Answer:
[tex]x = 2, x =-2, x = 1.73, x = -1.73[/tex]
Step-by-step explanation:
We are given the equation:
[tex]x^4+x^2=4x^3-12x+12\\x^4-4x^3+x^2+12x-12=0[/tex]
The attached image shows the graph for the given equation.
We can also factorize the given equation in the following manner.
By hit and trial method, we checked that 2 is a root of the given equation, that is,
[tex](2)^4-4(2)^3+(2)^2+12(2)-12 = 0[/tex]
Thus, we can write:
[tex](x-2)(x^3-2x^2-3x+6 ) = 0\\(x-2)(x^2(x-2)-3(x-2)) = 0\\(x-2)(x-2)(x^2-3) = 0\\(x-2)(x-2)(x-\sqrt3)(x+\sqrt3)=0\\x=2,x=-2, x = \sqrt3, x = -\sqrt3[/tex]
Thus, the roots of given equation are:
[tex]x = 2, x =-2, x = 1.73, x = -1.73[/tex]
