Question

d) If a and B are the zeroes of polynomial p(x) =
4x² – 5x - 1, then find the value of a² + B + aß?​

Answer 1

$\alpha \: and \: \beta \: are \: roots \: of \\ 4 {x}^{2} - 5x - 1 \\ \\ \alpha = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \alpha = \frac{ 5 + \sqrt{41} }{8} \\ \beta = \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \beta = \frac{5 - \sqrt{41} }{8} \\ \\ = > { \alpha }^{2} = \: \frac{66 + 10 \sqrt{41} }{64} = \frac{33 + 5 \sqrt{41} }{32} \\ = > { \beta }^{2} = \frac{66 - 10 \sqrt{41} }{64} = \frac{33 - 5 \sqrt{41} }{32} \\ = > \alpha \beta = ( \frac{5 + \sqrt{41} }{8} )( \frac{5 - \sqrt{41} }{8} ) \\ = > \alpha \beta = - 2 \\ \\ \\ so \\ \\ \: \: \: \: \: { \alpha }^{2} + { \beta }^{2} + \alpha \beta \\ = \frac{33 + 5 \sqrt{41} }{32} + \frac{33 - 5 \sqrt{41} }{32} + ( - 2) \\ = \frac{66}{32} - 2 = \frac{33}{16} - 2 = \frac{33 - 32}{16} = \frac{1}{16}$

Hence,

${ \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{1}{16}$