Question

How do you solve ${x^4} - 81 = 0$?

Answer 1

Hint: We will first use the identity given by ${a^2} - {b^2} = (a - b)(a + b)$, then we will get two quadratic equations and thus we can solve then individually to get the required roots.

Complete step-by-step answer:
We are given that we are required to solve ${x^4} - 81 = 0$.
We know that we have an identity given by the following expression:-
$ \Rightarrow {a^2} - {b^2} = (a - b)(a + b)$
Replacing a by ${x^2}$ and b by 9, we will then obtain the following equation:-
\[ \Rightarrow {\left( {{x^2}} \right)^2} - {9^2} = ({x^2} - 9)({x^2} + 9)\]
Simplifying the left hand side, we will then obtain the following equation:-
\[ \Rightarrow {x^4} - 81 = ({x^2} - 9)({x^2} + 9)\]
Putting this, we can write the given equation as the following:-
\[ \Rightarrow ({x^2} - 9)({x^2} + 9) = 0\]
This implies that either \[{x^2} - 9 = 0\] or \[{x^2} + 9 = 0\].
This implies that \[{x^2} = 9\] or \[{x^2} = - 9\].
This implies that \[x = \pm 3\] or \[x = \pm 3i\].

Thus we have the roots of the given equation ${x^4} - 81 = 0$ as 3, - 3, 3i and – 3i.

Note:
The students must note that after we got the product of two quadratics as the given expression, we have used a property named as “zero product property” which states that if we have: a.b = 0, then either a = 0 or b = 0 or both.
Using this, we obtained the roots of the given equations.
The students must note that the square of i is – 1.
The square of any real number can never be equal to any negative number. Therefore, when we got the square of x as – 9, then we got the imaginary roots that are 3i and – 3i.
The students must commit to memory the following formula:-
 ${a^2} - {b^2} = (a - b)(a + b)$
Zero Product Property: If a . b = 0, then either a = 0 or b = 0 or both.
The students, if confused about the roots of quadratics can also use the formula of quadratics if they are getting difficulty in finding the roots of the equations.