Question

please write a solution
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Answer 1

Answer:

x² + 4x + 4

Step-by-step explanation:

Given :

α & β are the zeroes of the polynomial f(x) = x² - 1,.

To Find :

Another polynomial whose zeroes are,

$\frac{2\alpha }{\beta}$ and

$\frac{2\beta }{\alpha }$.

Solution :

Polynomial of the form:

f(x) = ax² + bx + c ,. (where a ≠ 0),

In the polynomial,

The sum of the zeroes = $\frac{-b}{a}$

The product of the zeroes = $\frac{c}{a}$

In the given polynomial,.

Sum of the zeroes = $\frac{-b}{a}$

⇒ α + β = 0 ...(i)

Product of the zeroes = $\frac{c}{a}$

⇒ αβ = -1 ..(ii)

If the polynomial's zeroes were  $\frac{2\alpha }{\beta}$ and

$\frac{2\beta }{\alpha }$,.

Then,

let a = 1, for the 2nd polynomial,.

Sum of the zeroes = $\frac{-b}{a}$

$\frac{2\alpha }{\beta } + \frac{2\beta }{\alpha } = \frac{-b}{1}$

$\frac{2\alpha ^2 + 2\beta^2}{\alpha \beta} = -b$

$\frac{2\alpha ^2 + 2\beta^2}{-1} = -b$  (∵ αβ = -1 from (ii) )

$2\alpha ^2 + 2\beta^2 = b$

By adding 4αβ both the sides,

We get,

⇒ $2\alpha ^2 +4\alpha \beta + 2\beta^2 = b + 4\alpha \beta$

⇒ $2(\alpha ^2 +2\alpha \beta + \beta^2) = b + 4(-1)$

⇒ $2(\alpha + \beta)^2 = b - 4$

⇒ $2(0)^2 = b - 4$

⇒ $0 = b - 4$

⇒ $b = 4$  ..(iii)

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Product of the zeroes = $\frac{c}{a}$

$\frac{2\alpha }{\beta } \times \frac{2\beta }{\alpha } = \frac{c}{a}$

⇒ $2 \times 2 = c$ ⇒ c = 4  ..(iv)

Hence,

a = 1, b = 4 , c = 4,.

∴ The polynomial ax² + bx + c ⇒ x² + 4x + 4