Question

find the polynomial

Answer 1

$\textbf {Theory}$ :

We are provided with α and β as the zeroes of the polynomial -3x²-4x+1. We need to find a Quadratic polynomial such that its zeroes are $\dfrac{\alpha ^{2} } {\beta }$ and $\dfrac{\beta^{2} } {\alpha}$ .

For doing this, we will first need to find the values of α and β and then put them in the given case. For finding the values of α and β, we will use the relationship between coefficients and zeroes of a polynomial.

Lets start with this!

$\textbf {Solution}$ :

$\sf \alpha + \beta = -\frac{4}{3}\\ \\ \sf\alpha \beta = -\frac{1}{3} \\ \\ \\ \sf \: Sum \: of \: the \: zeroes\rightarrow \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} \\ \\ \sf= \frac{\alpha^3+\beta^3}{\alpha\beta} \\ \\ \sf = \frac{(\alpha + \beta)^3-3\alpha\beta (\alpha + \beta)}{\alpha\beta} \\ \\ \\ \sf= \frac{ \left(-\frac{4}{3}\right)^3 - \cancel{3} \left(-\frac{1}{\cancel{3}}\right) \left(-\frac{4}{3}\right)}{\left(-\frac{1}{3}\right) } \\ \\ \sf = \frac{ (\frac{ - 64}{27}) - \frac{4}{3} }{ \frac{ - 1}{3} } \\ \\ \sf = \frac{ - 64}{27} - \frac{4}{3} \times - 3 \\ \\ \sf = \frac{ - 64 - 36}{27} \times - 3 \\ \\ \sf = \frac{ - 100}{ \cancel{27}} \times \cancel{ - 3} \\ \\ \sf = \frac{100}{9} \\ \\ \\ \sf \: Product \: of \: the \: zeroes \rightarrow \: \frac{ \alpha ^{ \cancel 2} }{ \cancel\beta } \times \frac{ { \beta }^{ \cancel 2} }{ \cancel\alpha } \\ \\ \sf \: = \alpha \beta = \frac{ - 1}{3}$

Now, we know that

x² - Sx + P = 0 is the required polynomial when S represents the sum of the zeroes and P denotes to the product of the zeroes.

Putting the values we got, we get

$\sf {x}^{2} - sx + p \\ \\ \sf \: = {x}^{2} - \dfrac{100}{9} x + \dfrac{ - 1}{3} = 0 \\ \\ \sf= {x}^{2} - \dfrac{100}{9} x - \frac{1}{3} = 0 \\ \\ \sf= 9 {x}^{2} - 100x - 3 = 0$

$\textbf {Answer}$ :

Thus, the required polynomial is 9x² - 100x - 3.

Answer 2

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Α and β are the zeros of 3x²-4x +1 polynomial,

first of all we factorise 3x²-4x+1

3x² -4x + 1 

=3x² -3x -x +1 

=3x( x -1) -1(x -1)

=(3x -1)(x -1)

hence. (3x -1) and (x -1) are the factors of given polynomial .

so, x = 1/3 and 1 are the zeros of that polynomial.

hence, α = 1/3. and β = 1
or α = 1 and β. = 1/3
you can choose any one in both 

I choose α = 1. and β = 1/3 

now,
let any unknown. polynomial. whose zeros are α²/β and β²/α 

α²/β = (1)²/(1/3) = 3

β²/α = (1/3)²/1 = 1/9 

now, equation of unknown polynomial.

x²- ( sum of roots)x + product of roots 

= x²- ( α²/β + β²/α)x +(α²/β)(β²/α)

put α²/β = 3 and β²/α = 1/9

= x²- ( 3 +1/9)x + 3 × 1/9

= x² -28x/9 + 3/9

={ 9x² -28x + 3 }1/9

hence, 9x² -28x + 3 is answer 

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