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

Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.
Let p(x) be a polynomial with any number of terms any number of degree.
Now, zeroes of the polynomial will be the values of x at which p(x) = 0.
If p(x) = ax² + bx + c   is a quadratic polynomial (highest power is equal to 2) and its roots are α and β, then
Sum of the roots = α + β = - b/a
Product of roots = αβ = c/a
 
(i) p(x) = x² - 2x - 8
So, the zeroes will be the values of x at which p(x) = 0
Therefore,

⇒ x² - 4x + 2x - 8 = 0

(We will factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)
⇒ x(x - 4) + 2(x - 4) = 0
= (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 = 4 + (-2) = 2 

 

Hence, it is verified that, 

Product of zeroes =

Hence, it is verified that, 

(ii) 4s² - 4s + 1
= (2s)² - 2(2s)1 + 1²
As, we know (a - b)² = a² - 2ab + b², the above equation can be written as
= (2s - 1)²

The value of 4s² − 4s + 1 is zero when 2s − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4s² − 4s + 1 are  and .

Sum of zeroes =

Product of zeroes =

Hence Verified. 

(iii) 6x² - 3 - 7x
= 6x² - 7x - 3
(We will factorize 7 such that the product of the factors is equal to 18 and the difference is equal to - 7)
= 6x² + 2x - 9x - 3
= 2x(3x + 1) - 3(3x + 1)
= (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)    [As, x² - y² = (x - y)(x + y)]

 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
(We will factorize 1 in such a way that the product of factors is equal to 12 and the difference is equal to 1)
= 3x² - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x – 4 )(x + 1)

The value of 3x² − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0,

when  

Therefore, the zeroes of 3x² − x − 4 are and -1

Sum of zeroes =

Product of zeroes =

Hence, verified.