Question
Examine, whether the following numbers are rational or irrational:
$\big(\sqrt{2}+\sqrt{3}\big)^2$
$\big(\sqrt{2}+\sqrt{3}\big)^2$
✓
Answer
$\big(\sqrt{2}+\sqrt{3}\big)^2$
We have,
$\big(\sqrt{2}+\sqrt{3}\big)^2=2+2\sqrt{6}+3=5+\sqrt{6}$ [Since, (a + b)2 = a2 + 2ab + b2]
The sum of a rational number and an irrational number is an irrational number, so $\big(\sqrt{2}+\sqrt{3}\big)^2$ is an irrational number.
We have,
$\big(\sqrt{2}+\sqrt{3}\big)^2=2+2\sqrt{6}+3=5+\sqrt{6}$ [Since, (a + b)2 = a2 + 2ab + b2]
The sum of a rational number and an irrational number is an irrational number, so $\big(\sqrt{2}+\sqrt{3}\big)^2$ is an irrational number.
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