For what value of 𝑘 the system of equations 𝑘𝑥 + 3𝑦 = 1, 12𝑥 + 𝑘𝑦 = 2 have no solution?

                                                  (OR)

Find the value of 𝑘, for which the roots of the quadratic equation $2x^2+kx+8=0$.

Will they have equal roots ?

Solution:

The given pair of equations are 𝑘𝑥 + 3𝑦 − 1 = 0 and 12𝑥 + 𝑘𝑦 − 2 = 0

The pair of equations will have no solution if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$\text{ Where }{a}_1=k,a_2=12,b_1=3,b_2=k,c_1=-1,c_2=-2\text{, }$

$\begin{array}{r}\Rightarrow \frac{k}{12}=\frac{3}{k}\\ \Rightarrow k^2=36\\ \Rightarrow k=\pm 6\\ \Rightarrow \frac{3}{k}=\frac{-3}{-6}(k\ne 6)\\ \Rightarrow k=-6\end{array}$

Hence, the value of 𝑘 is − 6.

                                                    (OR)

The given equation has equal roots. 

The equation will have equal roots if discriminant (𝐷) = 0 𝑎 = 2, 𝑏 = 𝑘, 𝑐 = 8

$\begin{array}{r}a=2,b=k,c=8\\ \Rightarrow b^2-4ac=0\\ \Rightarrow k^2-4\left(2\right)\left(8\right)\\ \Rightarrow k^2=64\\ \Rightarrow k=\pm 8\end{array}$

Hence, the value of 𝑘 is ±8.