To determine the viability and number of solutions for the given system of equations:
[tex]\[ y = x^2 - 6x + 8.75 \][/tex]
[tex]\[ y = \frac{1}{4}x + 5 \][/tex]
we need to solve this system of equations.
1. Set the equations equal to each other to find the points where they intersect:
[tex]\[ x^2 - 6x + 8.75 = \frac{1}{4}x + 5 \][/tex]
2. Move all terms to one side of the equation to set it to zero:
[tex]\[ x^2 - 6x + 8.75 - \frac{1}{4}x - 5 = 0 \][/tex]
Combine like terms:
[tex]\[ x^2 - 6.25x + 3.75 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( x^2 - 6.25x + 3.75 = 0 \)[/tex] for [tex]\( x \)[/tex]:
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex].
[tex]\[ a = 1, \, b = -6.25, \, c = 3.75 \][/tex]
Using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-(-6.25) \pm \sqrt{(-6.25)^2 - 4(1)(3.75)}}{2(1)} \][/tex]
[tex]\[ x = \frac{6.25 \pm \sqrt{39.0625 - 15}}{2} \][/tex]
[tex]\[ x = \frac{6.25 \pm \sqrt{24.0625}}{2} \][/tex]
[tex]\[ x = \frac{6.25 \pm 4.9051}{2} \][/tex]
Therefore, the two solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{6.25 + 4.9051}{2} \approx 5.58 \][/tex]
[tex]\[ x = \frac{6.25 - 4.9051}{2} \approx 0.672 \][/tex]
4. Check the viability of both solutions (i.e., whether they are positive):
Both [tex]\( x = 5.58 \)[/tex] and [tex]\( x = 0.672 \)[/tex] are positive values of time.
Thus, the system has two solutions, and both are viable because they result in positive values of time.
Therefore, the correct statement that describes the number and viability of the system's solutions is:
D. The system has two solutions, and both are viable because they result in positive values of time.
