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

We have

α and β are the Zeros of the polynomial

p(x) = 3x² + 2x + 1 .

and we have to find a Polynomial whose Zeros are  $\sf \dfrac{ \alpha-1 }{ \alpha +1} \: \: and \: \: \dfrac{ \beta-1 }{\beta+1 }$

So Let's Begin :-)

$\large \bf \clubs \:  Solution :-$

As α and β are the Zeros of the Polynomial

3x² + 2x + 1

Hence ,

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

For The  Polynomial whose Zeros are

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

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

Putting Values from (i) & (ii)

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

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

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

Putting Values from (i) & (ii)

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

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

Hence The Required Polynomial is :

x² - (-2) x + 3

That is :

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

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

✏ 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 of the Form x² - Sx + P

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Answer 2

Given :-

3x² + 2x + 1

To Find :-

Find the polynomial whose zeroes are alpha - 1/alpha + 1 and beta - 1/beta +1​

Solution :-

On comparing the given equation with ax² + bx + c. We get

a = 3

b = 2

c = 1

Sum of zeroes = -b/a

Sum = -(2)/3

Sum = -2/3

Product of zeroes = c/a

Product = 1/3

Now,

New Sum of zeroes = α - 1/α + 1 + β - 1/β + 1

⇒ (α - 1)(β + 1) + (α + 1)(β - 1)/(α + 1)(β + 1)

⇒ αβ - 1 - β + α + αβ - α + αβ - α + β - 1/αβ + α + β + 1

⇒ (αβ + αβ) - (1 + 1)/αβ + (α + β) + 1

⇒ 2αβ - 2/αβ + (α + β) + 1

⇒ 2(1/3) - 2/(1/3) + (-2)/3 + 1

⇒ 2/3 - 2/(1 - 2 + 3/3)

⇒ (2 - 6/3)/(4 - 2/3)

⇒ (-4/3)/(2/3)

⇒ -4/3 × 3/2

⇒ -2

New Product of zeroes = (α - 1/α + 1)(β - 1/β + 1)

⇒ αβ - α - β + 1/αβ + α + β + 1

⇒ αβ - (α + β) + 1/αβ + (α + β) + 1

⇒ 1/3 - (-2/3) + 1/[1/3 + (-2/3) + 1]

⇒ 1/3 + 2/3 + 1/[1/3 - 2/3 + 1]

⇒ (1 + 2 + 3/3)/(1 - 2 + 3/3)

⇒ (6/3)/(2/3)

⇒ 6/3 × 3/2

⇒ 3

Standard form of quadratic polynomial = x² - (α + β)x + αβ

⇒ x² - (-2)x + (3)

⇒ x² + 2x + 3