Answer:
[tex]x^4-5x^2-36=0[/tex]
Step-by-step explanation:
We need to find equation that has exactly two real and two non real solutions
2 real and 2 non real solution means 4 solutions
x^3 equation has maximum of 3 solutions
So we ignore equation that has largest exponent x^3
We ignore options B and D
Let check option A
[tex]x^4 - 36x^2=0[/tex]
factor the left hand side
Factor out x^2
[tex]x^2(x^2 - 36)=0[/tex]
WE set each factor =0 and solve for x
x^2 =0 so x=0
x^2 - 36 =0 so x= +-6
So solutions are x=0, x=6 , x=-6. Only 3 real solutions we got
LEts check with option C
[tex]x^4-5x^2-36=0[/tex]
Factor left hand side
[tex] (x^2-9)(x^2+4)=0[/tex]
set each factor =0 and solve for x
x^2 -9 =0 so x= -3, + 3
x^2 + 4 =0 , x^2 = -4, so x= +2i, -2i
So we got two real solutions (-3,+3) and two non real solutions (-2i,+2i)
