Question

Simplify the following: \[\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)\]

Answer 1

Hint: We know that the square of any square root is always the number itself, for example let's say x is a number in which we applied square root and square simultaneously. Keep in mind that square root is basically \[\dfrac{1}{2}\] in the power of x
\[\therefore \sqrt {\left( {{x^2}} \right)} = {(x)^{2 \times \dfrac{1}{2}}} = x\] . Use this to solve the given equation.
Complete step by step answer:
First try to break the equation by multiplying them separately i,e., we can rewrite the given equation as
\[\begin{array}{l}
\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)\\
 = 3 \times 3 + \sqrt 3 \times 3 - 3 \times \sqrt 3 - \sqrt 3 \times \sqrt 3 \\
 = 9 + 3\sqrt 3 - 3\sqrt 3 - \sqrt {3 \times 3} \\
 = 9 + 3\sqrt 3 - 3\sqrt 3 - \sqrt {{{\left( 3 \right)}^2}} \\
 = 9 - 3\\
 = 6
\end{array}\]
 Therefore the solution for the given expression is 6.

Note: An alternative approach of doing the same question can be we can directly use an algebraic formula of \[{a^2} - {b^2} = (a + b)(a - b)\] if we put \[a = 3\& b = \sqrt 3 \] we will get it as
\[\begin{array}{l}
 = {3^2} - {\left( {\sqrt 3 } \right)^2}\\
 = 9 - 3\\
 = 6
\end{array}\]
Therefore in both the methods we are getting the same output.