Respuesta :
First, I will give the tickets the following names
Platinum tickets → P
Gold tickets → X
Silver tickets → S
Diamond tickets → D
You have this given information:
[tex]\begin{gathered} P=X \\ S+X=13 \\ D=X+13 \\ P\cdot X\cdot S\cdot D=3600 \end{gathered}[/tex]The last one is for the possible combinations of the independent variables
Now, you need to let the last equation just with x variable, so you have to replace the other variables as
[tex]\begin{gathered} P=X \\ S=13-X \\ D=13+X \end{gathered}[/tex]And then you will have the next equation
[tex]X\cdot X\cdot(13-X)\cdot(13+X)=3600[/tex]If you take a look (13-X)(13+X)=13²-X², replacing again
[tex]\begin{gathered} X^2\cdot(13^2-X^2)=3600 \\ X^2\cdot(169-X^2)=3600 \\ -X^4+169X^2=3600 \end{gathered}[/tex]At this point you will have to find the possible solutions for this equation
[tex]-X^4+169X^2-3600=0[/tex]You can substitute t for x²
[tex]-t^2+169t-3600=0[/tex]All equations of the form ax2+bx+c=0 can be solved using the quadratic formula:
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]If you replace a=-1, b=169 and c=-3600, then
[tex]\begin{gathered} t=\frac{-169\pm\sqrt[]{169^2-4(-1)(-3600)}}{2(-1)} \\ t=\frac{-169\pm119}{-2} \\ t=\frac{-169+119}{-2}=25\text{ and }t=\frac{-169-119}{-2}=144 \end{gathered}[/tex]Since t=x², then
[tex]\begin{gathered} X=\sqrt[]{t} \\ X=\sqrt[]{25}=5\text{ and }X=\sqrt[]{144}=12 \end{gathered}[/tex]Now the possible solutions for X are X=5 and X=12, and now you can evaluate the statements
For the first one, S=13-X, if X is 5 then S will be 8, and if X is 12 then S will be 1, so the first statement is false
The second one will be true because of the previous calculations we made
For the next 3 statements you can check the equation we use for the calculation of the gold tickets, as you can see, the one that matches is the third statement, this will be true. The other ones will be false then.
The answer will be: The second and the third statement are true
