Question

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Answer 1

we have,

$\alpha + \beta = 5/2$  and

$\alpha \beta =1$

therefore equation is ,

$x^{2} - 5/2 x + 1$

= [tex]1/2 ( 2x^{2} - 5x +1 ) [/tex]    (taking 1/2 common)


Now to find the zeros of $2 x^{2} - 5 x + 2$

$2 x^{2} -4 x - 1 x + 2$

= [tex]2 x(x-2) -1(x-2)

= (2x-1)(x-2) [/tex]

therefore zeros are 1/2 and 2

hope this helps.......plz mark as brainliest

Answer 2

Step-by-step explanation:

see the attachment Mate✌️.....!!!