Question

if alpha and beta be the zeros of polynomial 2 X square + 5 x + K such that Alpha square + beta square + alpha beta is equal to 21 upon 4 then K is equal to​

Answer 1

$putting \: values \: \\ \\ ( \frac{ - 5 - \sqrt{25 - 8k} }{4} ) {}^{2} + ({ \frac{ - 5 + \sqrt{25 - 8k} }{4} })^{2} + ( \frac{ - 5 - \sqrt{25 - 8k} }{4} )\times \frac{ - 5 + \sqrt{25 - 8k} }{4} = \frac{21}{4}$

We know that

$( {x + y})^{2} + ( {x - y})^{2} = 2 {x}^{2} + 2 {y}^{2}$

$= > { \alpha }^{2} + { \beta }^{2} =2( \frac{25 - 25 + 8k}{16} )$

$= > { \alpha }^{2} + { \beta }^{2} = \frac{2 \times 8k}{16} = k$

Again

$\alpha \beta = \frac{25 - 25 + 8k}{16} \\ \\ = > \alpha \beta = \frac{k}{2}$

$= > { \alpha }^{2} + { \beta }^{2} + \alpha \beta = k + \frac{k}{2} = \frac{21}{4} \\ \\ k + \frac{k}{2} = \frac{21}{4} \\ \\ = > \frac{2k + k}{2} = \frac{21}{4}$

$= > \frac{3k}{2} = \frac{21}{4}$

$= > k = \frac{21}{4} \times \frac{2}{3} \\ \\ = > k = \frac{7}{2}$