Question

$f (x) = {2x}^{2} + 5x + k \: polynomials \: zeros \: are \: \: \alpha \: and \\ \beta \: and \: it \: proves \: \alpha { }^{2} + \beta {}^{2} + \alpha \beta = \frac{21}{4} \: than \: find \: the \\ possible \: value \: of \: \ \: \: \: k \:$
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Answer 1

$f(x) = 2 {x}^{2} + 5x + k \\ there \: \: \: \: \: \: \: \: \: \: \: \: \: b = 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: c = k\: \: \\ the \: \: \: zeroes \: \: \: are \: \: \alpha \: \: and \ \: \beta \\ so \: \: \: \: \alpha + \beta = \frac{ - b}{a } = \frac{ - 5}{2} \\ and \: \: \\ \: \: \: \: { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{21}{4} \\ = > {( \alpha + \beta )}^{2} - \alpha \beta = \frac{21}{4} \\ = > {( \frac{ - 5}{2} )}^{2} - \alpha \beta = \frac{ 21}{4} \\ = > \frac{25}{4} - \alpha \beta = \frac{21}{4} \\ = > \alpha \beta = \frac{25}{4} - \frac{21}{4} \\ = > \alpha \beta = \frac{25 - 21}{4} \\ = > \alpha \beta = \frac{4}{4} \\ = > \alpha \beta = 1 \\ \\ we \: \: know \: \: that \: \: \\ \: \: \: 2 {x}^{2} + 5x + k \\ = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ = {x}^{2} - ( - \frac{5}{2} )x + 1 \\ = {x}^{2} + \frac{5}{2} x + 1 \\ = 2 {x}^{2} + 5x + 1 \\ so \: \\ \: \: 2 {x}^{2} + 5x + k = 2 {x}^{2} + 5x + 1 \\ then \: \: \: \: \: \: \: \: k = 1$