Question

Which of the following is not irrational? (a) (2-√3)2 (b)(√2+√3)2 (c) (√2-√3)(√2+√3) (d)27√7​

Answer 1

Answer: (c)

(√2-√3)(√2+√3)

Step-by-step explanation:

An irrational number is one that cannot be expressed as the ratio of two integers. In other words it cannot be expressed as $\frac{x}{y}$ where x and y are integers

$\sqrt{2}$ is irrational

$\sqrt{3}$ is irrational

In fact the square root of any prime number is irrational. So $\sqrt{5}$, $\sqrt{7}$ etc are irrational. But $\sqrt{9}$ is not irrational since it evaluates to 3 which can be expressed as $\frac{3}{1}$

Any expression that contains the square root of a prime number is also irrational

Looking at the choices we see that choices (a), (b) and (d) all evaluate to expressions containing square roots of primes

(a)  (2-√3)2 = 4 - 2√3  . Hence irrational

(b) √2+√3)2 = 2√2+2√3. Hence irrational

(d) 27√7​ is irrational

Let's look at choice (c)

(√2-√3)(√2+√3)

An expression $(a+b)(a-b)$  can be evaluated as $a^{2} - b^{2}$

Here a = √2, $a^{2}$ =$a = \sqrt{2}\\ a^2 = (\sqrt{2} )^2 = 2\\\\b = \sqrt{3} \\b^2 = (\sqrt{3} )^2 = 3\\\\a^2 - b^2 = 2-3 = -1$

This is a whole number(integer) and all integers are rational numbers

Hence correct answer is (c)