Question

If alpha and beta are zeroes of polynomial x2-2x+3 ,find a polynomial whose roots are alpha-1/alpha+1 and bita-1/bita+1

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If α and β are zeroes of polynomial x² - 2x + 3.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

A polynomial whose roots are α - 1/α + 1 & β - 1/β + 1.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have p(x) = x² - 2x + 3

As we know that given polynomial compared with ax² + bx + c

• a = 1
• b = -2
• c = 3

$\underline{\orange{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:(x)^{2} } }\\\\\\\mapsto\sf{\alpha +\beta =\dfrac{-(-2)}{1} }\\\\\\\mapsto\bf{\alpha +\beta =2}}$

$\underline{\orange{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:(x)^{2} } }\\\\\\\mapsto\sf{\alpha \times \beta =\dfrac{3}{1} }\\\\\\\mapsto\bf{\alpha \times \beta =3}}$

A/q

$\dag\:\underline{\underline{\bf{Sum\:of\:roots\::}}}}$

$\longrightarrow\tt{\dfrac{\alpha-1 }{\alpha +1 } +\dfrac{\beta -1}{\beta +1} }\\\\\\\longrightarrow\tt{\dfrac{(\alpha-1)(\beta +1)+(\beta -1)(\alpha +1) }{(\alpha +1)(\beta +1)} }\\\\\\\longrightarrow\tt{\dfrac{\alpha\beta \cancel{+\alpha -\beta}-1+\alpha \beta \cancel{+\beta -\alpha} -1}{\alpha \beta+\alpha+\beta +1} }\\\\\\\longrightarrow\tt{\dfrac{2\alpha\beta -2 }{\alpha \beta+\alpha +\beta +1 } }\\\\\\\longrightarrow\tt{\dfrac{2(3)-2}{3+2+1} }$

$\longrightarrow\tt{\dfrac{6-2}{6} }\\\\\\\longrightarrow\tt{\cancel{\dfrac{4}{6} }}\\\\\\\longrightarrow\tt{\green{\dfrac{2}{3} }}$

$\dag\:\underline{\underline{\bf{Product\:of\:roots\::}}}}$

$\longrightarrow\tt{\bigg(\dfrac{\alpha-1 }{\alpha+1 } \bigg)\bigg(\dfrac{\beta-1 }{\beta +1} \bigg)}\\\\\\\longrightarrow\tt{\dfrac{\alpha \beta-\alpha-\beta +1 }{\alpha\beta +\alpha+\beta +1 }} \\\\\\\longrightarrow\tt{\dfrac{3-(2)+1}{3+2+1} }\\\\\\\longrightarrow\tt{\dfrac{3-1}{6} }\\\\\\\longrightarrow\tt{\cancel{\dfrac{2}{6} }}\\\\\\\longrightarrow\tt{\green{\dfrac{1}{3} }}$

Now;

$\dag\:\underline{\underline{\pink{\bf{The\:required\:polynomial\::}}}}}$

$\longrightarrow\sf{x^{2} -(sum\:of\:roots)x+(product\:of\:roots)}\\\\\\\longrightarrow\sf{x^{2} -\dfrac{2}{3} x+\dfrac{1}{3} =0}\\\\\\\longrightarrow\sf{\green{3x^{2} -2x+1=0}}$

Answer 2

$\large{\underline{\bf{\pink{Answer:-}}}}$

✰The required polynomial is

$:\implies\bf \:p(x)=\:3x^2-2x+1$

$\large{\underline{\bf{\blue{Explanation:-}}}}$

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

✰ p(x) = x² - 2x + 3

✰ two root's of any polynomial is given as

$\bf(\frac{\alpha-1}{ \alpha + 1}) \: and \: ( \frac{ \beta - 1}{ \beta + 1})$

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the polynomial whose roots are $\bf(\frac{\alpha-1}{ \alpha + 1}) \: and \: ( \frac{ \beta - 1}{ \beta + 1})$

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

α and β are the zeroes of polynomial

=x² - 2x + 3.

${\pink{ \bf{ \alpha + \beta = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} }}}}$

$:\implies\bf = \frac{-(-2)}{1}$

$:\implies\bf=2$

${\pink{\bf{ \alpha\beta = \frac{ Constant\: term}{coefficient \: of \: {x}^{2} }}}}$

$:\implies\bf = \frac{3}{1}$

$:\implies\bf=3$

Now,

Sum of zeroes:-

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

$:\implies\bf\frac{(\alpha-1)(\beta+1)+(\beta-1)(\alpha+1}{(\alpha+1)(\beta+1)}$

$:\implies\bf \frac{\alpha\beta-\beta+\alpha-1+\alpha\beta-\alpha+\beta-1}{\alpha\beta+\alpha+\beta+1}$

$:\implies\bf \frac{2\alpha\beta-2}{\alpha\beta+(\alpha+\beta)+1}$

$:\implies\bf\frac{2(\alpha\beta-1) }{\alpha\beta+(\alpha+\beta)+1}$

By substituting α+β =2 and αβ = 3

we get ,

$:\implies\sf\frac{2(3-1)}{3+2+1}$

$:\implies\sf\frac{4}{6}$

Product of zeroes:-

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

$:\implies\sf\frac{\alpha\beta-\alpha-\beta+1}{\alpha\beta+\alpha+\beta+1}$

$:\implies\sf\frac{\alpha\beta-(\alpha+\beta)+1}{\alpha\beta+(\alpha+\beta)+1}$

$:\implies\sf\frac{3-2+1}{3+2+1}$

$:\implies\sf \frac{2}{6}$

So, the required polynomial is:-

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

$:\implies\bf\:(x^2 \frac{-2x}{3}+\frac{1}{3}$

$:\implies\bf \:3x^2-2x+1$

━━━━━━━━━━━━━━━━━━━