Question

if
$\alpha$
,
$\beta$
are zeroes of the polynomial p(x)=5x²+5x+1 then find the value of
$\alpha ^{2} + \beta ^{2}$

​

Answer 1

Given Equation :

• p(x) = 5x² + 5x + 1

Have to find the value of the ${\sf{ \alpha ^{2} + \beta ^{2} }}$

Here,

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

Relation between zeroes and coefficients

$\textsf{Sum of the Zeroes }$

$\longrightarrow\sf - {Coefficient \: of \: x/Coefficient \: of \: x^{2}}$

$:\implies \sf{ \alpha + \beta = \dfrac{- 5}{5}}$

$:\implies \sf{ \alpha + \beta = - 1 }$

_________________________

$\textsf{Product of the Zeroes }$

$\longrightarrow\sf{Constant \: Term/Coefficient \: of \: x^{2}}$

$:\implies \sf{ \alpha \times \beta = \dfrac{1}{5} }$

Now,

First we going to use the algebraic identity :

$\bullet\sf {a^{2} + b^{2} = ( a + b )^{2} - 2ab}$

(Putting values in the given Equation :

$\rightarrow \sf \alpha ^{2} + \beta ^{2}$

(-1)² - (2 × (-1) × 1/5)

$\rightarrow \sf 1 - \dfrac{2}{5}$

$\rightarrow \sf \dfrac{5 -2}{5}$

$\rightarrow \sf \dfrac{3}{5}$

Hence,

The value of α² + β² is :

${\boxed{\therefore{\sf{ \dfrac{3}{5}}}}}$

_________________________

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Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ 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

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀

Given that,

• $\alpha\: \& \:\beta$ are two zeroes of p(x) .

As, We know that ,

$\bf{\underline {\star \: Sum \:of\:zeroes \::}}$

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

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Cofficients\::}}$

⠀⠀⠀⠀$:\implies\sf { \alpha + \beta = \bigg( \dfrac{-5}{5}\bigg) }$

⠀⠀⠀⠀$:\implies\sf { \alpha + \beta = \bigg( \cancel {\dfrac{-5}{5}}\bigg) }$

⠀⠀⠀⠀$:\implies\bf { \alpha + \beta = -1 }$

And,

As, We know that,

$\bf{\underline {\star \: Product \:of\:zeroes \::}}$

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

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Cofficients\::}}$

$:\implies\bf { \alpha \times \beta = \bigg( \dfrac{1}{5}\bigg) }$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀Finding ⠀$\alpha ^{2} + \beta ^{2}$ .

$:\implies\sf { \alpha^2 + \beta^2 }$

As, We know that ,

• $Algebraic\:Indentity\::\:(a+b)^2 = a^2 + b^2 + 2ab$

Or ,

• $a^2 + b^2 = (a+b)^2 - 2ab$

Then ,

⠀⠀⠀⠀$:\implies\sf { \alpha^2 + \beta^2 = \bigg( \alpha + \beta\bigg)^2 - 2\alpha\beta }$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Found \: Values \::}}$

⠀⠀⠀⠀$:\implies\sf { \alpha^2 + \beta^2 = \bigg( \alpha + \beta\bigg)^2 - 2\alpha\beta }$

⠀⠀⠀⠀$:\implies\sf { \alpha^2 + \beta^2 = \bigg( -1\bigg)^2 - 2\times \dfrac{1}{5} }$

⠀⠀⠀⠀$:\implies\sf { \alpha^2 + \beta^2 = 1 - \dfrac{2}{5} }$

⠀⠀⠀⠀$:\implies\sf { \alpha^2 + \beta^2 = \dfrac{5 - 2}{5} }$

⠀⠀⠀⠀$:\implies\bf { \alpha^2 + \beta^2 = \dfrac{3}{5} }\qquad \longrightarrow\bf{AnswEr}$

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