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$aandb$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)$$\left(a+b\right)$use have $a^2+ab-ab-b^2$
$\left(+ab\right)$and $\left(-ab\right)$ cancel each other because negative numbers and positive numbers cancel each other.
So we have
$a^2-b^2$ … (i)
In the question $(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})$
$$\begin{gathered} a=\sqrt{2} \\ b=\sqrt{3} \end{gathered}$$
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.