Question

if a and ß are the zeroes of the quadratic polynomial P(x)= 5x^2 + x-1 find the value of a^2+ ß^2.​

Answer 1

Step-by-step explanation:

hope my ans is correct...

Answer 2

Answer:

11/25

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Step-by-step explanation:

According to the question,

⇒ p(x) = 5x² + x - 1

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Where,

• a (Coefficient of x²) = 5

• b (Coefficient of x) = 1

• c (Constant term) = -1

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Also, α and β are the zeroes of p(x).

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We know that,

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$\sf \Longrightarrow Sum \ of \ the \ roots = - \dfrac{Coefficient \ of \ x}{Coefficient \ of \ x^2}$

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$\sf \Longrightarrow \alpha + \beta = - \dfrac{Coefficient \ of \ x}{Coefficient \ of \ x^2}$

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$\sf \Longrightarrow \alpha + \beta = - \dfrac{b}{a}$

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$\sf \Longrightarrow \alpha + \beta = - \dfrac{1}{5}$

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We also know that,

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$\sf \Longrightarrow Product \ of \ the \ roots = \dfrac{Constant \ term}{Coefficient \ of \ x^2}$

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$\sf \Longrightarrow \alpha \beta = \dfrac{Constant \ term}{Coefficient \ of \ x^2}$

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$\sf \Longrightarrow \alpha \beta = \dfrac{c}{a}$

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$\sf \Longrightarrow \alpha \beta = \dfrac{-1}{5}$

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Finding the value of α² + β²:

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

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Using the identity a² + b² = (a + b)² - 2ab we get,

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$\sf \Longrightarrow (\alpha+ \beta)^{2} - 2\alpha \beta$

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On substituting the values we've gotten for α + β and αβ we get,

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$\sf \Longrightarrow \Bigg\{-\dfrac{1}{5}\Bigg\}^{2} - \bigg\{2 \times -\dfrac{1}{5}\bigg\}$

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$\sf \Longrightarrow \dfrac{1}{25} - \bigg\{-\dfrac{2}{5}\bigg\}$

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$\sf \Longrightarrow \dfrac{1}{25} + \dfrac{2}{5}$

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On taking LCM we get,

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$\sf \Longrightarrow \dfrac{1 + 10}{25}$

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$\sf \Longrightarrow \dfrac{\textsf{\textbf{11}}}{\textsf{\textbf{25}}}$

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∴ The value of α² + β² is 11/25.