Q.1 Find the value of 𝑐 for which the pair of equations 𝑐𝑥 − 𝑦 = 2 and 6𝑥 − 2𝑦 = 3 will have infinitely many solutions.

OR

Q.2 Find the roots of the quadratic equation $\sqrt{3}{x}^2-2x-\sqrt{3}=0$.

Here, we have been given the equations 𝑐𝑥 − 𝑦 = 2 and 6𝑥 − 2𝑦 = 3 and we need to find the value of 𝑐.

It is known that the system of equations 𝑎₁𝑥 + 𝑏₁𝑦 + 𝑐₁ = 0  and 𝑎₂𝑥 + 𝑏₂𝑦 + 𝑐₂ = 0 has infinitely many solutions when $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

By comparison, we can conclude, 𝑎₁ = 𝑐, 𝑎₂ = 6, 𝑏₁ =− 1, 𝑏₂ =− 2, 𝑐₁ =− 2 𝑎𝑛𝑑 𝑐₂ =− 3

$\begin{array}{r}\text{ Now, }\frac{a_1}{a_2}=\frac{c}{6}\\ \frac{b_1}{b_2}=\frac{-1}{-2}\\ \Rightarrow \frac{b_1}{b_2}=\frac{1}{2}\\ \text{ Also, }\frac{c_1}{c_2}=\frac{-2}{-3}\\ \Rightarrow \frac{c_1}{c_2}=\frac{2}{3}\end{array}$

From the above results, we can observe that, $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

This result is not in accordance with equation (𝑖). Hence, there is no value of 𝑐 to obtain infinitely many solutions for the given set of equations.

OR

We have been given the quadratic equation $\sqrt{3}{x}^2-2x-\sqrt{3}=0$  and we need to find its roots.

Factorising by splitting the middle term, we get

Taking terms common,

$\begin{array}{r}\Rightarrow \sqrt{3}x(x-\sqrt{3})+1(x-\sqrt{3})=0\\ \Rightarrow (x-\sqrt{3})(\sqrt{3}x+1)=0\\ \Rightarrow x-\sqrt{3}=0\text{ or, }\\ (\sqrt{3}x+1)=0\\ \Rightarrow x=\sqrt{3}\text{ or }\frac{-1}{\sqrt{3}}\end{array}$