Question

If alpha beta are the zeros of px is equal to x square - 2 x + 3 - 1 + 1

Answer 1

$\bold \red \star \: \underline{Given:-} \begin{cases} \sf{ \underline{Zero \: of :-}} \\ \sf{ \alpha \: \: and \: \: \beta } \\ \sf{ - 1 \: \: and \: \: 1} \end{cases}$

$\\ \\ \bold \red \star \: \large\underline{Solution:-}$

$Our \: \gray{\underline{Equation }}\: Given \\ \\P( \pink{x}) = {x}^{2} - 2x + 3. \\ \\ \{ \therefore \: \: where \: as \: \: P( \pink{x}) = - 1 .\}$

$\underline{Case} \: \red{1 .}\\ \\ {x}^{2} - 2x + 3. \\ \\ \implies \: { - 1}^{2} - 2( - 1) + 3. \\ \\ \implies \: 1 + 2 + 3. \\ \\ \implies \: \red{6.}$

_____________________________________

$\underline{Case} \: \blue{2.}$

$P( \pink{x}) = {x}^{2} - 2x + 3. \\ \\ \{ \therefore \: \: where \: as \: \: P( \pink{x}) = 1 .\}$

${x}^{2} - 2x + 3. \\ \\ \implies \: {1}^{2} - 2(1) + 3. \\ \\ \implies \: 1 - 2 + 3. \\ \\ \implies \: \blue{2.}$