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Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

(i) p(x) = x2 + 2√(2x) - 6

(ii) q(x) = √3x2 - 10x - 7√3

(iii) f(x) = x2 - (√3 + 1)x + √3

(iv) g(x) = a(x2 + 1) - x(a2 + 1)

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(i) p(x) = x2 + 2√(2x) - 6

P (x) = x2 + 3√2x - √2x - 6

For zeros of p(x), p(x) = 0

⇒ x (x + 3√2) -√2 (x + 3√2) = 0

⇒ (x - √2) (x + 3√2) = 0

x = √2, -3√2

Therefore zeros of the polynomial are √2 & -3√2

Sum of zeros = √2 - 3√2 = - 2√2 = - 2√2 = \(\frac{-coefficient\,of\,x}{coefficient\,of\,x^2}\)

Product of zeros = √2 × - 3√2 = - 6 = - 6 = \(\frac{constant\,term}{coeffixcient\,of\,x^2}\)

(ii) q(x) = √3x2 + 10x + 7√3

⇒ √3x2 + 10x + 7√3

⇒ √3x2 + 7x + 3x + 7√3

⇒ √3x (x + \(\frac{7}{\sqrt3}\)) + 3 (x + \(\frac{7}{\sqrt3}\))

⇒ (√3x + 3) (x + \(\frac{7}{\sqrt3}\))

For zeros of Q(x), Q(x) = 0

(√3x + 3) (x + \(\frac{7}{\sqrt3}\)) = 0

X = \(\frac{-3}{\sqrt3}\)\(\frac{7}{\sqrt3}\)

Therefore zeros of the polynomial are, \(\frac{-3}{\sqrt3}\)\(\frac{7}{\sqrt3}\)

Sum of zeros \(\frac{-3}{\sqrt3}\)\(\frac{-7}{\sqrt3}\)\(\frac{-10}{\sqrt3}\)\(\frac{-coefficient\,of\,x}{coefficient\,of\,x^2}\)

Product of zeros \(\frac{-3}{\sqrt3}\) × \(\frac{-7}{\sqrt3}\) = 7 = \(\frac{constant\,term}{coeffixcient\,of\,x^2}\)

(iii) f(x) = x2 - (√3 + 1)x + √3

f(x) = x2 - (√3 + 1)x + √3

f(x) = x2 - √3x - x + √3

f(x) = x(x - √3) -1(x - √3)

f(x) = (x - 1) (x - √3)

For zeros of f(x), f(x) = 0

(x - 1) (x - √3) = 0

X = 1, √3

Therefore zeros of the polynomial are 1 & √3

Sum of zeros = 1 + √3 = √3 + 1 = \(\frac{-coefficient\,of\,x}{coefficient\,of\,x^2}\)

Product of zeros = 1 × √3 = √3 = \(\frac{constant\,term}{coeffixcient\,of\,x^2}\)

(iv) g(x) = a(x2 + 1) - x(a2 + 1)

g(x) = ax2 - a2x – x + a

g(x) = ax2 - (a2 + 1)x + a

g(x) = ax(x - a) -1(x - a)

g(x) = (ax - 1) (x - a)

For zeros of g(x), g(x) = 0

(ax - 1) (x - a) = 0

X = \(\frac{1}{a}\), a

Therefore zeros of the polynomial are \(\frac{1}{a}\) & a

Sum of zeros

 \(\frac{-coefficient\,of\,x}{coefficient\,of\,x^2}\)

Product of zeros\(\frac{1}{a}\) × a = 1 = 1 = \(\frac{constant\,term}{coeffixcient\,of\,x^2}\)

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