# [✓] Find multiple solutions for one equation?

GROUPS:
 When -5<=a<=5, the equation x^5-(a^2+4)x^3+x^2+a^2 x+2a+1=0 has 3, 4 or 5 solutions, depending on the value of a. Use a dynamic plot of the graph of the left side of this equation to determine the values of a for which the equation has: 3, 4, and 5 solutions. Can anyone help show me how to find these solutions?
 Marco Thiel 1 Vote Here are some comments:i) this is a fifth order polynomial and has 5 solutions (in the complex numbers). Reduce[x^5 - (a^2 + 4) x^3 + x^2 + a^2 x + 2 a + 1 == 0, x] ii) I suppose you are looking for real solutions?What about this: sols = Table[{b, Length[x /. Normal[Solve[x \[Element] Reals && x^5 - (a^2 + 4) x^3 + x^2 + a^2 x + 2 a + 1 == 0 /. a -> b, x]]]}, {b, -5, 5, 0.01}]; iii) Of course, there are lots of ways of getting this: Table[{b, Length[Select[x /. Normal[Solve[x^5 - (a^2 + 4) x^3 + x^2 + a^2 x + 2 a + 1 == 0 /. a -> b, x]], Im[#] == 0 &]]}, {b, -5, 5, 0.001}] Alternatively, you can work with Reduce. Cheers,Marco
 Did you mean something like this? Manipulate[ Plot[x^5 - (a^2 + 4) x^3 + x^2 + a^2 x + 2 a + 1, {x, -3, 3}, PlotRange -> {-40, 40}], {{a, 0}, -5, 5}]