Answer:
a) Have 3 solutions (0; -0.38...; -2.618...)
b) Have 3 solutions (0; 5; 1)
c) Have 1 real solution and 2 complex
Step-by-step explanation:
a) x^3+3x^2+x=x(x^2+3x+1)=0, so if x=0 the equation is zero too, and if x^2+3x+2=0 the eq is zero again. To find the roots of x^2+3x+2, I have to use the quadratic formula, [tex](-b+-\sqrt{b^{2}-4ac } ) /2a[/tex], with a=1, b=3 and c=1 the solutions are [tex]\frac{-3-\sqrt{5} }{2}[/tex] and [tex]\frac{-3+\sqrt{5} }{2}[/tex]
b) x^4-6x^3+5x^2=x^2(x^2-6x+5), so if x^2=0 the equations 0, then x=0 is a double root, if I want to know the value of x for x^2-6x+5=0, then I use the quadratic formula again with a=1, b=-6 and c=5, and the solutions are 5 and 1
c) I couldn't calculate this with the analytical methods, so I had to do an aproximation and the unique real solution was approximately 0.87
