Question

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

Answer 1

Given,

f(x) = x^2 - 2x - 3 and zeroes are alpha-1/alpha+1 and beta-1/beta+1.

To Find,

Quadratic polynomial = ?

Solution,

f(x) = x^2 - 2x - 3

a = 1, b = -2, c = 3

(α - 1)/α + 1 and (β - 1)/β + 1.

Sum of zeroes will be,

==> [(α - 1)(β + 1) + (β - 1)(α + 1)]/[(α + 1)(β + 1)]

==> (αβ + α - β - 1 + αβ + β+ β - α - 1)/(3 + 2 + 1)

==> 3 - 1 + 3 - 1/6

==> 4/6

==> 2/3

Product of zeroes will be,

==> [(α - 1)/α + 1] * [(β - 1)/β + 1)]

==> 3 - (α + β) + 1/3 + 2 + 1

==> 3 - 2 + 1/6

==> 2/6

==> 1/3

Required polynomial,

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

==> x² - (2/3)x + 1/3

==> 3x² - 2x + 1

#BeBrainly

Answer 2

Appropriate Question:

$\sf{If \: alpha \: and \: beta \: are \: the \: zeroes \: of \: the \: q uadratic \: polynomial \: p(x) = x^2-2x+3, \: find \: a \: polynomial \: whose \: zeroes \: are \: \frac{ \alpha - 1}{ \alpha + 1} , \: \frac{ \beta - 1}{ \beta + 1} }$

Solution:

$\sf{We \: have , \: p(x)= {x}^{2} - 2x + 3 }$

Here,

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

And

$\sf{ \alpha \beta = \frac{c}{a} } \\ \\ \alpha \beta = \frac{3}{1} \\ \\ \alpha \beta = 3$

Now,

$\sf{sum \: (S) \: \: of \: zeroes} = \frac{ \alpha - 1}{ \alpha + 1} + \frac{ \beta - 1}{ \beta + 1} \\ \\ \sf{= \frac{( \alpha -1)( \beta +1)+( \beta -1)( \alpha +1)}{( \alpha +1)( \beta +1)} } \\ \\ \sf{ = \frac{ \alpha \beta+ \alpha - \beta -1+ \alpha \beta + \beta - \alpha -1 }{ \alpha \beta + \alpha + \beta +1} } \\ \\ \sf{= \frac{2 \alpha \beta -2}{ \alpha \beta +( \alpha + \beta )+1} }$

By, Putting Values

$\sf{ = { \frac{2(3) - 2}{3+2+1} }} \\ \\ \sf{ = \frac{6 - 2}{6} }\\ \\ \sf{= \frac{4}{6} } \\ \\ \sf{ = \frac{2}{3} }$

$\sf{Also, \: Product \: (P) \: Of \: Zeroes = \frac{ \alpha -1}{ \alpha +1} \times \frac{ \beta -1}{ \beta +1} } \\ \\ \sf{= \frac{( \alpha -1)( \beta -1)}{ ( \alpha +1)( \beta +1)} } \\ \\ \sf{= \frac{ \alpha \beta - \alpha - \beta +1}{ \alpha \beta + \alpha + \beta +1}} \\ \\ \sf{ = \frac{ \alpha \beta -( \alpha + \beta )+1}{ \alpha \beta +( \alpha + \beta )+1 } }$

By, Putting Values

$\sf{ = \frac{3 - 2 + 1}{3 + 2 + 1} } \\ \\ \sf{ = \frac{2}{6} } \\ \\ \sf{ = \frac{1}{3} }$

$\sf{\therefore \: Required \: quadratic \: polynomial, \: p(x)=k\{ {x}^{2}-Sx+P\} }$

$: \implies \sf{p(x) = k \bigg\{ {x}^{2} - \frac{2}{3}x + \frac{1}{3} \bigg \}}$

Final Answer:

$: \leadsto \mathfrak{p(x) = k \bigg\{ {x}^{2} - \frac{2}{3}x + \frac{1}{3} \bigg\} }$