Question

if alfa and bita are zeros of 5x^2+x-2 then find (alfa upon bita+bita upon alfa​)

Answer 1

Answer:

-21/10

Step-by-step explanation:

See attached pic.

Answer 2

SOLUTION

5x^2+ x-2

$\alpha \: and \: \beta \\ \\ = > \alpha + \beta = \frac{ - b}{a} = \frac{ - 1}{5} \\ \\ = > \alpha \beta = \frac{c}{a} = \frac{ - 2}{5} \\ \\ = > \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{ { \alpha }^{2} + \beta {}^{2} }{ { \alpha \beta }} = \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ = > \frac{ (\alpha \beta ) {}^{2} }{ \alpha \beta } - 2 \\ \\ = > \frac{ \frac{ (\frac{ - 1}{5}) {}^{2} }{ - 2} }{5} - 2 \\ = > \frac{ 1}{25} \times \frac{5}{ - 2} - 2 \\ = > \frac{1}{ - 10} - 2 \\ \\ = > \frac{1 - 20}{ - 10} = \frac{ - 19}{ - 10}$

hope it helps