is $3$. The area of the rectangle $y_2=3(x + 5)=3x+15$.

Step3: Set up the system of equations

Since $y_1 = y_2$, we have the equation $3x^{2}+2x=3x + 15$. Rearrange it to the standard quadratic form $3x^{2}-x - 15=0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 3$, $b=-1$, $c=-15$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values: $x=\frac{1\pm\sqrt{(-1)^{2}-4\times3\times(-15)}}{2\times3}=\frac{1\pm\sqrt{1 + 180}}{6}=\frac{1\pm\sqrt{181}}{6}$.
One solution is $x=\frac{1+\sqrt{181}}{6}>0$ and the other is $x=\frac{1 - \sqrt{181}}{6}<0$. Since side - lengths cannot be negative, only one of the two solutions is viable.

Answer:

D. The system has two solutions, but only one is viable because the other results in negative side lengths.