Question

sol??????????????????​

Answer 1

Step-by-step explanation:

$given \: \\ k {x}^{2} + 4x + 4 \\ { \alpha }^{2} + { \beta }^{2} = 24 \\ now \\ \alpha \beta = \frac{c}{a } \\ = > \alpha \beta = \frac{4}{k} \\ \\ \alpha + \beta = - \frac{b}{a} \\ = > \alpha + \beta = - \frac{4}{k} \\ ( {\alpha + \beta})^{2} = \frac{16}{ {k}^{2} } \\ = > {k}^{2} ( { \alpha }^{2} + { \beta }^{2} + 2\alpha \beta) = 16 \\ = > {k}^{2} (24 + \frac{8}{k} ) = 16 \\ = > {k}^{2} ( \frac{24k + 8}{k} ) = 16 \\ = > 24 {k}^{2} + 8k = 16 \\ = > 6 {k}^{2} + 2k - 4 = 0 \\ = > 6 {k}^{2} +6k - 4k - 4 = 0 \\ = > 6k(k + 1) - 4(k + 1) = 0 \\ = > (k + 1)(6k - 4) = 0 \\ either \: k = - 1 \: or \: k = \frac{2}{3}$