Question

If a & ß are the roots of x² – 2x + 4 = 0, than
the value of a^6+B^6 is k then the value
k/64

Answer 1

EXPLANATION.

α and β are the roots of the equation.

⇒ x² - 2x + 4 = 0.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-2)/1 = 2.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 4/1 = 4.

To find :

⇒ α⁶ + β⁶

As we know that,

By apply forceful factorization in the equation, we get.

⇒ α⁶ + β⁶ = (α³ + β³)(α³ + β³) - 2α³β³.

⇒ α⁶ + α³β³ + α³β³ + β⁶ - 2α³β³.

⇒ α⁶ + β⁶ + 2α³β³ - 2α³β³.

⇒ α⁶ + β⁶.

Hence proved.

This equation is correct.

Whenever you apply forceful factorization always check the equation, we get.

⇒ (α³ + β³)(α³ + β³) - 2α³β³ = k.

⇒ (α³ + β³)² = k.

⇒ [(α + β)(α² - αβ + β²)]² = k.

⇒ [(α + β){(α + β)² - 2αβ - αβ}]² = k.

⇒ [(α + β){(α + β)² - 3αβ}]² = k.

Put the value in the equation, we get.

⇒ [(2){(2)² - 3(4)}]² = k.

⇒ [(2){4 - 12}]² = k.

⇒ [2(-8)]² = k.

⇒ [-16]² = k.

⇒ 256 = k.

To find :

value of = k/64 = 256/64 = 4.

value of : k/64 = 4.

                                                                                                                       

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

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

As, We know that ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Sum \:of \:zeroes \:\:\red {\:( \:\alpha \: + \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{\:-(Cofficient \:of \:x\:)\:\:}{Cofficient \:of \:x^2 \:}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \bigg( \alpha +\:\beta \: \bigg) \:=\:\dfrac{-(\:Cofficient \:of \:x\:)}{Cofficient \:of \:x^2 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{-(\:-2\:)}{1 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{2\:}{1 \:}\\\\\qquad \dashrightarrow \sf \alpha \:+ \beta \: \:=\:2\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \:+ \beta \: \:=\:2\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Product \:of \:zeroes \:\:\red {\:( \:\alpha \: \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:\beta \: \bigg) \:=\:\dfrac{\:Constant \:Term\:\:}{Cofficient \:of \:x^2 \:}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:\beta \: \bigg) \:=\:\dfrac{\:Constant\:Term\:}{Cofficient \:of \:x^2 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:\dfrac{\:4\:}{1 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:\cancel{\dfrac{\:4\:}{1 \:}}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:4\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \: \beta \: \:=\:4\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding value of (i) α⁶ + β⁶ :

$\qquad \dashrightarrow \sf \alpha ^6 + \beta ^6 \:$

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

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

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

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{Algebraic \:Indentity \:\:: ( a + b )^3 \:= \:a^3 + b^3+ 3ab ( a + b ) }\bigg\rgroup$

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

$\qquad \dashrightarrow \sf (\alpha^2 + \beta^2) ^3 = \alpha^6 + 3\alpha^2 \beta^2( \alpha^2 + \beta ^2 ) \:$

$\qquad \dashrightarrow \sf (\alpha^2 + \beta^2) ^3 = \alpha^6 + 3\alpha^2 \beta^2( \alpha^2 + \beta ^2 ) \:$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{Algebraic \:Indentity \:\:: a^2 + b^2 \:= \: ( a + b )^2 - 2ab }\bigg\rgroup$

$\qquad \dashrightarrow \sf (\alpha^2 + \beta^2) ^3 = \alpha^6 + \beta^6 + 3\alpha^2 \beta^2( \alpha^2 + \beta ^2 ) \:$

$\qquad \dashrightarrow \sf \{ (\alpha + \beta)^2 - 2\alpha\beta \} ^3 = \alpha^6 +\beta^6+ 3\alpha^2 \beta^2 \{(\alpha + \beta)^2 - 2\alpha\beta \} \:$

$\qquad \dashrightarrow \sf \{ (\alpha + \beta)^2 - 4\alpha\beta \} ^3 = \alpha^6 +\beta^6+ 3(\alpha\beta)^2 \{(\alpha + \beta)^2 - 2\alpha\beta \} \:$

$\qquad \dashrightarrow \sf \{ (2)^2 - 4(2) \} ^3 = \alpha^6 +\beta^6+3(\alpha\beta)^2 \{(\alpha + \beta)^2 - 2\alpha\beta \} \:$

$\qquad \dashrightarrow \sf \{ (2)^2 - 2(4) \} ^3 = \alpha^6 +\beta^6+ 3(\alpha\beta)^2 \{ (2)^2 - 2(4) \} \:$

$\qquad \dashrightarrow \sf \{ 4 - 8 \} ^3 = \alpha^6 +\beta^6+ 3(\alpha\beta)^2 (4 - 8 ) \:$

$\qquad \dashrightarrow \sf \{ -4 \} ^3 = \alpha^6 +\beta^6+3(4)^2 (-4 ) \:$

$\qquad \dashrightarrow \sf -64 = \alpha^6 + \beta^6+3(16) (-4 ) \:$

$\qquad \dashrightarrow \sf \alpha^6 + \beta^6 = 3(16)(-4) + 64 \:$

$\qquad \dashrightarrow \sf \alpha^6 + \beta^6 = 192 - 64 \:$

$\qquad \dashrightarrow \sf \alpha^6 + \beta^6 = 128 \:$

$\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \alpha^6 + \beta^6 = 128\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding value of (ii) k/64 :

$\qquad \dashrightarrow \sf \dfrac{k}{64}\:$

⠀⠀⠀⠀⠀⠀Given that,

• Value of k = α⁶ + β⁶ : 128

$\qquad \dashrightarrow \sf \dfrac{k}{64}\:$

$\qquad \dashrightarrow \sf \dfrac{k}{64}\:=\:\dfrac{128}{64}\:$

$\qquad \dashrightarrow \sf \dfrac{k}{64}\:=\:\cancel {\dfrac{128}{64}}\:$

$\qquad \dashrightarrow \sf \dfrac{k}{64}\:=\:2\:$

$\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{\dfrac{k}{64}\:=\:2 }}}}}\:\:\bigstar$

$\qquad \therefore \:\: \underline{\sf Hence , \: The \:value \:of \:\:k/64 \:\:is \:\pmb{\bf 2 }.}$