Question

p(x)=4x²‐6x+1 find alfa³+beta³​

Answer 1

Question

p(x)=4x²‐6x+1 find alfa³+beta³

Solution

$sum \: of \: zeros = \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha + \beta = \frac { - (- 6)}{ 4} \\ \\ \alpha + \beta = \frac{6}{4} \\ \\ on \: cubing \: both \: sides \\ \\ {( \alpha + \beta )}^{3} = ( { \frac{6}{4} })^{3} \\ \\ { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta ) = { \frac{216}{64} }^{} - - - - (1)\\ \\ \\ as \: we \: know \: \\ product \: ofzeros = \alpha \beta = \frac{1}{4} \\ \\ by \: putting \: value \: in \:$

$equation \: 1 \: \\ \\ { \alpha }^{3} + { \beta }^{3} + 3 \frac{1}{4} ( \frac{6}{4} ) = \frac{216}{64} \\ \\ { \alpha }^{3} + { \beta }^{3} + \frac{18}{16} = \frac{216}{64} \\ \\ { \alpha }^{3} + { \beta }^{3} = \frac{216}{64} - \frac{18}{16} \\ \\ { \alpha }^{3} + { \beta }^{3} = \frac{216 - 72}{64} \\ \\ { \alpha }^{3} + { \beta }^{3} = \frac{144}{64} \\ \\ { \alpha }^{3} + { \beta }^{3} = 2.25$