Question

If α and β are the zeroes of the polynomial x then find the value

2 − 4x + 1

of α +
1
—
α

Answer 1

Given :- If α and β are the zeroes of the polynomial x² - 4x + 1 = 0 then find the value ,

• 1/α + 1/β ?

Solution :-

comparing the given polynomial x² - 4x + 1 = 0 with ax² + bx + c = 0 , we get,

• a = 1
• b = (-4)
• c = 1 .

now, we know that, for polynomial ax² + bx + c = 0,

• sum of roots = (-b/a)
• product of roots = (c/a) .

then, for for polynomial x² - 4x + 1 = 0 we have,

• α + β = (-b/a) = -(-4/1) = 4
• α * β = c/a = 1 /1 = 1 .

therefore,

→ 1/α + 1/β

taking LCM of denominator,

→ (β + α) / αβ

putting values,

→ 4 / 1

→ 4 (Ans.)

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Answer 2

$\textbf{Given:}$

$\mathsf{\alpha\;and\;\beta\;are\;zeroes\;of\;the\;polynomial\;x^2-4x+1}$

$\textbf{To find:}$

$\textsf{The value of}\;\mathsf{\alpha+\dfrac{1}{\alpha}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{x^2-4x+1}$

$\mathsf{Sum\;of\;the\;zeroes=\dfrac{-b}{a}}$

$\implies\mathsf{\alpha+\beta=4}$

$\mathsf{Product\;of\;the\;zeroes=\dfrac{-b}{a}}$

$\implies\mathsf{\alpha\,\beta=1}$

$\implies\mathsf{\dfrac{1}{\alpha}=\beta}$

$\mathsf{Now,}$

$\mathsf{\alpha+\dfrac{1}{\alpha}}$

$\mathsf{=\alpha+\beta}$

$\mathsf{=4}$

$\implies\boxed{\mathsf{\alpha+\dfrac{1}{\alpha}=4}}$

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