Question

Q. The number(1+√3)² is
(a) natural number
(b) irrational number
(c) rational number
(d) integer​

Answer 1

Answer:

$(1 + \sqrt{3} ) {}^{2} = 1 {}^{2} + ( \sqrt{3} ) {}^{2} + 2 \times \sqrt{3} \times 1 \\ = 1 + 3 + 2 \sqrt{3} = 4 + 2 \sqrt{3}$

It is an irrational number

Answer 2

Given:

(1+√3)²

To find:

The type of number

Solution:

The given number is an 'irrational number'. (Option b)

We can determine the number's type by obtaining its value using the given identity.

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

The given number=(1+√3)²

On solving, we get

=$1^{2}$+$\sqrt{3} ^{2}$ +2×1×√3

=1+3+2√3

=4+2√3

=2(2+√3)

Now, since the number contains √3, which is irrational, the number becomes irrational.

The sum of a number that is rational and another that is irrational is irrational.

Therefore, the given number is an 'irrational number'.