Question

If alpha and beta are the zeros of polynomial p(x)=3x^2+2x+1,find the polynomial whose zeroes are alpha-1/alpha+1 and beta-1/beta+1​

Answer 1

$\large \bf \clubs \:  Given :-$

α and β are the Zeros of the polynomial

p(x) = 3x² + 2x + 1 .

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$\large \bf \clubs \:  To  \: Find :-$

The Polynomial whose Zeros are 

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

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$\large \bf \clubs \:   Main  \:  Concepts :-$

✏ 1》For a qudratic Polynomial of the Form  ax² + bx + c :-

Sum of Zeros =  $\sf-\dfrac{b}{a}$

Product of Zeros = $\sf\dfrac{c}{a}$

✏ 2》 A Quadratic Polynomial whose sum and product of Zeros are S and P respectively is given by x² - Sx + P

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$\large \bf \clubs \:  Solution :-$

As α and β are the Zeros of the Polynomial

3x² + 2x + 1

Hence ,

$\pmb{ \alpha  +  \beta  = - \dfrac{2}{3} }  \:  \:  -  -    - (1)\\  \bf and \\  \pmb{ \alpha  \beta  = \dfrac{1}{3} } \:  \:  -  -  - (2)$

For The  Equation whose Zeros are

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

$\sf Sum\:of\:Zeros=\dfrac{ \alpha - 1 }{ \alpha + 1 }   + \frac{ \beta - 1}{ \beta + 1} \\ \\ =\dfrac{(\alpha - 1)(\beta + 1) + (\alpha + 1)(\beta - 1)}{(\alpha + 1)(\beta + 1)} \\\\ \:  =  \:  \: \dfrac{\alpha \beta - 1 - \beta + \alpha + \alpha \beta - \alpha + \beta - 1}{\alpha \beta + \alpha + \beta + 1}\\\\ \:  =  \:  \: \dfrac{2\alpha \beta - 2}{\alpha \beta + \alpha + \beta + 1} \\\\ \:  =  \:  \: \dfrac{2(\alpha \beta - 1)}{\alpha \beta + (\alpha + \beta) + 1}$

Using Equations (1) and (2) :

$= \dfrac{2 \times \bigg(\dfrac{1}{3} - 1\bigg) }{\dfrac{1}{3} - \dfrac{2}{3} + 1} \\\\  = \dfrac{2 \times \bigg(\dfrac{1 - 3}{3} \bigg) }{\dfrac{1 - 2 + 3}{3} }\\\\= \dfrac{2 \times \bigg( - \cancel \dfrac{ 2}{3} \bigg) }{ \cancel\dfrac{2}{3} }\\\\= - 2$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf S = - 2} }}}$

$\sf Product\:of\:Zeros= \dfrac{\alpha - 1}{\alpha + 1} \times \dfrac{\beta - 1}{\beta + 1}\\\\= \dfrac{\alpha \beta - \alpha - \beta + 1}{\alpha \beta + \alpha + \beta + 1}\\\\=  \dfrac{\alpha \beta -( \alpha + \beta) + 1}{\alpha \beta + \alpha + \beta + 1}$

Using Equations (1) and (2) :

$\dfrac{\dfrac{1}{3} + \dfrac{2}{3} + 1}{\dfrac{1}{3} - \dfrac{2}{3} + 1}\\\\ = \dfrac{\dfrac{1 + 2 + 3}{3} }{\dfrac{1 - 2 + 3}{3} }\\\\= \cancel\dfrac{6}{2} \\ \\ = 3$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf P =3 } }}}$

Hence The Řequired Polynomial is :

x² - (-2) x + 3

That is :

$\LARGE \underline{\underline{\bf \pink{ {x}^{2} + 2x - 3}}}$

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Answer:

Given:-

$Polynomial \: 3 {x}^{2} + 2x + 1 \: whose \: zeros \: are \: \alpha \: and \: \beta$

$find \: the \: plynomial \: whose \: zeros \: are \: ( \frac{1 - \alpha }{1 + \alpha } ) \: and \: ( \frac{1 - \beta }{1 + \beta }$)

$\alpha + \beta = \frac{ - b}{a} = \frac{ - 2}{3}$

$\alpha \beta = \frac{b}{a} = \frac{1}{3}$

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

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

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

• Beta and alpha gets cancelled.

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

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

• Here 3 gets cancelled.

$= 2( \frac {2}{3} ) \div \frac{3 - 2 + 1}{3}$

$= 2 (\frac{1 - 1}{3} ) \div 1 - \frac{ 2}{3} + \frac{1}{3}$

$= \alpha + \beta = 2$

#BrainLock.