How do you solve \[{{x}^{2}}-81=0\]?

Hint: Write 81 as the product of its prime factors and if factors are repeated then write it in the exponent form. Try to write the factors with their exponent equal to 2. Now, apply the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to simplify the given quadratic equation and substitute each term equal to 0 to get the two roots of the equation.

Complete step by step answer:
Here, we have been provided with the quadratic equation \[{{x}^{2}}-81=0\] and we are asked to solve it. That means we have to find the values of x.
Now, as we can see that the given equation is quadratic in nature, so we must have two roots or values of x. Since, the coefficient of x in the above equation is 0, so we will not apply the middle term split method. Here, let us apply the factorization method using the algebraic identity: - \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\].
Now, writing 81 as the product of its primes, we get,
\[\Rightarrow 81=3\times 3\times 3\times 3\]
This can be written as: -
\[\Rightarrow 81={{3}^{2}}\times {{3}^{2}}\]
Using the formula of ‘exponents and powers’ given as: - \[{{a}^{m}}\times {{b}^{m}}={{\left( a\times b \right)}^{m}}\], we get,
\[\begin{align}
  & \Rightarrow 81={{\left( 3\times 3 \right)}^{2}} \\
 & \Rightarrow 81={{9}^{2}} \\
\end{align}\]
Now, substituting this converted form of 81 in the quadratic equation \[{{x}^{2}}-81=0\], we get,
\[\Rightarrow {{x}^{2}}-{{9}^{2}}=0\]
Using the algebraic identity: - \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\], we get,
\[\Rightarrow \left( x+9 \right)\left( x-9 \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( x+9 \right)=0\] or \[\left( x-9 \right)=0\]
\[\Rightarrow x=-9\] or \[x=9\]
Hence, the roots of the given quadratic equations are 9 and -9.

Note:
One may note that whatever method we may apply we will need the prime factorization of 81 at one step. So, you must know how to write a number as the product of its primes. Note that you can also use the discriminant method to get the answer. But it will not be a much different process that we have done above as you will still need the prime factorization.