Question

the value of k for which the system of equations x²-4=0 & 2x+ky =3 has no solutio pi ln is​

Answer 1

Answer:

k ∈ $R -${${\frac{4}{3}$}

Step-by-step explanation:

For no solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$x^2+0y - 4 = 0 \\(a_1 = 1 , b_1 = 0 , c_1=-4)\\\\2x+ky-3=0\\(a_2 = 2 , b_2 = k , c_2=-3)$

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\\\\\frac{1}{2} = \frac{0}{k} \neq \frac{-4}{-3}\\\\from \frac{1}{2} = \frac{0}{k} :$

k ∈ R

$also, k \neq \frac{4}{3}$

∴ k ∈ $R -${${\frac{4}{3}$}