Question

Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.

(ii) 4s2 – 4s +1
(iii) 6x2 – 3-7x
(iv) 4u2 + 8u
(v) - 15
(vi) 3x2 - X-4​

Answer 1

Answer:

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Answer 2

Answer:

ii) 4s2

- 4s +1= (2s-1)2 The value of 4s2 - 4s

+1 is zero when 2s -1=0, i.e., s = 1/2

Therefore, the zeroes of 4s2 - 4s+1 are

1/2 and 1/2.Sum of zeroes = 1/2 +1/2=1

--4)/4 -(Coefficient of s)/Coefficient

of s2 Product of zeroes = 1/2 x 1/2 = 1/4

Constant term/Coefficient of s2

(iv) 4u2+8u

8u 4u2 +8u +0 = 4u(u + 2) The value of

4u2 +8u is zero when 4u = 0 or u + 2 0,

i.e., u = O or u = - 2 Therefore, the zeroes

of 4u2+8u are 0 and - 2. Sum of zeroes

0+(-2) =-2 =-(8)/4= -(Coefficient of u)/

Coefficient of u2 Product of zeroes = 0x

(vi) 3x2 -X - 4 (3x - 4) (x + 1) The

value of 3x2 - x -4 is zero when 3x - 4

O and x +1=0,i.e., when x =4/3 or x=-1

Therefore, the zeroes of 3x2 x- 4 are

4/3 and -1.Sum of zeroes = 4/3 +(-1) = 1/3

-(-1/3=-(Coefficient of x)/Coefficient of

x2 Product of zeroes = 4/3 x (-1) = -4/3 =

Constant term/Coefficient of x2.