Question

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Answer 1

Answer:

Step-by-step explanation:

sum of the zeroes of polynomial = -b/a

a) 2x² - 3x + 6

sum of the zeroes of polynomial = -b/a

here : b = -3 and a = 2

-(-3/2) = 3/2

b)  -x² + 3x - 3

sum of the zeroes of polynomial = -b/a

here : b = 3 and a = -1

-3/-1 = 3

c) √2x² - 3/√2 x + 1 = 0

sum of the zeroes of polynomial = -b/a

here : b = -3/√2 and a = √2

-(-3/√2)/√2

3/√2/√2 = 3√2/√2 = 3

d) 3x² - 3x + 3 = 0

sum of the zeroes of polynomial = -b/a

here : b = -3 and a = 3

-(-3)/3

3/3 = 1

SO, FROM THE ABOVE SOLUTION , WE OBSERVED THAT THE EQUATIONS WHICH HAS SUM OF IT'S ROOT AS 3  ARE:

-x² + 3x - 3  , and √2x² - 3/√2 x + 1