Question

9. Classify the following as rational or irrational

(a) (√3+√2)(√3-√2)​

Answer 1

Given: (√3 + √2)(√3 - √2)

To find: Classify whether the given expression is rational or irrational.

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Understanding the question: There is a expression given to us and we've to classify it as rational or irrational, We'll use a suitable law of exponent and simplify it !

▪︎ As a, b can be classified, we can use any law of exponent according to the question.

✇ Now, By using the law of exponent given below. Then let's find out whether the given question is rational or irrational —

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$\star\:{\underline{\boxed{\pmb{\sf{\Big( \sqrt{a} - \sqrt{b} \Big) \Big( \sqrt{a} + \sqrt{b} \Big) = \Big( \sqrt{a} \Big)^2 - \Big( \sqrt{b} \Big)^2}}}}}$

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$\begin{gathered}\twoheadrightarrow \sf \big(\sqrt{3} - \sqrt{2} \Big) \Big( \sqrt{3} + \sqrt{2} \Big) \sf \\\\\\\twoheadrightarrow\sf \Big( \sqrt{a} - \sqrt{b} \Big) \Big( \sqrt{a} + \sqrt{b} \Big) = \Big( \sqrt{a} \Big)^2 - \Big( \sqrt{b} \Big)^2\\\\\\\twoheadrightarrow\sf \big(\sqrt{3} - \sqrt{2} \Big) \Big( \sqrt{3} + \sqrt{2} \Big) = \Big( \sqrt{3} \Big)^2 - \Big( \sqrt{2} \Big)^2\\\\\\\twoheadrightarrow\sf 3 - 2\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{\dfrac{1}{1}}}}}}\;\bigstar\end{gathered}$

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Therefore,

• (√3 + √2)(√3 - √2) is rational number as it can be expressed in p/q form.(1/1)

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$\therefore{\underline{\pmb {\sf{Hence, (\sqrt{3} - \sqrt{2}) (\sqrt{3} + \sqrt{2} ) \: is \: a \: rational \: number, Respectively.}}}}$

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⠀⠀⠀⠀⠀✇ $\underline{ \pmb{ \frak{ \red{Additional \: information...}}}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto \sqrt{\dfrac{a}{b}} \: = \dfrac{\sqrt{a}}{\sqrt{b}}}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto (a+\sqrt{b}) (a-\sqrt{b}) = a^2 - b}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto \sqrt{ab} \: = \sqrt{a} \sqrt{b}}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b}) (\sqrt{c} - \sqrt{d}) = \sqrt{ac} + \sqrt{as} + \sqrt{bc} + \sqrt{bd}}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto (\sqrt{a} + \sqrt{b})^{2} \: = a + 2\sqrt{ab}+b}}$

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• $\; \; \; \; \; \; \;{\sf{\leadsto ( {a} - {b})^{2} \: = {a}^{2} + 2{ab}+ {b}^{2} }}$