Question

3+root2/ 4root2
Rationalize denominator

Answer 1

Answer:

$\frac{3}{4}$

Step-by-step explanation:

$\frac{3 + \sqrt{2}}{4\sqrt{2}}$

⇒$\frac{3 + \sqrt{2}}{4\sqrt{2}} \times \frac{4 \sqrt{2}}{4 \sqrt{2}}$

⇒$\frac{12 \times 2}{16 \times 2}$

⇒$\frac{12 \times \cancel{2}}{16 \times \cancel{2}}$

⇒$\frac{12}{16}$

⇒$\frac{\cancel{12}}{\cancel{16}}$

⇒$\frac{6}{8}$

⇒$\frac{\cancel{6}}{\cancel{8}}$

⇒$\frac{3}{4}$

Explaination of answer:

• Here, we've to rationalise the denomination. We can't write the root $\sqrt{2}$ in the denominator. So, we'll rationalise it.

• (Or) in simple words, we'll write in the form of $\frac{p}{q}$ where, p and q both are integers and q ≠ 0.

• As we know that, $\sqrt{2}$ is not a integer, it is a irrational number as it's decimal expansion is non - terminating and non - recurring (non - repeating).

• As we can see in the in step - 2 we've multiplied the denominator in the numerator as well as denominator, as we are equalising the denominator.

• So that, the root and root will be get cancelled and we'll get a rationalised value.

• And thus, we're abled to rationalise the denominator.

Answer 2

$\huge{\underline{\red{\mathcal{Hey, Mate..!}}}}$      

$\huge{\orange{AnSwEr:-}}$

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