Question

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

Answer 1

Hint: Here we can solve this polynomial using the formula of a difference between two perfect squares ${a^2} - {b^2}$ can be factored into $\left( {a + b} \right)\left( {a - b} \right)$. On doing some simplification we get the required answer.

Complete step-by-step solution:
Given the equation ${x^4} - 81 = 0$, it is the fourth degree polynomial equation so we have four roots.
Simplifying ${x^4} - 81 = 0$ first we reorder the terms,
$ \Rightarrow - 81 + {x^4} = 0$
Solving for variable $'x'$ ,
Move all terms containing $x$ to the left, all terms to the right
Add $'81'$ to each side of the equation.
$ \Rightarrow - 81 + 81 + {x^4} = 0 + 81$
Combine like terms: $ - 81 + 81 = 0$
$ \Rightarrow 0 + {x^4} = 0 + 81$
We get,
$ \Rightarrow {x^4} = 0 + 81$
Combine like terms: $0 + 81 = 81$
$ \Rightarrow {x^4} = 81$
Reorder the terms:
$ \Rightarrow {\left( {{x^2}} \right)^2} = {\left( 9 \right)^2}$
Combine the above term we get,
$ \Rightarrow {\left( {{x^2}} \right)^2} - {\left( 9 \right)^2} = 0$
Using the formula of a difference between two perfect squares ${a^2} - {b^2}$ can be factored into $\left( {a + b} \right)\left( {a - b} \right)$ we get,
$ \Rightarrow \left( {{x^2} + 9} \right)\left( {{x^2} - 9} \right) = 0$
Simplifying the term we get,
$ \Rightarrow \left( {{x^2} + 9} \right)\left( {{x^2} - {3^2}} \right) = 0$
Again we using the formula of a difference between two perfect squares ${a^2} - {b^2}$ can be factored into $\left( {a + b} \right)\left( {a - b} \right)$ we get,
$ \Rightarrow \left( {{x^2} + 9} \right)\left( {x - 3} \right)\left( {x + 3} \right) = 0$
Now set the factor $\left( {{x^2} + 9} \right)$ equal to zero and attempt to solve:
Simplifying ${x^2} + 9 = 0$ ,
Move all terms containing $x$ to the left, all other terms to the right
Add $' - 9'$ to each side of the equation, we get,
$ \Rightarrow 9 + ( - 9) + {x^2} = 0 + ( - 9)$
Combine like terms: $9 + ( - 9) = 0$
$ \Rightarrow 0 + {x^2} = 0 + ( - 9)$so we get,
$ \Rightarrow {x^2} = - 9$
Simplifying, ${x^2} = - 9$
The solution to this equation could not be determined.
But we can simplify this factor by using complex number $'i'$ ,
We can change the number like this,
$ \Rightarrow {x^2} = {\left( {3i} \right)^2}$
Because ${i^2} = - 1$ ,
Reorder the term like this we get,
$ \Rightarrow {x^2} - {\left( {3i} \right)^2} = 0$
Now we using the formula of a difference between two perfect squares ${a^2} - {b^2}$ can be factored into $\left( {a + b} \right)\left( {a - b} \right)$ we get,
$ \Rightarrow \left( {x + 3i} \right)\left( {x - 3i} \right) = 0$
Therefore we have the factors of the given equation that is $\left( {x + 3i} \right)\left( {x - 3i} \right)\left( {x + 3} \right)\left( {x - 3} \right) = 0$

And then the roots of the equation is $x = 3\,,\, - 3\,,\,3i\,,\, - 3i$

Note: We can simply use this formula to solve the given fourth degree polynomial equation.
That is $\left( {{a^4} - {b^4}} \right) = \left( {{a^2} + {b^2}} \right)\left( {a + b} \right)\left( {a - b} \right)$
Simply substitute from our equation ${x^4} - 81 = {x^4} - {3^4}$ , $a = x$ and $b = 3$ because $81 = {3^4}$ we get,
 $\left( {{x^4} - {3^4}} \right) = \left( {{x^2} + {3^2}} \right)\left( {x + 3} \right)\left( {x - 3} \right)$