How do you solve for the equation? x^2+5x+2=0 and x^4+12x^2-64=0

Question

How do you solve for the equation?

x^2+5x+2=0

 

and

 

x^4+12x^2-64=0

Answer 1

x^2+5x+2=0; here the way of solving is The quadratic solution which gives us x=(-b±√(b^2 - 4ac))/2a having: a=1; b=5; c=2 Then x1= (-5+√17)/2= -0.8768/2 = 0.4384 and x2= (-5-√17)/2 = 4.56

For the second equation, x^4+12x^2-64=0, I will factor it in:

(X^2+16)*(x^2-4) =0 Then if the result is zero one of the factors is zero, then we may begin with (x^2+16) = 0 that will, gives x^2 = -16, then x1=+√(-16) and the second root will be x2= -√(-16) But it will lead us to complex numbers knowing that √(-1) =i so this answers are: x1= 4i and x2= -4i The equation is grade 4 so we need two more roots that will be obtained as: (x^2 -4)=0 => x^2=4; x3= 2 and x4=-2

This solves the two equations