Question

Write whether $\dfrac{{2\sqrt {45} + 3\sqrt {20} }}{{2\sqrt 5 }}$ on simplification gives as rational or an irrational.

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Hint:Here we need to factorize the root numbers which were in the given mathematical expression.To find a factorization of the root number we use LCM method.
On some simplification, we get the solution and find whether it is rational or irrational.
According to the definition of rational number, A number is said to be rational number if it is in the form of$\;\dfrac{p}{q},{\text{ q}} \ne 0$ and $p,{\text{ q}} \in I$ where I is the set of integers)

It is given that the expression, $\dfrac{{2\sqrt {45} + 3\sqrt {20} }}{{2\sqrt 5 }}$
Take an expression as equation (1)
Now we can write, $\dfrac{{2\sqrt {45} + 3\sqrt {20} }}{{2\sqrt 5 }}..... \to (1)$
Then, we find factors of the root value $45,20$ by using LCM (Least Common Multiple) Method.
On finding factors,
$45 = 3 \times 3 \times 5$
Similarly, we can find factors for
$\;20 = 2 \times 2 \times 5$
Substituting the value of $45$ in the root can be expressed as
We are getting,
$\therefore \sqrt {45} = \sqrt {3 \times 3 \times 5} = 3\sqrt 5$
Similarly, we do this process for $20$
Substituting the value of $20$ in the root can be expressed as
We are getting,
$\therefore \sqrt {20} = \sqrt {2 \times 2 \times 5} = 2\sqrt 5$
Substituting the values of $\sqrt {45}$ and $\sqrt {20}$ in the equation $(1)$
We will get,$\dfrac{{2 \times 3\sqrt 5 + 3 \times 2\sqrt 5 }}{{2\sqrt 5 }}$
Now take common factor $2\sqrt 5$ in the numerator part, we get
$= \dfrac{{2\sqrt 5 \left( {3 + 3} \right)}}{{2\sqrt 5 }}$
On some simplification we will get the answer
=6, which is the rational number.
Hence, the given expression is a rational number on simplification.

Note:
Definition of irrational number:A number is said to be rational number if it is in the form of $\;\dfrac{p}{q},{\text{ q}} \ne 0$ and $p,{\text{ q}} \in I$ where I is the set of integers).In this question, it is in factor formation, so we can say this expression is not irrational on simplification.