Question

let's do this question ✌️✌️​

Answer 1

Answer

x² - 4x - 5

Explanation

Given that $\sf \alpha$ and $\sf \beta$ are the zeroes of the quadratic polynomial f(x) = x² - x + 2.

To find

A polynomial whose zeroes are $\sf 2\alpha + 1$ and $\sf 2\beta + 1$

We know that sum of zeroes = -b/a and product of zeroes = c/a for a quadratic polynomial.

Comparing the given polynomial f(x) = x² - x - 2 with the standard quadratic equation ax² + bx + c, we get

a = 1

b = -1

c = -2

So, we get

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

$\sf \alpha + \beta = \frac{-(-1)}{1}$

$\sf \alpha + \beta = 1$

Again, we get

$\sf \alpha\beta = \frac{c}{a}$

Thus,

$\sf \alpha\beta = -2$

Now, we will find the sum of $\sf2\alpha + 1$ and $\sf 2\beta +1$

$\sf 2\alpha + 2\beta + 1 + 1$

$\sf 2(\alpha + \beta) + 2$

$\sf 2(1) + 2$

= 4

So, sum of zeroes in this case is 4

And, product of these zeroes would be

$\sf (2\alpha + 1)(2\beta + 1)$

$\sf 4\alpha\beta + 2\alpha + 2\beta + 1$

$\sf 4(-2) + 2(\alpha + \beta) + 1$

$\sf -8 + 2(1) + 1$

$\sf -8 + 3$

= -5

Noq, quadratic polynomial is given as

$\sf k(x^2 - (sum\ of\ zeroes)x + product\ of\ zeroes)$

Where, k is some constant

So, a polynomial having zeroes $\sf 2\alpha + 1$ and $\sf 2\beta + 1$ would be

k(x² - 4x - 5)

For k = 1, we get

x² - 4x - 5.