ANSWER
a. 2 real roots, 2 imaginary roots
Step-by-step explanation
The given equation is
![{x}^(4) - 64 = 0](https://img.qammunity.org/2020/formulas/mathematics/high-school/ij85mw73om7uzjwtn8ni2jla6s3l8qb3it.png)
We rewrite as difference of two squares,
![( {x}^(2) )^(2) - {8}^(2) = 0](https://img.qammunity.org/2020/formulas/mathematics/high-school/g0uelg6woucz7bozoggyg0g525lb0yn9a2.png)
We factor using difference of two squares to get;
![( {x}^(2) - 8)( {x}^(2) + 8) = 0](https://img.qammunity.org/2020/formulas/mathematics/high-school/fvkvyil94jbizw8h38biv57qej4ixm5ism.png)
We now use the zero product property to get:
![{x}^(2) = 8 \: or \: {x}^(2) = - 8](https://img.qammunity.org/2020/formulas/mathematics/high-school/15ccxvxca5tbmokkz7pz930bfqhhw9dg00.png)
Take the square root of both sides to get;
![{x} = \pm √(8) \: or \: {x}^(2) = \pm √( - 8)](https://img.qammunity.org/2020/formulas/mathematics/high-school/4jppuguspcvoktntu01t6fmel33fekprtf.png)
![{x} = \pm 2√(2) \: or \: {x} = \pm 2√( 2) i](https://img.qammunity.org/2020/formulas/mathematics/high-school/wqavhz7uhnh5n9wk9tflt4o0mt3n9bikn3.png)
![{x} = - 2√(2) \: or \: {x} = 2√( 2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/3niy1bego6om1uqwuly09p2sets22nxuho.png)
are two real roots.
![{x} = - 2√(2)i \: or \: {x} = 2√( 2) i](https://img.qammunity.org/2020/formulas/mathematics/high-school/ys2z5ff2og36c3l1p61qc7dorcsnub1tnf.png)
are two imaginary roots.
The correct answer is A.