Question

If alpha and beta
are zeros of polynomial
x² - 2x+3 Find polynomial with
zeros
alpha-1/alpha+1
and beta-1/beta+1
​

Answer 1

Solution :

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

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

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

The polynomial with zeroes α - 1/α + 1 and 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

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

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

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

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

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

Now;

$\underline{\underline{\bf{The\:required\;polynomial\::}}}}}}$

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

Answer 2

Answer:

$\implies$ 3x² - 2x + 1 = 0

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

$If \: \alpha \: \: and \: \beta are \: \: the \: \: zeroes \: \: of \: \: quadratic \: \: polynomial \: x {}^{2} - 2x \: + \: 2$

$The \: polynomial \: with \: zeroes \: a \: \: \alpha -( 1 \div \alpha ) + 1 \: amd \: \beta - \: (1 \div \beta ) + 1$

$\large\underline\mathrm{To \: find}$

Explanation:

• We have p(x) = x² - 2x + 3
• The given polynomial compared with ax² + bx + c

$\implies$ a = 1

$\implies$ b = -2

$\implies$ c = 3

Sum of the zeroes:

$\alpha + \beta = - b \div a \\ \alpha + \beta = - ( - 2) \div 1 \\ \alpha + \beta = 2$

Product of the zeroes:

$\alpha \times \beta = c \div a \\ \alpha \times \beta = 3 \div 1 \\ \alpha \times \beta = 3$

Sum of zeroes:

$( \alpha - 1 \: )( \: \alpha + 1 \: ) + ( \: \beta - 1 \: )( \beta + 1) \\ ( \: \alpha - 1 \: )( \: \beta + 1 \: ) + ( \beta + 1 \: )( \alpha + 1 \: ) \div ( \: \alpha + 1 \: )( \beta \: + 1 \: ) \\ (\alpha \beta + \alpha + \beta - 1 + \alpha \beta + \beta - \alpha - 1) \div \alpha \beta + \alpha + \beta + 1 \\ (2 \alpha \beta - 2) \div (3 + 2 + 1) \\ (6 - 2) \div 6 \\ 4 \div 6 \\ 2 \div 3$

Product of zeroes:

$( \alpha - 1 \div \alpha + 1)( \beta - 1 \div \beta + 1) \\ (\alpha \beta - \alpha - \beta + 1) \div ( \alpha \beta + \alpha + \beta + 1) \\ (3 - 2 + 1) \div (3 + 2 + 1) \\ (1 + 1) \div 6 \\ 2 \div 6 \\ 1 \div 3$

Now,

$\implies$ x² - (2/3)x + (1/3) = 0

$\implies$ x² - 2x/3 + 1/3 = 0

$\implies$ 3x² - 2x + 1 = 0