Question

α and β are the zeroes of the polynomial x2 – px + q, then find the value of (i) /+ /() αβ3 + βα3 (iii) α3β2 + α2 β3​

Answer 1

Question:-

If α & β are the roots of the polynomial 3x² + 2x - 6 , then find the value of

i) α - β

ii) α² + β²

iii) α³ + β ³

iv) 1/α + 1/β

Answer:-

Given:

α & β are the roots of 3x² + 2x - 6.

On comparing the given polynomial with the standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let,

• a = 3
• b = 2
• c = - 6

We know that,

Sum of the roots = - b/a

⟹ α + β = - 2/3 -- equation (1)

Product of the roots = c/a

⟹ αβ = - 6/3 = - 2 -- equation (2)

We have to find:

i) α - β

We know that,

(a - b)² = (a + b)² - 4ab

So,

⟹ (α - β)² = (α + β)² - 4αβ

Substituting the values from equations (1) & (2) we get,

⟹ (α - β)² = (- 2/3)² - 4( - 2)

⟹ (α - β)² = 4/9 + 8

⟹ (α - β)² = (4 + 72)/9

⟹ (α - β)² = 76/9

⟹ (α - β) = √(76/9)

⟹ (α - β) = 2√19/3

Simple Method:-

If α , β are the roots of ax² + bx + c = 0 ;

• (α - β) = √(b² - 4ac) / |a|

⟹ α - β = √2² - 4(3)( - 6) / |3|

⟹ α - β = √4 + 72 / 3

⟹ α - β = √76/3

⟹ α - β = 2√19/3

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ii) α² + β²

We know,

a² + b² = (a + b)² - 2ab

So,

⟹ α² + β² = (α + β)² - 2αβ

⟹ α² + β² = ( - 2/3)² - 2( - 2)

⟹ α² + β² = 4/9 + 4

⟹ α² + β² = (4 + 36)/9

⟹ α² + β² = 40/9

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iii) α³ + β³

We know,

a³ + b³ = (a + b)(a² + b² - ab)

⟹ α³ + β³ = (α + β)(α² + β² - αβ)

⟹ α³ + β³ = (- 2/3) [ 40/9 - (- 2) ]

⟹ α³ + β³ = (- 2/3) (40/9 + 2)

⟹ α³ + β³ = (- 2/3) (40 + 18) / 9

⟹ α³ + β³ = (- 2/3) (58/9)

⟹ α³ + β³ = - 116/27

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iv) 1/α + 1/β

Taking LCM we get,

⟹ (β² + α²) / αβ

⟹ (40/9) / ( - 2)

⟹ (40/9) * ( - 1/2)

⟹ 1/α + 1/β = - 20/9.

Answer 2

$\qquad \mathbb{ GIVEN \:\:POLYNOMIAL \:\::\: }\sf 3x^2 + 2x - 6 \:$

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Sum \:of \:zeroes \:\:\red {\:( \:\alpha \: + \beta )}\::}}}\\\\\quad \dashrightarrow \sf \bigg( \alpha +\:\beta \: \bigg) \:=\:\dfrac{-(\:Cofficient \:of \:x\:)}{Cofficient \:of \:x^2 \:} \\\\\qquad \dashrightarrow \sf \alpha \:+ \beta \: \:=\:\dfrac{-\:2\:}{3 \:}\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \:+ \beta \: \:=\:\dfrac{-\:2\:}{3 \:}\: }}}}}\:\:\bigstar \\\\\qquad\underline {\boxed {\pmb{ \:\maltese \;Product \:of \:zeroes \:\:\red {\:( \:\alpha \: \beta )}\::}}}\\\qquad \\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:\dfrac{\:-6\:}{3 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:\cancel{\dfrac{\:-6\:}{3 \:}}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:-2\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \: \beta \: \:=\:-2\: }}}}}\:\:\bigstar$

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⠀⠀⠀⠀⠀¤ Finding value of (i) α - β :

