Question

Find the nature of the product $(\sqrt 2 - \sqrt 3 )\,\,(\sqrt 3 + \sqrt 2 ).$

Answer 1

Hint: Firstly, We will consider two equation$\left( {a - b} \right)\,\,\left( {a + b} \right)$.Further we will multiply the equation$\left( {a - b} \right)\,\,\left( {a + b} \right) = {a^2} - {b^2}$.Thereafter we will replace the value of$a and b $instead of $a = \sqrt 2 and = \sqrt 3 $to get the answer.


Complete step by step solution
 Let us consider two equation
$\left( {a - b} \right)\,\,\left( {a + b} \right)$ in product.
$\left( {a - b} \right)\,\,\left( {a + b} \right)$
$ \Rightarrow \,a \times a$ (product of a)
$ = {a^2}$
$ \Rightarrow a \times \left( { + b} \right)$ (product of a with$ + b)$
$ = ab$
$ \Rightarrow \, - b \times a$ (product of $ - b$ with a)
$ = - \,ab$
$ \Rightarrow - \,b \times + b$ (product of $ - b$with$ + b)$
$ = - \,{b^2}$
Solve by $\left( {a - b} \right)$$\left( {a + b} \right)$use have ${a^2} + ab - ab - {b^2}$
$\left( { + \,ab} \right)$and $\left( { - \,\,ab} \right)$ cancel each other because negative numbers and positive numbers cancel each other.
So we have
${a^2} - {b^2}$ … (i)
In the question $(\sqrt 2 - \sqrt 3 )\,\,(\sqrt 2 + \sqrt 3 )$
$
  a = \sqrt 2 \\
  b = \sqrt 3 \\
 $
Put the value of a and b in equation (i)
${(\sqrt 2 )^2} - {(\sqrt 3 )^2}$
The power and square root are cancelled each other. So we have
$2 - 3 = - 1$
$ = - \,1$ ans.


Note: Students must know that the roots of the equation are an irrational number. Then they exist in pairs. If one root is $(\sqrt 2 - \sqrt 3 )\,$ and other will be $(\sqrt 2 + \sqrt 3 ).$And the product of $(\sqrt 2 - \sqrt 3 )\,\,(\sqrt 2 + \sqrt 3 ) = - \,1$.Their product is real number.