(i) 6x² + 11x + 5 = 0
6x2 + 11x + 5 = 6x2 + 6x + 5x + 5
= 6x(x +1) + 5(x +1)
= (x +1) (6x +5)
∴ zeroes of polynomial equation 6x2 +11x +5 are { −1, −5/6 }
Now, Sum of zeroes of this given polynomial equation = −1+( −56 ) = −11/6
But, the Sum of zeroes of any quadratic polynomial equation is given by = −coeff.of x / coeff.ofx2 = −11/6
And Product of these zeroes will be = −1×−5/6 = 5/6
But, the Product of zeroes of any quadratic polynomial equation is given by = constant term / coeff.of x2 = 5/6
Hence the relationship is verified.
(ii) 4s2 – 4s + 1
4s2 – 4s + 1 = 4s2 – 2s – 2s + 1
= 2s (2s – 1) –1(2s – 1)
= (2s – 1) (2s – 1)
∴ zeroes of the given polynomial are: {1/2,1/2}
∴ Sum of these zeroes will be = = 1.
But, The Sum of zeroes of any quadratic polynomial equation is given by = −coeff. of s / coeff.of s2 = −4/4 = 1
And the Product of these zeroes will be = 1/2 × 1/2 =1/4
But, Product of zeroes in any quadratic polynomial equation is given by = constant term/ coeff.of s2 = 1/4.
Hence, the relationship is verified.
(iii) 6x2 – 3 – 7x
6x2 – 7x – 3 = 6x2 – 9x + 2x – 3
= 3x (2x – 3) +1(2x – 3)
= (3x + 1) (2x – 3)
∴ zeroes of the given polynomial are: – (−1/3,3/2)
∴ sum of these zeroes will be = −1/3 + 3/2 =7/6
But, The Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x2 = 7/6
And Product of these zeroes will be = −1/3 × 3/2=−1/2
Also, the Product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of x2 = −3/6 = −1/2
Hence, the relationship is verified.
(iv) 4u2 + 8u
4u2 + 8u = 4u (u+2)
Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0
Hence, the zeroes of the above polynomial equation will be (0, −2)
∴ Sum of these zeroes will be = −2
But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.of u / coeff.of u2 = −8/4 = −2
And product of these zeroes will be = 0 × −2 = 0
But, the product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of u2 = −0/4 = 0
Hence, the relationship is verified.
(v) t2 – 15
t2 – 15 = (t+ √15) (t − √15)
Therefore, zeroes of the given polynomial are: – {√15, −√15}
∴ sum of these zeroes will be = √15 − √15 = 0
But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x2 = −0/1 = 0
And the product of these zeroes will be = (√15) × (−√15) = −15
But, the product of zeroes in any quadratic polynomial equation is given by
= constant term / coeff.of t2 = −15/1 = −15
Hence, the relationship is verified.
(vi) 3x2 – x – 4
3x2 − x − 4 = 3x2 – 4x + 3x − 4
= x (3x – 4) +1(3x – 4)
= ( x + 1) (3x – 4)
∴ zeroes of the given polynomial are: – {−1, 4/3 }
∴ sum of these zeroes will be = −1 +4/3 = 13
But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x2= −(−1)/3 = 1/3
And the Product of these zeroes will be = {−1 × 4/3 }= −4/3
But, the Product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of x2 = −4/3
Hence, the relationship is verified.