Question

If alpha and beta are quadratic equation 9x²-24x+8=0then find .1) alpha plus beta plus three alpha beta.2)alpha cube plus beta cube

Answer 1

Answer

1) α+ β+ 3αβ =16/3

2) α³+ β³ = 320/27

Solution

Given ,

α and β are the roots of the equation 9x² - 24x + 8 = 0

We have from sum of the roots

$\sf sum \: of \: the \: roots = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ \\ \sf \implies \alpha + \beta = - \dfrac{ - 24}{9} \\ \\ \sf \implies \alpha + \beta = \frac{8}{3} \longrightarrow(i)$

And the product of the roots

$\sf product \: of \: the \: roots = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \\ \sf \implies \alpha \beta = \dfrac{8}{9} \longrightarrow(ii)$

Calculating (1) we have ,

$\leadsto \alpha + \beta + 3 \alpha \beta \\ \\ \sf \leadsto \dfrac{8}{3} + 3 \times \dfrac{8}{9} \\ \\ \sf \leadsto \dfrac{8}{3} + \dfrac{8}{3} \\ \\ \sf \leadsto2 \times \frac{8}{3} \\ \\ \sf \leadsto \dfrac{16}{3}$

Calculating (2) we have :

$\leadsto \alpha { }^{3} + { \beta }^{3} \\ \\ \sf \leadsto ( \alpha + \beta )( \alpha {}^{2} - \alpha \beta + \beta {}^{2} ) \\ \\ \sf \leadsto ( \alpha + \beta )( \alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta - 3 \alpha \beta ) \\ \\ \sf \leadsto \{ \alpha + \beta \} \{ (( \alpha + \beta ) {}^{2} - 3 \alpha \beta \} \\ \\ \sf \leadsto \dfrac{8}{ 3} \times \{ ( \dfrac{8}{3} ) {}^{2} - 3 \times \dfrac{8}{9} \} \\ \\ \ \leadsto \sf \dfrac{8}{3} \times \dfrac{40}{9} \\ \\ \sf \leadsto \frac{320}{27}$