Question

if alpha,beta are the zeroes of quadratic polynomial p(x)=x^2-5x+6. find a quadratic polynomial whose roots are alpha-1/alpha+1,beta-1/beta+1 .. please answer me..​

Answer 1

$\large\bf\underline{Given:-}$

• Quadratic polynomial :- x² - 5x + 6

$\large\bf\underline {To \: find:-}$

• Quadratic polynomial whose roots are :- α- 1/α + 1 , β - 1/β +1

$\huge\bf\underline{Solution:-}$

• p(x) = x² - 5x + 6

↣ x² - 5x + 6

↣ x² - 3x - 2x + 6

↣ x(x - 3) -2(x - 3)

↣ (x - 2)(x -3)

↣ x = 2 or x = 3

• Let α = 2 and β = 3

Now , finding value of α- 1/α + 1 and β - 1/β +1

$\dashrightarrow \tt \: \frac{ \alpha - 1}{ \alpha + 1} \\ \\ \tt \: putting \: value \: of \: \alpha = 2 \\ \\ \dashrightarrow \tt \: \frac{2 - 1}{2 + 1} \\ \\ \dashrightarrow \tt \: \frac{1}{3}$

Now,

$\dashrightarrow \tt \: \frac{ \beta - 1}{ \beta + 1} \\ \\ \tt \:putting \: value \: of \: \beta = 3 \\ \\ \dashrightarrow \tt \: \frac{3 - 1}{3 + 1} \\ \\ \dashrightarrow \tt \: \cancel \dfrac{2}{4} \\ \\ \dashrightarrow \tt \: \frac{1}{2}$

Now,

Roots of required polynomial are 1/3 and 1/2

≫Sum of zeroes = 1/3 + 1/2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = (2+3)/6

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 5/6

≫ Product of zeroes = ⅓× ½

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀ = 1/6

Required polynomial

= x² -(sum of zeroes)x + product of zeroes

= x² -(⅚)x + 1/6

= x² - 5x/6 + 1/6

= 6x² - 5x + 1 = 0

Hence,

▶️ Required polynomial is 6x² - 5x + 1.

$\rule{200}3$

Answer 2

$\sf\large{\underline{\red{Question:-}}}$

if alpha,beta are the zeroes of quadratic polynomial p(x)=x^2-5x+6. find a quadratic polynomial whose roots are alpha-1/alpha+1,beta-1/beta+1.

$\sf\large{\underline{\red{Given:-}}}$

• Quadratic polynomial = $\sf x^3-5x+6$
• Roots are = $\sf \alpha, \beta$

$\sf\large{\underline{\red{To\:Find:-}}}$

• $\sf \large\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}=?$

$\sf\large{\red{\underline{Solution:-}}}$

➡️ By splitting method.

$\sf→ x^2-5x+6=0\\\sf→ x^2-3x-2x+6=0\\\sf→ x(x-3)-2(x-3)=0\\\sf→ (x-3)(x-2)=0\\\sf→ x-3=0, x-2=0\\\sf→ x=3\: and\: x=2$

Here,

$\sf{\fbox{\red{\alpha=3 \:and\: \beta=2}}}$

$\sf→ \frac{\alpha-1}{\alpha+1}=\frac{3-1}{3+1} \\\sf{\fbox{\red{\alpha}= \frac{1}{2}}}$

$\sf→\frac{\beta-1}{\beta+1}= \frac{2-1}{2+1}\\\sf→{\fbox{\red{\beta}=\frac{1}{3}}}$

Now,

we know the sum of zeroes = $\sf \alpha+\beta$

$\sf→ \alpha+\beta=\frac{1}{2}+\frac{1}{3}\\\sf→ \alpha+\beta=\frac{3+2}{6}\\\sf→\alpha+\beta= \frac{5}{6}$

Product of the zeroes = $\sf \alpha\beta$

$\sf →\alpha\beta= \frac{1}{2}×\frac{3}{2}\\\sf→ \alpha\beta=\frac{1}{6}$

Hence,

Required polynomial ↓

$\sf→ x^2-(\alpha+\beta)x+(\alpha\beta)\\\sf→ x^2-\frac{5}{6}x+\frac{1}{6}\\\sf→ \frac{x^2-5+1}{6}\\\sf→6x^2-5x+1$

$\sf{\red{\fbox{\underline{\underline{\blue{\star\:\: 6x^2-5x+1}}}}}}$