$\qquad \dashrightarrow \sf \alpha - \beta \:\\\\\qquad \dag\:\:\bigg\lgroup \sf{Algebraic \:Indentity \:\:: \:( a - b )^2 = ( a + b )^2 - 4ab }\bigg\rgroup \\\\\qquad \dashrightarrow \sf \alpha - \beta \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta)^2 \:= \: ( \alpha + \beta )^2 - 4 \alpha \beta \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta)^2 \:= \: \bigg( \dfrac{-2}{3} \bigg)^2 - 4 (-2) \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta)^2 \:= \: \bigg( \dfrac{-2}{3} \bigg)^2 + 8 \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta)^2 \:= \: \bigg( \dfrac{4}{9} \bigg) + 8 \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta)^2 \:= \: \bigg( \dfrac{4 + 72 }{9} \bigg) \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta)^2 \:= \: \bigg( \dfrac{76 }{9} \bigg) \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta) \:= \: \sqrt{ \dfrac{76 }{9}} \:\\\\\qquad \dashrightarrow \sf (\alpha - \beta) \:= \: \dfrac{2\sqrt{19} }{3} \:\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:(\alpha - \beta) \:= \: \dfrac{2\sqrt{19} }{3}\: }}}}}\:\:\bigstar$

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⠀⠀⠀⠀⠀⠀¤ Finding value of (ii) α² + β² :

$\qquad \dashrightarrow \sf \alpha^2 + \beta^2 \:\\\\\qquad \dag\:\:\bigg\lgroup \sf{Algebraic \:Indentity \:\:: \:a^2+ b^2= ( a + b )^2 - 4ab }\bigg\rgroup \\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2 \:\\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2 \:= \: ( \alpha + \beta )^2 - 2 \alpha \beta \:\\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2\:= \: \bigg( \dfrac{-2}{3} \bigg)^2 - 2 (-2) \:\\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2\:= \: \bigg( \dfrac{-2}{3} \bigg)^2 + 4 \:\\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2\:= \: \bigg( \dfrac{4}{9} \bigg) + 4 \:\\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2\:= \: \bigg( \dfrac{4+ 36}{9} \bigg) \:\\\\\qquad \dashrightarrow \sf \alpha^2 + \beta^2\:= \: \bigg( \dfrac{40}{9} \bigg) \:\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \ \alpha^2 + \beta^2\:= \: \dfrac{40}{9} \: }}}}}\:\:\bigstar$

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⠀⠀⠀⠀⠀¤ Finding value of (iii) α³ + β³ :

$\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:\\\\\qquad \dag\:\:\bigg\lgroup \sf{Algebraic \:Indentity \:\:: \:a^3+ b^3= ( a + b ) ( a^2 + b^2 - ab }\bigg\rgroup \\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: ( \alpha + \beta ) ( \alpha^2 + \beta^2 - \alpha \beta ) \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \bigg( \dfrac{-3}{2} \bigg) ( \alpha^2 + \beta^2 - (-2) ) \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \bigg( \dfrac{-2}{3} \bigg) ( \alpha^2 + \beta^2 + 2 ) \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \bigg( \dfrac{-2}{3} \bigg) \bigg( \dfrac{40}{9} +2 \bigg) \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \bigg( \dfrac{-2}{3} \bigg) \bigg( \dfrac{40 + 18 }{9} \bigg) \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \bigg( \dfrac{-2}{3} \bigg) \bigg( \dfrac{58 }{9} \bigg) \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \dfrac{-2}{3} \times \dfrac{58 }{9} \:\\\\\qquad \dashrightarrow \sf \alpha^3 + \beta^3 \:= \: \dfrac{-116 }{27} \:\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \ \alpha^2 + \beta^2\:= \: \dfrac{-116}{27} \: }}}}}\:\:\bigstar$

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⠀⠀⠀⠀⠀¤ Finding value of (iv) 1/α + 1/β :

$\qquad \dashrightarrow \sf \dfrac{1}{\alpha } + \dfrac{1}{\beta} \\\\\qquad \dashrightarrow \sf \dfrac{\alpha^2 + \beta^2 }{\alpha \beta } \\\\\qquad \dashrightarrow \sf \dfrac{\alpha^2 + \beta^2 }{-2 } \\\\\qquad \dashrightarrow \sf \dfrac{\alpha^2 + \beta^2 }{-2 } \\\\\qquad \dashrightarrow \sf \dfrac{40/9 }{-2 } \\\\\qquad \dashrightarrow \sf \dfrac{-20 }{9 } \\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\dfrac{1}{\alpha } + \dfrac{1}{\beta} = \dfrac{-20 }{9 } \: }}}}}\:\:\bigstar$