How do you solve $ {x^4} - 8{x^2} - 9 $ ?

Hint: To determine the factors of the above equation assume $ {x^2}\,as\,t $ use the Splitting up the middle term method to find the roots that later put back the value of $ t $ what we assumed.

Complete step-by-step answer:
Given a quadratic equation $ {x^4} - 8{x^2} - 9 $ let it be $ f(x) $
 $ f(x) = {x^4} - 8{x^2} - 9 $
Rewriting the equation
 $ f(x) = {\left( {{x^2}} \right)^2} - 8({x^2}) - 9 $
Let us assume $ {x^2}\,as\,t $ ,now our above equation becomes
 $ f(x) = {t^2} - 8t - 9 $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
a becomes $ 1 $
b becomes $ - 8 $
And c becomes $ - 9 $
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of $ {t^2} $ and the constant term which comes to be
 $ = - 9 \times 1 = - 9 $
Now the second Step is to find the 2 factors of the number $ 9 $ such that the weather addition or subtraction of those numbers is equal to the middle term or coefficient of $ t $ and the product of those factors results in the value of constant .
So if we factorize $ 9 $ ,the answer comes to be $ - 9 $ and $ 1 $ as $ - 9 + 1 = - 16 $ that is the middle term . and $ 9 \times 1 = 9 $ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained, so equation $ f(x) $ becomes
 $ f(x) = {t^2} - 9t + t - 9 $
Now taking common from the first 2 terms and last 2 terms
 $ f(x) = t(t - 9) + 1(t - 9) $
Finding the common binomial parenthesis, the equation becomes
 $ f(x) = (t + 1)(t - 9) $
Putting back the value of t
 $ f(x) = ({x^2} + 1)({x^2} - 9) $
Using the formula $ \left( {{A^2} - {B^2}} \right) = (A + B)(A - B) $ in the second part
 $ f(x) = ({x^2} + 1)(x - 3)(x + 3) $
Now equating our equation to 0 to find the solution
 $ 0 = ({x^2} + 1)(x - 3)(x + 3) $
Thus,
 $
  x + 3 = 0 \\
   \Rightarrow x = - 3 \\
  x - 3 = 0 \\
   \Rightarrow x = 3 \\
  and \\
  {x^2} + 1 = 0 \\
   \Rightarrow {x^2} = - 1 \\
   \Rightarrow x = \pm \sqrt { - 1} \;
  $
We know that $ - 1 = {i^2} $
 $ \Rightarrow x = \pm i $
Value of x can be $ 3, - 3,i, - i $
Therefore, the solution to the equation $ {x^4} - 8{x^2} - 9 $ can be $ 3, - 3,i, - i $ .
So, the correct answer is “ $ 3, - 3,i, - i $ .”.

Note: You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
 $ x_1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} $ and $ x_2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} $
$x_1$ , $x_2$ are root to quadratic equation $ a{x^2} + bx + c $
 Hence the factors will be $ (x - (x_1))\,and\,(x - (x_2))\, $ .
One must be careful while calculating the answer as calculation error may come.
Don’t forget to compare the given quadratic equation with the standard one every time.