if
[tex] \alpha[/tex]
,
[tex] \beta [/tex]
are zeroes of the polynomial p(x)=5x²+5x+1 then find the value of<

2 thoughts on “if <br />[tex] \alpha[/tex]<br />,<br />[tex] \beta [/tex]<br />are zeroes of the polynomial p(x)=5x²+5x+1 then find the value of<”

1. ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN⠀ POLYNOMIAL – p(x) = 5x² + 5x + 1 :

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   ⠀⠀⠀⠀⠀The Cofficients of Polynomial p(x) = 5x² + 5x + 1 :

   • The Cofficient of x² = 5
   • The Cofficient of x = 5
   • Constant term = 1

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   ⠀⠀⠀⠀

   Given that,

   • [tex] \alpha\: \& \:\beta [/tex] are two zeroes of p(x) .

   As, We know that ,

   [tex]\bf{\underline {\star \: Sum \:of\:zeroes \::}}\\[/tex]

   • [tex]\bf{ \alpha + \beta = \bigg( \dfrac{-(Cofficient \:of\:x^2)}{Cofficient \:of\:x}\bigg) }\\\\[/tex]

   ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Cofficients\::}}\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha + \beta = \bigg( \dfrac{-5}{5}\bigg) }\\\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha + \beta = \bigg( \cancel {\dfrac{-5}{5}}\bigg) }\\\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\bf { \alpha + \beta = -1 }\\\\[/tex]

   And,

   As, We know that,

   [tex]\bf{\underline {\star \: Product \:of\:zeroes \::}}\\[/tex]

   ⠀⠀⠀⠀[tex]\bf{ \alpha \times \beta = \bigg( \dfrac{Constant\:Term}{Cofficient \:of\:x^2}\bigg) }\\\\[/tex]

   ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Cofficients\::}}\\[/tex]

   [tex]:\implies\bf { \alpha \times \beta = \bigg( \dfrac{1}{5}\bigg) }\\\\[/tex]

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   ⠀⠀⠀⠀⠀⠀⠀⠀Finding ⠀[tex] \alpha ^{2} + \beta ^{2} [/tex] .

   [tex]:\implies\sf { \alpha^2 + \beta^2 }\\\\[/tex]

   As, We know that ,

   • [tex]Algebraic\:Indentity\::\:(a+b)^2 = a^2 + b^2 + 2ab [/tex]

   Or ,

   • [tex] a^2 + b^2 = (a+b)^2 – 2ab [/tex]

   Then ,

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha^2 + \beta^2 = \bigg( \alpha + \beta\bigg)^2 – 2\alpha\beta }\\\\[/tex]

   ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: the \: Found \: Values \::}}\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha^2 + \beta^2 = \bigg( \alpha + \beta\bigg)^2 – 2\alpha\beta }\\\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha^2 + \beta^2 = \bigg( -1\bigg)^2 – 2\times \dfrac{1}{5} }\\\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha^2 + \beta^2 = 1 – \dfrac{2}{5} }\\\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\sf { \alpha^2 + \beta^2 = \dfrac{5 – 2}{5} }\\\\[/tex]

   ⠀⠀⠀⠀[tex]:\implies\bf { \alpha^2 + \beta^2 = \dfrac{3}{5} }\qquad \longrightarrow\bf{AnswEr}\\\\\\[/tex]

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2. Given Equation :

   • p(x) = 5x² + 5x + 1

   Have to find the value of the [tex] {\sf{ \alpha ^{2} + \beta ^{2} }} [/tex]

   Here,

   • Coefficient of x² → 5
   • Coefficient of x = 5
   • Constant term = 1

   Relation between zeroes and coefficients

   [tex] \textsf{Sum of the Zeroes } [/tex]

   [tex]\longrightarrow\sf – {Coefficient \: of \: x/Coefficient \: of \: x^{2}}[/tex]

   [tex]:\implies \sf{ \alpha + \beta = \dfrac{- 5}{5}}\\ \\ [/tex]

   [tex]:\implies \sf{ \alpha + \beta = – 1 }\\ \\ [/tex]

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   [tex]\textsf{Product of the Zeroes } [/tex]

   [tex]\longrightarrow\sf{Constant \: Term/Coefficient \: of \: x^{2}}[/tex]

   [tex]:\implies \sf{ \alpha \times \beta = \dfrac{1}{5} } \\ \\ [/tex]

   Now,

   First we going to use the algebraic identity :

   [tex]\bullet\sf {a^{2} + b^{2} = ( a + b )^{2} – 2ab} \\ [/tex]

   (Putting values in the given Equation :

   [tex]\rightarrow \sf \alpha ^{2} + \beta ^{2} \\ \\ [/tex]

   (-1)² – (2 × (-1) × 1/5)

   [tex]\rightarrow \sf 1 – \dfrac{2}{5} \\ \\ [/tex]

   [tex]\rightarrow \sf \dfrac{5 -2}{5} \\ \\ [/tex]

   [tex]\rightarrow \sf \dfrac{3}{5} \\ \\ [/tex]

   Hence,

   The value of α² + β² is :

   [tex]{\boxed{\therefore{\sf{ \dfrac{3}{5}}}}} \\ \\ [/tex]

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