Question

if α and β are the zeroes of the polynomials x²-px+q then find the values of each of the following.
i)α\β+β\α
ii)αβ³+βα³
iii)α³β² + α²β³​

Answer 1

If a and b are zeroes of x^2 - px + q, then using the relation between zeroes and coefficients, we get that

Sum of zeroes = a + b = - (-p) = p

Product of zeroes = ab = q

Square on both sides of a + b, we get

a² + b² + 2ab = p² → a² + b² + 2q = p²

a² + b² = p² - 2q

(i) a/b + b/a = (a² + b²)/ab

= (p² - 2q)/q

(ii) ab³ + ba³ = ab(b² + a²)

= q(p² - 2q)

(iii) a³b² + a²b³ = a²b²(a + b)

= (ab)²(a + b)

= q²p

= pq²

Answer 2

Given that , The α and β are the zeroes of the polynomials x²- px + q .

Exigency To Find : The value of :

• α\β + β\α
• αβ³+βα³
• α³β² + α²β³

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline{\pmb{\mathbb{\bigstar \:\:\:POLYNOMIAL \:\::}{\: x^2 \:-\: px\: + \:q \:\:}}}$

As , We know that ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Sum \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha + \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\} \: \:=\:\dfrac{ - \:( Cofficient \:of \: x \:)\:}{Cofficient \:of \:x^2 \:} \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ - \:( Cofficient \:of \: x \:)\:}{Cofficient \:of \:x^2 \:}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ - \:( - p \:)\:}{1 \:}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:\dfrac{p}{1}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:p \:\\\\\dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha \:+\:\beta \: \:\:=\:\:p \:\:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Product \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\} \: \:=\:\dfrac{ \:Constant \:Term\:}{Cofficient \:of \:x^2 \:} \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ Constant \:Term\:}{Cofficient \:of \:x^2 \:}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ q\:}{1 \:}\\\\\dashrightarrow \sf \: \alpha \:\:\beta \: \:\:=\:\:\dfrac{q}{1}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:q \:\\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha \:\:\beta \: \:\:=\:\:q \:\:}}}}}\:\:\bigstar$

━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline{\pmb{{\maltese \:\purple { \:\: \alpha / \beta \:+ \:\beta / \alpha \:\:}}}}$

$\qquad \dashrightarrow \sf \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:\\\\\qquad \dashrightarrow \sf \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:=\: \dfrac{\alpha^2 + \beta ^2 }{\alpha\beta } \:$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \:Algebraic \: Indentity\:\::\: a^2 + b^2 \:=\:( a + b )^2 - 2ab }\bigg\rgroup \\\\\qquad \dashrightarrow \sf \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:=\: \dfrac{\alpha^2 + \beta ^2 }{\alpha\beta } \:\\\\\qquad \dashrightarrow \sf \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:=\: \dfrac{(\alpha + \beta )^2 - 2 \alpha \beta }{\alpha\beta } \:\\\\\qquad \dashrightarrow \sf \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:=\: \dfrac{ (p )^2 - 2(q) }{q } \:\\\\\qquad \dashrightarrow \sf \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:=\: \dfrac{p^2 - 2q }{q } \:\\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha \:}\:=\: \dfrac{p^2 - 2q }{q }\:\:}}}}}\:\:\bigstar$

━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline{\pmb{{\maltese \:\purple { \:\: \alpha^3 \beta^2 \:+ \:\beta^3 \alpha^2 \:\:}}}}$

$\qquad \dashrightarrow \alpha\beta^3 + \beta \alpha^3 \:\\\\\qquad \dashrightarrow \alpha\beta^3 + \beta \alpha^3 = \alpha\beta (\beta^2 + \alpha^2 ) \:$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \:Algebraic \: Indentity\:\::\: a^2 + b^2 \:=\:( a + b )^2 - 2ab }\bigg\rgroup \\\\\qquad \dashrightarrow \sf\alpha\beta^3 + \beta \alpha^3 = \alpha\beta (\beta^2 + \alpha^2 ) \:\\\\\qquad \dashrightarrow \sf \alpha\beta^3 + \beta \alpha^3 = \alpha\beta \Big\{ (\alpha + \beta )^2 - 2 \alpha \beta \Big\} \:\\\\\qquad \dashrightarrow \sf \alpha\beta^3 + \beta \alpha^3 = q \Big\{ (p )^2 - 2(q) \Big\} \:\\\\\qquad \dashrightarrow \sf \alpha\beta^3 + \beta \alpha^3 = q \Big\{ p^2 - 2q \Big\} \:\\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha\beta^3 + \beta \alpha^3 = q \{ p^2 - 2q \} \:\:}}}}}\:\:\bigstar$

━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline{\pmb{{\maltese \:\purple { \:\: \alpha \beta^3 \:+ \:\beta \alpha^3 \:\:}}}}$

$\qquad \dashrightarrow \sf \alpha^3\beta^3 + \beta^3 \alpha^2 \:\\\\\qquad \dashrightarrow \sf \alpha^3\beta^2 + \beta^3 \alpha^2 \:=\: \alpha^2 \beta^2( \alpha \beta ) \:\\\\\qquad \dashrightarrow \sf\alpha^3\beta^2 + \beta^3 \alpha^2 \:=\: (\alpha \beta)^2( \alpha \beta ) \:\\\\\qquad \dashrightarrow \sf \alpha^3\beta^2 + \beta^3 \alpha^2 \:=\: (q)^2(p ) \:\\\\\qquad \dashrightarrow \sf \alpha^3\beta^2 + \beta^3 \alpha^2 \:=\: q^2p \:\\\\\qquad \dashrightarrow \sf \alpha^3\beta^2 + \beta^3 \alpha^2 \:=\: pq^2 \:\\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha^3\beta^2 + \beta^3 \alpha^2 \:=\: pq^2\:\:}}}}}\:\:\bigstar$