Question

$if \alpha \: \beta are \: the \: zeros \: of \: p(x) = 2x {}^{2} - 3x + 1 \: then \: find \: the \: polyominal \: where \: zeros \: are \ \alpha { \beta }^{2} and \alpha {}^{2} \beta$
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Answer 1

Answer:

It is the correct answer.

Step-by-step explanation:

Hope this attachment helps you.

Answer 2

Step-by-step explanation:

Given,

$\sf\alpha,\beta$ are the zeroes of p(x) = $\sf 2x^{2} - 3x + 1$

To Find :-

The Quadratic Polynomial whose roots are :-

$\alpha { \beta }^{2} , { \alpha }^{2} \beta$

Solution :-

To need obtain the roots of the polynomial we need to equate that polynomial to zero :-

2x² - 3x + 1 = 0

2x² - 2x - x + 1 = 0

2x(x - 1) - 1(x - 1) = 0

(x - 1) (2x - 1) = 0

(x - 1) = 0 , (2x - 1) = 0

x = 1 , 2x = 1

x = 1 , x = 1/2

$\alpha = 1, \beta = \dfrac{1}{2}$

$\sf\alpha(\beta)^{2} = 1\times \left(\dfrac{1}{2}\right)^2$

= $\sf\dfrac{1}{4}$

$\sf(\alpha)^{2}\beta = (1)^{2}\times \dfrac{1}{2}$

= $\sf\dfrac{1}{2}$

Required Polynomial :-

$\sf x^{2}-\left(\dfrac{1}{4}+\dfrac{1}{2}\right)x + \left(\dfrac{1}{4}\times \dfrac{1}{2}\right)$ = 0

= $\sf x^{2} - \dfrac{3x}{4} + \dfrac{1}{8}$ = 0

= $\sf 8x^{2} - 6x + 1 = 0$