Question

if alpha
$\alpha and \beta are the zeroes \: of \: the \: polynomial \: 2 x { + 5x + 1 \: }^{2} then \: the \: value \: of \alpha + \beta + \alpha \times \beta$


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Answer 1

Answer

α + β + αβ = - 2

Explanation

Compare given equation 2x² + 5x + 1 with ax² + bx + c , we get ,

• a = 2 , b = 5 , c = 1

also given α and β are zeroes of polynomial .

Let ,

Sum of zeroes , α + β = -b/a = -(5)/2

⇒ α + β = - ⁵/₂ ... (1)

Product of zeroes , αβ = c/a = (1)/(2)

⇒ αβ = ¹/₂ ... (2)

Now our required ,

⇒ α + β + αβ

⇒ ( α + β ) + ( αβ )

⇒ ( - ⁵/₂ ) + ( ¹/₂ ) [ From (1) & (2) ]

⇒ ( ⁻ ⁵ ⁺ ¹/₂ )

⇒ ⁻⁴/₂

⇒ - 2

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Answer 2

$\large{\red{\underline{\underline{\bold{Given:-}}}}}$

• A quadratic polynomial 2x² + 5x + 1

• $\sf \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \: given \: quadratic \: polynomial$

$\large{\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• $\sf the \: value \: of \: \alpha + \beta + \alpha \beta$

$\large{\green{\underline{\underline{\bold{Solution:-}}}}}$

For a quadratic polynomial ax² + bx + c, if the zeroes are α and β

• $\sf Sum \: of \: zeroes( \: \alpha + \beta ) = \dfrac{ - b}{a}$

• $\sf Product \: of \: zeroes( \: \alpha \beta ) = \dfrac{ c}{a}$

Compare 2x² + 5x + 1 with ax² + bx + c

• a = 2
• b = 5
• c = 1

$\sf sum \: of \: zeroes( \: \alpha + \beta ) = \dfrac{ - b}{a} = \dfrac{ - 5}{2}$

$\sf product \: of \: zeroes( \: \alpha \beta ) = \dfrac{ c}{a} = \dfrac{1 }{2}$

Now:-

$\sf \implies \alpha + \beta + \alpha \beta$

$\sf \implies \dfrac{ - 5}{2} + \dfrac{1}{2}$

$\sf \implies \dfrac{ - 5 + 1}{2}$

$\sf \implies \dfrac{ - 4}{2}$

$\sf \implies -2$

Hence;

$\large \boxed{ \sf \purple{ \implies \alpha + \beta + \alpha \beta = - 2}}$