If alpha and beta are the zeros of quadratic polynomal 2x2-5x+7 , find a polynomial whose zeros are 3alpha+2beta and - Maths - Polynomials

Question

If alpha and beta are the zeros of quadratic polynomal 2x2-5x+7
, find a polynomial whose zeros are 3alpha+2beta and
2alpha+3beta

Gursheen kaur. · Accepted Answer

$L e t t h e g i v e n q u a d r a t i c p o l y n o m i a l be f(x)= 2 x^{2} - 5 x + 7 g i v e n : α & β a r e z e r o s o f g i v e n q u a d r a t i c p o l y n o m i a l . t h e r e f o r e, s u m o f r o o t s = \frac{- (- 5)}{2} α + β = \frac{5}{2} ..... (i) a n d p d t o f r o o t s = \frac{7}{2} α β = \frac{7}{2} ..... (i i)$

now, when zeros of polynomial are (3α+2β) and (2α+3β), then

sum of roots= (3α+2β) + (2α+3β) =5α+5β=5(α+β)

$= 5 (α + β) [f r o m (i)] = 5 * \frac{5}{2} = \frac{25}{2} a n d p d t o f r o o t s = (3 α + 2 β) (2 α + 3 β) = 6 α^{2} + 9 α β + 4 α β + 6 β^{2} [f r o m (i i)] = 6 (α^{2} + β^{2}) + 13 α β = 6 [{(α + β)}^{2} - 2 α β] + 13 α β = 6 [{(\frac{5}{2})}^{2} - 2 * \frac{7}{2}] + 13 * \frac{5}{2} [f r o m (i) & (i i)] = 6 (\frac{25}{4} - 7) + \frac{65}{2} = 6 (\frac{25 - 28}{4}) + \frac{65}{2} = 6 (\frac{- 3}{4}) + \frac{65}{2} = \frac{- 9}{2} + \frac{65}{2} = \frac{- 9 + 65}{2} = \frac{56}{2} = 28$

thus, the polynomial obtained is

$g (x) = x^{2} - (s u m o f z e r o s) x + p d t o f z e r o s = x^{2} - \frac{25}{2} x + 28$

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If alpha and beta are the zeros of quadratic polynomal 2x2-5x+7 , find a polynomial whose zeros are 3alpha+2beta and 2alpha+3beta