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
Answers
Step-by-step explanation:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ans. (i)
Comparing given polynomial with general form of quadratic polynomial ,
We get a = 1, b = -2 and c = -8
We have,
= x(x−4)+2(x−4) = (x−4)(x+2)
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
(x−4)(x+2) = 0
⇒ x = 4, −2 are two zeroes.
Sum of zeroes = 4 + (– 2) = 2 =
=> =
Product of zeroes = 4 × (−2) = −8
=
(ii)
Here, a = 4, b = -4 and c = 1
We have,
=
=2s(2s−1)−1(2s−1)
= (2s−1)(2s−1)
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
⇒ (2s−1)(2s−1) = 0
⇒ s =
Therefore, two zeroes of this polynomial are
Sum of zeroes = = 1 =
=
Product of Zeroes =
(iii)
Here, a = 6, b = -7 and c = -3
We have,
= 3x(2x−3)+1(2x−3) = (2x−3)(3x+1)
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
⇒ (2x−3)(3x+1) = 0
⇒ x =
Therefore, two zeroes of this polynomial are
Sum of zeroes =
Product of Zeroes =
(iv)
Here, a = 4, b = 8 and c = 0
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
⇒ 4u(u+2) = 0
⇒ u = 0,−2
Therefore, two zeroes of this polynomial are 0, −2
Sum of zeroes = 0−2 = −2
= =
Product of Zeroes = 0
=
(v)
Here, a = 1, b = 0 and c = -15
We have, ⇒ ⇒ t =
Therefore, two zeroes of this polynomial are
Sum of zeroes =
Product of Zeroes =
(vi)
Here, a = 3, b = -1 and c = -4
We have, =
= x(3x−4)+1(3x−4) = (3x−4)(x+1)
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
⇒ (3x−4)(x+1) = 0
⇒ x =
Therefore, two zeroes of this polynomial are
Sum of zeroes =
Product of Zeroes =