Question

check whether root 2 + root 3 whole square is rational or irrational

Answer 1

Given:

$(\sqrt{2}+\sqrt{3})^{2}$

We have to check whether $(\sqrt{2}+\sqrt{3})^{2}$ is a rational or irrational.

Solution:

∴ $(\sqrt{2}+\sqrt{3})^{2}$

Using the algebraic identity:

$(a+b)^{2} =a^{2} +2ab+b^{2}$

= $(\sqrt{2})^{2}$ + $(\sqrt{3})^{2}$ + $2(\sqrt{2})(\sqrt{3})$

= 2 + 3 + 2$\sqrt{6}$

= 5 + 2$\sqrt{6}$

Suppose 5 + 2$\sqrt{6}$ is a rational.

5 + 2$\sqrt{6}$ = $\dfrac{a}{b}$ Where, a and b are integers

⇒ 2$\sqrt{6}$ = $\dfrac{a}{b}$ - 5

⇒ 2$\sqrt{6}$ = $\dfrac{a-5b}{b}$

Dividing both sides by 2, we get

⇒ $\sqrt{6}$ = $\dfrac{a-5b}{2b}$

Here, $\dfrac{a-5b}{2b}$ is a rational number.

∴ $(\sqrt{2}+\sqrt{3})^{2}$ is an irrational.

Thus, $(\sqrt{2}+\sqrt{3})^{2}$ is an irrational.

Answer 2

Answer:

it is irrational guys

mark as brainliest to get the proof