Question

if alpha and beta are zeros of 3x2 -4x+1 find polynomial whose zeros are alpha square/beta and beta square/alpha

Answer 1

Answer:

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Answer 2

Solution :

We have quadratic polynomial p(x) = 3x² - 4x + 1 & zero of the polynomial p(x) = 0

$\underline{\underline{\tt{Using\:\:by\:\:factorization\:\:method\::}}}$

$\longrightarrow\sf{3x^{2} -4x + 1=0}\\\\\longrightarrow\sf{3x^{2} -3x -x +1=0}\\\\\longrightarrow\sf{3x(x-1) -1(x-1) = 0}\\\\\longrightarrow\sf{(x-1) (3x-1) = 0}\\\\\longrightarrow\sf{x-1=0\:\:\:Or\:\:\:3x-1=0}\\\\\longrightarrow\sf{x=1\:\:Or\:\:3x=1}\\\\\longrightarrow\bf{x=1\:\:Or\:\:x=1/3}$

∴ α = 1 & β = 1/3 are the zeroes of the given polynomials .

According to the question we have another two zeroes are α = α²/β & β = β²/α.

Now;

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

$\mapsto\tt{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\\\mapsto\tt{\dfrac{\alpha^{2} }{\beta} + \dfrac{\beta ^{2}}{\alpha} }\\\\\\\mapsto\tt{\dfrac{\alpha^{3} + \beta ^{3} }{\alpha\beta }} \\\\\\\mapsto\tt{\dfrac{(\alpha+\beta)^{3}-3\alpha \beta (\alpha + \beta ) }{\alpha\beta }} \\\\\\\mapsto\tt{\dfrac{\bigg(1+\dfrac{1}{3}\bigg)^{3} - 3\times 1\times \dfrac{1}{3} \bigg(1+\dfrac{1}{3} \bigg)}{1\times 1/3} }$

$\mapsto\tt{\dfrac{(1)^{3} +\bigg(\dfrac{1}{3}\bigg)^{3} + 3\times 1\times 1/3 \bigg(1+1/3\bigg)- 3\times 1\times \dfrac{1}{3} \bigg(1+\dfrac{1}{3} \bigg)}{ 1/3} }\\\\\\\mapsto\tt{\dfrac{1 + \dfrac{1}{27} + \cancel{3 /3} (4/3) - \cancel{\dfrac{3}{3} }\bigg(\dfrac{3+1}{3} \bigg)}{ 1/3} }\\\\\\\mapsto\tt{\dfrac{1 +\dfrac{1}{27} + {1 \bigg(\dfrac{4}{3} \bigg)-1 }\bigg(\dfrac{4}{3} \bigg)}{ 1/3} }$

$\mapsto\tt{\dfrac{1 +\dfrac{1}{27} +\cancel{ {\dfrac{4}{3} -} \dfrac{4}{3} } }{ 1/3} }\\\\\\\mapsto\tt{\dfrac{1 +\dfrac{1}{27} }{ 1/3} }\\\\\\\mapsto\tt{\dfrac{\dfrac{27+1}{27} }{ 1/3} }\\\\\\\mapsto\tt{\dfrac{\dfrac{28}{27} }{ 1/3} }\\\\\\\mapsto\tt{\dfrac{28}{\cancel{27}} \times \dfrac{\cancel{3}}{1} }\\\\\\\mapsto\bf{\dfrac{28}{9} }$

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

$\mapsto\tt{\alpha \times \beta = \dfrac{-c}{a}=\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\\\mapsto\tt{\dfrac{\cancel{\alpha} ^{2} }{\cancel{\beta} } \times \dfrac{\cancel{\beta} ^{2} }{\cancel{\alpha} }} \\\\\\\mapsto\tt{\alpha \beta }\\\\\\\mapsto\tt{1\times 1/3 }\\\\\\\mapsto\bf{1/3}$

Now;

The required polynomial are ;

$\longrightarrow\sf{x^{2} - (sum\:of\:zeroes)x + (product\:of\:zereos)}\\\\\longrightarrow\sf{x^{2} - (28/9)x + (1/3)}\\\\\longrightarrow\bf{9^{2} -28x + 3}$