Question

Find a quadratic polynomial having zeros as-3 by alpha and 2 by alpha

Answer 1

Step-by-step explanation:

$Sum \: of \: zeros = \frac{ - 3}{ \alpha } + \frac{ 2}{ \alpha } = \frac{ - 1}{ \alpha } \\ Product \: of \: zeros = \frac{ - 3}{ \alpha } \times \frac{ 2}{ \alpha } = \frac{ - 6}{ { \alpha }^{2} } \\ required \: quadratic \: polynomial \\ is \: given \: as\colon \\ {x}^{2} - (Sum \: of \: zeros)x + Product \: of \: zeros \\ = {x}^{2} - ( \frac{ - 1}{ \alpha } )x + ( \frac{ - 6}{ { \alpha }^{2} } ) \\ = {x}^{2} + \frac{x}{ \alpha } - \frac{6}{ { \alpha }^{2} } \\ = { \alpha }^{2} {x}^{2} + \alpha x - 6$

Answer 2

Answer:

here is your answer

x²+x/alpha-6/alpha²

Step-by-step explanation:

the zeroes given

-3/alpha and 2/ alpha

since x should be equal to either 3/alpha or 2/ alpha

let the factors of polynomial

(x+3/alpha) (x-2/alpha)

x²-2x/alpha+3x/alpha-6/alpha²

x²+x/alpha-6/alpha²

hence the quadratic eqn which we need is x²+x/alpha-6/alpha²

hope it helped u

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