if alpha and beta are the zeroes of the quadratic polynomial f(x)=x2+P

1.	if alpha and beta are the zeroes of the quadratic polynomial f(x)=x2+Px+q form a polynomial whose zeroes are (alpha+beta)square and (alpha-beta) square..
Answer» *SOLUTION ☺️* $= > we \: have \: polynomial \: f(x) = {x}^{2} + px + q \\ = > roots \: are \: \alpha \: \: and \: \: \beta \\ = > we \: know \: from \: relationship \: between \: zeros \: and \: coefficient \: \\ = > sum \: of \: zeros = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ = > ( \alpha + \beta ) {}^ \: \: \: = - p...............(1) \\ = > taking \: whole \: square \: on \: both \: side \: we \: get \\ = > {( \alpha + \beta )}^{2} = {p}^{2} ..............(2) \\ = > { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = {p}^{2} \\ > { \alpha }^{2} { \beta }^{2} + 2 \alpha \beta - 2 \alpha \beta + 2 \alpha \beta = {p}^{2} \\ = > { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta + 4 \alpha \beta = {p}^{2} \\ = > ( { \alpha - \beta )}^{2} + 4 \alpha \beta = {p}^{2} ................(3) \\ and \\ = > products \: of \: zeros = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ so \\ = > \alpha \beta = q \: \: (substitute \: that \: value \: in \: equation \: (3) \: we \: get) \\ = > ( \alpha - \beta ) {}^{2} + 4(q) = {p}^{2} \\ = > {( \alpha - \beta )}^{2} + 4q = {p}^{2} \\ = > {( \alpha - \beta )}^{2} = {p}^{2} - 4q..................(4) \\ = > now \:we \: add \: equation \: (2) \: and \: (4) \: get \\ = > {( \alpha + \beta )}^{2} + {( \alpha - \beta )}^{2} = {p}^{2} + {p}^{2} - 4q \\ = > 2 {p}^{2} - 4q \\ = > and \: we \: multiply \: equation \: (2) \: and \: (4) \: get \\ = > {( \alpha + \beta )}^{2} \times {( \alpha - \beta )}^{2} = {p}^{2} ( {p}^{2} - 4q) \\ = > {p}^{4} - 4 {p}^{2} q \\ = > as \: we \: know \: formula \: for \: polynomial \: when \: sum \: of \: zeros \: and \: product \: of \: zeros \: we \: know \\ = > polynomial = k( {x}^{2} - (sum \: of \: zeros) \times (product \: of \: zeros)) \\ = > here \: k \: is \: any \: non \: zero \: real \: number \\ = > substitute \: value \: get \\ = > quadratic \: polynomial \: = k( {x}^{2} - (2 {p}^{2} - 4q)x + (2 {p}^{4} - 4p {}^{2} q) \\ \\ = > {x}^{2} - (2 {p}^{2} - 4q)x + (2p {}^{4} - 4p {}^{2}q) \: \: \: \: \: \: \: \: \: \: (\: taking \: k = 1)$ *hope* *it HELPS* ✔️☺️

if alpha and beta are the zeroes of the quadratic polynomial f(x)=x2+Px+q form a polynomial whose zeroes are (alpha+beta)square and (alpha-beta) square..

Answer»

SOLUTION ☺️

$= > we \: have \: polynomial \: f(x) = {x}^{2} + px + q \\ = > roots \: are \: \alpha \: \: and \: \: \beta \\ = > we \: know \: from \: relationship \: between \: zeros \: and \: coefficient \: \\ = > sum \: of \: zeros = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ = > ( \alpha + \beta ) {}^ \: \: \: = - p...............(1) \\ = > taking \: whole \: square \: on \: both \: side \: we \: get \\ = > {( \alpha + \beta )}^{2} = {p}^{2} ..............(2) \\ = > { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = {p}^{2} \\ > { \alpha }^{2} { \beta }^{2} + 2 \alpha \beta - 2 \alpha \beta + 2 \alpha \beta = {p}^{2} \\ = > { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta + 4 \alpha \beta = {p}^{2} \\ = > ( { \alpha - \beta )}^{2} + 4 \alpha \beta = {p}^{2} ................(3) \\ and \\ = > products \: of \: zeros = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ so \\ = > \alpha \beta = q \: \: (substitute \: that \: value \: in \: equation \: (3) \: we \: get) \\ = > ( \alpha - \beta ) {}^{2} + 4(q) = {p}^{2} \\ = > {( \alpha - \beta )}^{2} + 4q = {p}^{2} \\ = > {( \alpha - \beta )}^{2} = {p}^{2} - 4q..................(4) \\ = > now \:we \: add \: equation \: (2) \: and \: (4) \: get \\ = > {( \alpha + \beta )}^{2} + {( \alpha - \beta )}^{2} = {p}^{2} + {p}^{2} - 4q \\ = > 2 {p}^{2} - 4q \\ = > and \: we \: multiply \: equation \: (2) \: and \: (4) \: get \\ = > {( \alpha + \beta )}^{2} \times {( \alpha - \beta )}^{2} = {p}^{2} ( {p}^{2} - 4q) \\ = > {p}^{4} - 4 {p}^{2} q \\ = > as \: we \: know \: formula \: for \: polynomial \: when \: sum \: of \: zeros \: and \: product \: of \: zeros \: we \: know \\ = > polynomial = k( {x}^{2} - (sum \: of \: zeros) \times (product \: of \: zeros)) \\ = > here \: k \: is \: any \: non \: zero \: real \: number \\ = > substitute \: value \: get \\ = > quadratic \: polynomial \: = k( {x}^{2} - (2 {p}^{2} - 4q)x + (2 {p}^{4} - 4p {}^{2} q) \\ \\ = > {x}^{2} - (2 {p}^{2} - 4q)x + (2p {}^{4} - 4p {}^{2}q) \: \: \: \: \: \: \: \: \: \: (\: taking \: k = 1)$

hope it HELPS ✔️☺️