Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 - 2x - 8
(ii) 4s2 - 4s + 1
(iii) 6x2 - 3 - 7x
(iv) 4u2 + 8u
(v) t2 - 15
(vi) 3x2 - x - 4
(i) x2 - 2x - 8
= (x - 4)(x + 2)
The value of x2 - 2x - 8 is zero when x − 4 = 0 or x + 2 = 0,
i.e, x = 4 or x = −2
Therefore, The zeroes of x2 - 2x - 8 are 4 and −2
Sum of zeroes = 
Hence, it is verified that, 
Product of zeroes = 
Hence, it is verified that, 
(ii) 4s2 - 4s + 1
= (2s - 1)2
The value of 4s2 − 4s + 1 is zero when 2s − 1 = 0, when, s = 1/2
Therefore, the zeroes of 4s2 − 4s + 1 are 1/2 and 1/2
Sum of zeroes = 
Product of zeroes = 
Hence Verified.
(iii) 6x2 - 7x - 3
= (3x + 1)(2x - 3)
The value of 6x2 − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0,
i.e. 
Therefore, the zeroes of 6x2 − 3 − 7x are 
.
Sum of zeroes = 
Product of zeroes =
Hence, verified.
(iv) 4u2 + 8u = 4u2 + 8u + 0
= 4u (u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0,
i.e., u = 0 or u = −2
Therefore, the zeroes of 4u2 + 8u are 0 and −2.
Sum of zeroes = 
Product of zeroes = 
(v) t2 – 15
= t2 – (√15)2
= (t - √15)(t + √15)
The value of t2 − 15 is zero when (t - √15) = 0 or (t + √15) = 0,
i.e., when t = √15 or t = -√15
Therefore, the zeroes of t2 − 15 are √15 and -√15.
Sum of zeroes = 
Product of zeroes = 
Hence verified.
(vi) 3x2 – x – 4
= (3x – 4 )(x + 1)
The value of 3x2 − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0,
when x = 4/3 or x = −1
Therefore, the zeroes of 3x2 − x − 4 are 4/3 and −1.
Sum of zeroes = 
Product of zeroes = 
Hence, verified.