Question

Please reply the fastest

Answer 1

Answer:

alpha+ bitha = (alpha)^-1 + (bitha)^-1.

Answer 2

Question:

If α and β are the zeroes of a polynomial x² + 5x + 5, find the value of α⁻¹ + β⁻¹.

Solution:

We've been given a polynomial x² + 5x + 5. With the help of the relation between the coefficients and the zeroes, we'll find the value of α⁻¹ + β⁻¹.

$\sf \underline{\underline{p(x) = ax^2 + bx + c}}\\ \\ \\\sf 'a' \ is \ the \ coefficient \ of \ x^2.\\ \\\sf 'b' \ is \ the \ coefficient \ of \ x.\\ \\\sf 'c' \ is \ the \ constant \ term.\\ \\ \\\sf \Longrightarrow \ \alpha + \beta = -\dfrac{b}{a}\\ \\ \\\sf \Longrightarrow \ \alpha \times \beta = \dfrac{c}{a}$

ATQ,

p(x) =  x² + 5x + 5.

Here, a = 1, b = 5 and c = 5.

Sum of the zeroes:

$\Longrightarrow \sf \alpha + \beta = -\dfrac{b}{a}\\ \\ \\\Longrightarrow \sf \alpha + \beta = -\dfrac{5}{1}\\ \\ \\\Longrightarrow \sf \alpha + \beta = -5 \ \ \ \longmapsto \ \ \ Eq(1)$

Product of the zeroes:

$\Longrightarrow \sf \alpha \beta = \dfrac{c}{a}\\ \\ \\\Longrightarrow \sf \alpha \beta = \dfrac{5}{1}\\ \\ \\\Longrightarrow \sf \alpha \beta = 5 \ \ \ \longmapsto \ Eq(2)$

To Find:

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

Substitute Eq(1) and (2) above.

$\Longrightarrow \sf \ \dfrac{-5 \ \ }{5}\\ \\ \\\Longrightarrow \sf \ -1$

Answer:

α⁻¹ + β⁻¹ = -1