Question

In the following equations, find which variables x, y, z etc. represent rational or irrational numbers:

(i) x² = 5 

(ii) y² = 9 

(iii) z² = 0.04 

(iv) u² = \(\frac{17}{4}\)

(v) v² = 3 

(vi) w² = 27

(vii) t² = 0.4

Answer 1

(i) We have,

x² = 5 

Taking square root on both sides,

\(= \sqrt{x}^2 = \sqrt5\)

= x = \(\sqrt5\)

\(\sqrt5\)

is not a perfect square root, so it is an irrational number.

(ii) We have,

y² = 9 

y = \(\sqrt9\)

= 3 = \(\frac{3}{1}\)

\(\sqrt9\) can be expressed in the form of \(\frac{p}q\) , so it is a rational number.

(iii) We have, 

z² = 0.04 

Taking square root on both the sides, we get,

\(\sqrt{z}^2\) = \(\sqrt{0.04}\)

\(z = \sqrt{0.04}\)

\(= 0.2 = \frac{2}{10}\)

\(= \frac{1}{5}\)

z can be expressed in the form of \(\frac{p}q\) , so it is a rational number.

(iv) We have,

\(u^2 = \frac{17}{4}\)

Taking square root on both the sides, we get

\(\sqrt{u}^2\) \(= \frac{\sqrt{17}}{\sqrt4}\)

u = \(\frac{\sqrt{17}}{\sqrt2}\)

Quotient of an rational number is irrational, so u is an irrational number.

(v) We have,

v ² = 3 

Taking square roots on both the sides, we get,

\(\sqrt{v}^2\) = \(\sqrt3\)

v = \(\sqrt3\)

\(\sqrt3\) is not a perfect square root, so v is an irrational number.

(vi) We have,

w² = 27 

Taking square roots on both the sides, we get,

\(\sqrt{w}^2\) = \(\sqrt{27}\)

w = \(\sqrt3\times\sqrt3\times\sqrt3\) = \(\sqrt[3]{3}\)

Product of a rational number and an irrational number is irrational number. So, it is an irrational number.

(vii) We have,

t² = 0.4 

Taking square roots on both the sides, we get,

\(\sqrt{t}^2\) = \(\sqrt{0.4}\) = \(\frac{\sqrt4}{\sqrt{10}}\)

\(= \frac{2}{\sqrt{10}}\)

Since, quotient of a rational number and an irrational number is irrational number, so t is an irrational number.