Question

α, β are roots of y²-2y-7=0 find 1) α²+β²
2) α³+β³​

Answer 1

$\huge\boxed{\fcolorbox{red}{ink}{SOLUTION:}}$

$\huge\boxed{\fcolorbox{purple}{ink}{GIVEN:}}$

α, β are roots of y2 – 2y –7 = 0 

$\huge\boxed{\fcolorbox{pink}{ink}{TO FIND:}}$

1) α²+β²

$\huge\fbox\pink{✯Answer✯}$

$a= 1,b= -2 ,c= -7$

$\alpha + \beta = \frac{ - b}{ \alpha } = \frac{- ( - 2)}{1} = 2$

$\alpha \beta = \frac{c}{ \alpha } = \frac { - 7}{1} = - 7$

${ \alpha }^{2} + { \beta }^{2} = ( \alpha + { \beta }^{2}) - 2 \alpha \beta$

$= ( {2)}^{2} - 2( - 7)$

$= 4 + 14$

$= 18$

$\huge\boxed{\fcolorbox{purple}{ink}{GIVEN:}}$

α, β are roots of y2 – 2y –7 = 0 

$\huge\boxed{\fcolorbox{pink}{ink}{TO FIND:}}$

1) α²+β²

β³=(α+β)(α²+β²+αβ)----(1)

$\huge\fbox\pink{✯Answer✯}$

Putting all values in (1)

we get

α³+β³=(2) × (18+7)

=2 × 25=50

Answer 2

${ \huge{ \color{aquamarine}{ \fcolorbox{magenta}{black}{ \textsf{ \textbf{Answer}}} \: { \purple\uparrow}}}}$

1st answer in first picture.

2nd answer in 2nd and 3rd picture.