Question

3s^2-6s+2 find the zeroes​

Answer 1

Answer:

Answer:

\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } + 2\bigg( \frac{1}{ \alpha } + \frac{1}{ \beta } \bigg) + 3 \alpha \beta = 8 \\ \\ \\

Given that

3 {s}^{2} - 6s + 4 \\ \\

equation has two zeros,namely

\alpha \: \: and \: \: \beta \\ \\

from coefficient and zeros relation

\alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a} \\ \\

here

a = 3 \\ \\ b = - 6 \\ \\ c = 4 \\ \\

\alpha + \beta = \frac{6}{3} = 2 \\ \\ \alpha \beta = \frac{4}{3} \\ \\

To find

\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } + 2\bigg( \frac{1}{ \alpha } + \frac{1}{ \beta } \bigg) + 3 \alpha \beta \\ \\ \\ = > \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } + 2\bigg( \frac{ \alpha + \beta }{ \alpha \beta } \bigg) + 3 \alpha \beta \\ \\

we don't have the value of

{ \alpha }^{2} + { \beta }^{2} \\ \\

to find that

\alpha + \beta = 2 \\ \\ square \: both \: side \\ \\ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = 4 \\ \\ { \alpha }^{2} + { \beta }^{2} = 4 - 2 \alpha \beta \\ \\ { \alpha }^{2} + { \beta }^{2} = 4 - 2 \times \frac{4}{3} \\ \\ { \alpha }^{2} + { \beta }^{2} = 4 - \frac{8}{3} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{ 4 }{3} \\ \\

Now put all these values

\frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } + 2\bigg( \frac{ \alpha + \beta }{ \alpha \beta } \bigg) + 3 \alpha \beta \\ \\ \\ = > \frac{ \frac{4}{3} }{ \frac{4}{3} } + 2\bigg( \frac{2}{ \frac{4}{3} } \bigg) + 3 \times \frac{4}{3} \\ \\ \\ = > 1 + \frac{4 \times 3}{4} + 4 \\ \\ \\ = > 1 + 3 + 4 \\ \\ = > 8 \\ \\

Hope it helps you.