Question

How do you solve $ {x^4} - 8{x^2} - 9 = 0? $ ?

Answer 1

Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factorising an expression is to take out any common factors which the terms have. So if we were asked to factor the expression $ {x^2} + x $ , since $ x $ goes into both terms, we would write $ x(x + 1) $ . Here we will use identities which will help us to factorise an algebraic expression easily i.e. $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ and $ {a^2} - {b^2} = (a + b)(a - b) $ .

Complete step by step solution:
Since we cannot factorise it normally so we will simplify the equation. We can rewrite $ {x^2} $ as $ y $ . So the equation will be $ {y^2} - 8y - 9 = 0 $ .
We will now simplify the equation normally by middle term factorisation: $ {y^2} - 8y - 9 = 0 $ . It gives us
 $ {y^2} - 9y + y - 9 = 0 \\
\Rightarrow y(y - 9) + 1(y - 9) = 0
 $ .
We can now write it as taking the common factors together we have: $ (y - 9)(y + 1) $ . We will now substitute $ {x^2} $ in the place of $ y $ . So we have $ ({x^2} - 9)({x^2} + 1) $ . We can see that $ ({x^2} - 9) $ is a difference of two squares, so by applying the above difference of squares formula we can write it as
 $ \Rightarrow {x^2} - 9 = ({x^2} - {3^2}) $ . Hence we have $ (x - 3)(x + 3) $ .
By putting this value in the previous equation: $ (x - 3)(x + 3)({x^2} + 1) $ .Now we will calculate the values of $ x $ . Since $ x - 3 = 0 \Rightarrow x = 3 $ . From another factor we have $ x + 3 = 0\\
\Rightarrow x = - 3
 $ .
In the third part
$ {x^2} + 1 = 0 \\
\Rightarrow {x^2} = - 1 $
WE will take the both negative and positive root which means that $ x = \pm \sqrt { - 1} $ .
Hence the answer of the given equation is $ x = \pm i,3 $ and $ - 3 $ .
So, the correct answer is “ $ x = \pm i,3 $ and $ - 3 $ ”.

Note: We should keep in mind while solving these expressions that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give the wrong answer. We also know that iota i.e. $ i $ is equal to the value of $ \sqrt { - 1} $ . These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeros.