Answer:

Step-by-step explanation:
Zeros of a quadratic equation are -1 and 5.
Sum of zeros = -1 + 5 = 4
Product of zeros = (-1)(5) = -5
The quadratic equation will be :

So, the required quadratic equation is
.
Relative extrema occur where the derivative is zero (at least for your polynomial function).
So taking the derivative we get
<span>20<span>x3</span>−3<span>x2</span>+6=0
</span><span>
This is a 3rd degree equation, now if we are working with complex numbers this equation is guaranteed to have 3 solutions by the fundamental theorem of algebra. But the number of real roots are 1 which can be found out by using Descartes' rule of signs. So the maximum number of relative extrema are 1.</span>
A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2. Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k.
V1 - velocity first train
v2 - velocity second train
v2 > v1
v2 - v1 = 17 mph
We know, that:
![s_1+s_2=210 \ [miles] \\ \\ t_1=t_2=2h](https://tex.z-dn.net/?f=s_1%2Bs_2%3D210%20%5C%20%5Bmiles%5D%20%5C%5C%20%5C%5C%20t_1%3Dt_2%3D2h)
So:

NOw we've got simple system of equations:
![+\begin{cases} v_2-v_1=17 \\ v_2+v_1=105\end{cases} \\ \\ 2v_2=122 \qquad /:2 \\ \\ v_2=61 \qquad [mph] \\ \\ v_2-v_1=17 \\ \\ 61-v_1=17 \\ \\ v_1=44](https://tex.z-dn.net/?f=%2B%5Cbegin%7Bcases%7D%20v_2-v_1%3D17%20%5C%5C%20v_2%2Bv_1%3D105%5Cend%7Bcases%7D%20%5C%5C%20%5C%5C%202v_2%3D122%20%5Cqquad%20%2F%3A2%20%5C%5C%20%5C%5C%20v_2%3D61%20%5Cqquad%20%5Bmph%5D%20%5C%5C%20%5C%5C%20v_2-v_1%3D17%20%5C%5C%20%5C%5C%2061-v_1%3D17%20%5C%5C%20%5C%5C%20v_1%3D44)
Velocities of these trains are 61mph and 44mph