Question

please help.................... ​

Answer 1

SOLUTION:-

Given:

If alpha & beta are the zeroes of polynomial are x² +4x+3.

To find:

The polynomial whose zeroes are 1+alpha/beta & 1+beta/alpha.

Explanation:

We have,

Let p(x)= x² +4x +3

&

alpha & beta are the zeroes of p(x).

Sum of the zeroes:

$\alpha + \beta = \frac{ - b}{a} = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

Therefore,

Compare with the given quadratic polynomial Ax² + Bx +C.

• A= 1
• B= 4
• C= 3

$\alpha + \beta = \frac{ - 4}{1} \\ \\ \alpha + \beta = - 4$

&

Product of the zeroes:

$\alpha \beta = \frac{c}{a} = \frac{constant \: term }{coefficient \: of \: x} \\ \\ \alpha \beta = \frac{3}{1} \\ \\ \alpha \beta = 3$

So,

We have zeroes of required quadratic polynomial.

$1 + \frac{ \alpha }{ \beta } \: \: and \: \: 1 + \frac{ \beta }{ \alpha }$

Sum of the roots of required polynomial:

$(1 + \frac{ \alpha }{ \beta } ) + (1 + \frac{ \beta }{ \alpha } ) \\ \\ Sum = 2 + \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ Sum = 2 + \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ Sum = 2 + \frac{( - 4) {}^{2} - 2 \times 3 }{3} \\ \\ Sum = 2 + \frac{16 - 6}{3} \\ \\ Sum = 2 + \frac{10}{3} \\ \\ Sum = \frac{6 + 10}{3} \\ \\ Sum = \frac{16}{3}$

&

Product of roots of required polynomial:

$(1 + \frac{ \alpha }{ \beta } )(1 + \frac{ \beta }{ \alpha } ) \\ \\ Product = 2 + \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ Product = 2 + \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ Product = 2 + \frac{( - 4) {}^{2} - 2 \times 3}{3} \\ \\ Product = 2 + \frac{16 - 6}{3} \\ \\ p</p><p>Product = 2 + \frac{10}{3} \\ \\ Product = \frac{6 + 10}{3} \\ \\ Product = \frac{16}{3}$

Now, required polynomial are;

f(x)= k(x² - Sx + P)

[k is non-zero real number].

f(x)= k(x² - 16/3x + 16/3]

f(x)= k/3 [3x² -16x +16]

Taking k= 3,

f(x)= 3x² - 16x +16.