Question

find the zeroes of the polynomial √3x^2+10x+7√3 and verify the relationship between zeroes and coefficients

Answer 1

Answer:

zeros are -√3 and -7/√3

Step-by-step explanation:

f(x) = √3x² + 10x + 7√3

= √3x² + 7x + 3x + 7√3

= x(√3x + 7).√3(√3x + 7)

(x + √3)(√3x + 7)

x = -√3 and √3x = -7

x = -√3 and -7/√3

verification

α + β = -b/a

-√3 + (-7√3) = -10/√3

(-3 - 7)/√3 = -10/√3

-10/√3 = -10/√3

αβ = c/a

(-√3) × (-7/√3) = 7√3/√3

7√3/√3 = 7√3/√3

may it helps you

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Answer 2

Answer:

P(x) = √3x2+10x+7√3

= √3x+3x+7x+7√3

= (√3x+7)(x+√3)

= √3x+7 = 0 and x+√3 = 0

= x = -7/√3 and x = -√3

verification by α and β,

let α = -7/√3 and β = -√3

sum of zeros = α+β = -7/√3+(-√3)

= -7/√3-√3

= -7-3/√3

= -10/√3

product of zeroes = αβ

= -7/√3 . -√3

= 7.

verification by coefficients,

sum of zeros = -b/a

= -10/√3.

product of zeros ,

= c/a

= 7√3/√3

= 7.

hence relationship is verified.