Question

If a and b are the zeros of the polynomial x2-4x+3 find the polynomial whose zeros are 1+β/α and 1+α/β​

Answer 1

Answer:

$\boldsymbol{ \implies \: {x}^{2} - \dfrac{16}{3} x + \dfrac{16}{3} }$

Explanation:

Given polynomial,

$\boldsymbol{\implies \: {x}^{2} - 4x + 3}$

On comapring with the general form of equation i.e., $\boldsymbol{{ax}^{2} - bx + c}$

We get,

• a = 1
• b = -4
• c = 3

By quadratic formula,

$\boldsymbol{\implies x = \dfrac{ - b \pm \sqrt{ {(b)}^{2} - 4ac} }{2a} }$

Substitute the values,

$\boldsymbol{\implies x = \dfrac{ - ( - 4 )\pm \sqrt{ {( - 4)}^{2} - 4(1)(3)} }{2(1)} }$

$\boldsymbol{\implies x = \dfrac{ 4 \pm \sqrt{ 16 - 12} }{2} }$

$\boldsymbol{\implies x = \dfrac{ 4 \pm \sqrt{ 4} }{2} }$

$\boldsymbol{\implies x = \dfrac{ 4 \pm 2 }{2} }$

$\boldsymbol{\implies x = \dfrac{ 4 - 2 }{2} \: { \sf \: or} \: \dfrac{ 4 + 2}{2} }$

$\boldsymbol{\implies x = \dfrac{ 2 }{2} \: { \sf \: or} \: \dfrac{6}{2} }$

$\boldsymbol{ \implies \: x = 1 \: { \rm \: or} \: 3}$

$\sf \therefore \: \alpha = \: 1 \: and \: \beta = 3$

Now,

Given zeros,

$\sf = 1 + \dfrac{ \beta }{ \alpha } \: \: and \: \: 1 + \dfrac{ \alpha }{ \beta }$

Substituting the values,

$\sf = \bigg( 1 + \dfrac{3}{1} \bigg) \: and \: \bigg(1 + \dfrac{1}{3} \bigg)$

$\sf = ( 1 + 3) \: and \: \bigg( \dfrac{4}{3} \bigg)$

$\sf = 4 \: and \: \dfrac{4}{3}$

Then,

Sum of the new zeros is

$= 4 + \dfrac{4}{3} \\ = \dfrac{16}{3} \: \: \: \: \:$

And also,

Product of zeros is

$= 4 \times \dfrac{4}{3} \\ = \dfrac{16}{3} \: \: \: \: \: \:$

Required polynomial is given by,

$\boldsymbol{\implies \: {x}^{2} - {\sf(sum \: of \: the \: zeros)}x + \sf \: product \: of \: zeros}$

$\boldsymbol{\implies \: {x}^{2} - {\sf\dfrac{16}{3} }x + \sf \: \dfrac{16}{3} }$