Go To: Exercise 2.1 Exercise 2.3 Exercise 2.4
NCERT Solutions Maths Class 10 Exercise 2.2
1. 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
Solution:
(i) x2 − 2x − 8
We have, x2 − 2x – 8 = x2 − 4x + 2x − 8
= x(x − 4) + 2(x − 4) = (x − 4)( x + 2)
To
find the zeroes of this polynomial, equate the given equation to 0.
x2
− 2x – 8 = 0
(x − 4)(x + 2) = 0
⇒ x =
4, −2
Therefore,
two zeroes of this polynomial are 4 and −2.
Comparing
given polynomial with general form ax2 + bx + c, we get
a =
1, b = −2 and c = −8
Sum
of zeroes = 4 + (– 2) = 2 = −(−2)/1
= −b/a = − Coefficient of x/Coefficient of x2
Product
of zeroes = 4 × −2 = −8
= −8/1 = c/a = Constant
term/Coefficient of x2
(ii) 4s2
− 4s + 1
We
have, 4s2 − 4s + 1 = 4s2 − 2s − 2s
+ 1
=2s(2s−1)−1(2s−1)
=
(2s−1)(2s−1)
To
find the zeroes of this polynomial, equate the given equation to 0.
4s2 − 4s + 1 = 0
⇒ (2s − 1)(2s − 1) = 0
⇒ s = 1/2,
1/2
Therefore,
two zeroes of this polynomial are 1/2 and 1/2.
Comparing
given polynomial with general form ax2 + bx + c, we get
a = 4, b = -4 and c = 1
Sum
of zeroes = 1/2 + 1/2 = 1 = −(−1)/1 × 4/4 = −(−4)/4
= −b/a = −Coefficient of x/Coefficient of x2
Product
of zeroes = 1/2 × 1/2 = 1/4
= c/a = Constant term/Coefficient
of x2
(iii) 6x2 – 3 −
7x
We
have, 6x2
– 3 − 7x = 6x2
− 7x − 3
= 6x2 − 9x + 2x − 3
= 3x(2x − 3) + 1(2x − 3) = (2x − 3)(3x + 1)
To
find the zeroes of this polynomial, equate the given equation to 0.
6x2 – 3 − 7x = 0
⇒ (2x − 3)(3x + 1) = 0
⇒ x = 3/2,
−1/3
Therefore,
two zeroes of this polynomial are 3/2 and −1/3.
Comparing
given polynomial with general form ax2 + bx + c, we get
a =
6, b = −7 and c = −3
Sum
of zeroes = 3/2 + −1/3 = (9−2)/6
= 7/6 = −(−7)/6
= −b/a = −Coefficient of x/Coefficient of x2
Product
of zeroes = 3/2 × −1/3
= −1/2 = −3/6
= c/a = Constant
term/Coefficient of x2
(iv) 4u2 + 8u
We
have, 4u2 + 8u = 4u(u + 2)
To
find the zeroes of this polynomial, equate the given equation to 0.
4u2
+ 8u = 0
⇒ 4u(u + 2) = 0
⇒ u =
0, −2
Therefore,
two zeroes of this polynomial are 0 and −2.
Comparing
given polynomial with general form ax2 + bx + c, we get
a =
4, b = 8 and c = 0
Sum
of zeroes = 0−2 = −2
= −2/1
× 4/4 = −8/4
= −b/a
= −Coefficient of x/Coefficient of x2
Product
of zeroes = 0 × −2 = 0
= 0/4 = c/a = Constant term/Coefficient of x2
(v) t2 – 15
We
have, t2 – 15 = 0
⇒ t2 = 15
⇒ t = ±√15
Therefore,
two zeroes of this polynomial are √15 and −√15.
Comparing
given polynomial with general form ax2 + bx + c, we get
a =
1, b = 0 and c = −15
Sum
of zeroes = √15 + (−15) = 0 = 0/1
= −b/a = −Coefficient of x/Coefficient of x2
Product
of Zeroes = √15 × (−√15) = −15 = −15/1
= c/a = Constant term/Coefficient of x2
(vi) 3x2 – x −
4
We
have, 3x2 – x − 4
= 3x2 − 4x + 3x − 4
= x(3x − 4) + 1(3x − 4)
= (3x − 4)(x + 1)
To
find the zeroes of this polynomial, equate the given equation to 0.
3x2 – x − 4 = 0
⇒ (3x − 4)(x + 1) = 0
⇒ x = 4/3,
−1
Therefore,
two zeroes of this polynomial are 4/3 and −1.
Comparing
given polynomial with general form ax2 + bx + c, we get
a =
3, b = −1 and c = −4
Sum
of zeroes = 4/3 + (−1) = (4−3)/3
= 1/3 = −(−1)/3
= −b/a =
−Coefficient of x/Coefficient of x2
Product
of zeroes = 4/3 × (−1) = −4/3
= c/a = Constant term/Coefficient of x2
2. Find a quadratic polynomial each with the
given numbers as the sum and product of its zeroes respectively.
(i) 1/4, −1 (ii) √2, 13
(iii) 0, √5 (iv) 1, 1
(v) −1/4, 1/4 (vi) 4, 1
Solution: (i) 1/4, −1
Let the
quadratic polynomial be ax2 + bx
+ c.
Let α and β be two zeroes of above quadratic polynomial.
Then,
sum of zeroes = α + β = 1/4 =
−b/a
And
product of zeroes = α × Î² = −1 = −1/1
× 4/4 = −4/4 = c/a
Thus,
a = 4, b = −1, c = −4
Therefore, the quadratic
polynomial which satisfies the above conditions is 4x2 – x – 4.
(ii) √2, 1/3
Let the
quadratic polynomial be ax2 + bx
+ c.
Let α and β be two zeroes of above quadratic polynomial.
Then,
sum of zeroes = α + β = √2 = 3√2/3
= −b/a
And
product of zeroes = α × Î² = 1/3 = 1/3 = c/a
Thus,
a = 3, b = −3√2, c =1
Therefore, quadratic
polynomial which satisfies the above conditions is 3x2 − 3√2x + 1.
(iii) 0, √5
Let the
quadratic polynomial be ax2 + bx
+ c.
Let α and β be two zeroes of above quadratic polynomial.
Then,
sum of zeroes = α + β = 0 = 0/1
= −b/a
And
product of zeroes = α × Î² = √5 = √5/1 = c/a
Thus, a
= 1, b = 0, c = √5
Therefore, the quadratic
polynomial which satisfies the above conditions is x2 + √5.
(iv) 1, 1
Let the
quadratic polynomial be ax2 + bx
+ c.
Let α and β be two zeroes of above quadratic polynomial.
Then,
sum of zeroes = α + β = 1 =
−(−1)/1 = −b/a
And
product of zeroes = α × Î² =
1 = 1/1 = c/a
Thus, a = 1, b = −1, c
= 1
Therefore, the quadratic
polynomial which satisfies the above conditions is x2 – x + 1.
(v) −1/4, 1/4
Let the
quadratic polynomial be ax2 + bx
+ c.
Let α and β be two zeroes of above quadratic polynomial.
Then,
sum of zeroes = α + β = −1/4 =
−b/a
And
product of zeroes = α × Î² = 1/4 = c/a
Thus, a
= 4 , b = 1 , c = 1
Therefore, the quadratic
polynomial which satisfies the above conditions is 4x2 + x + 1.
(vi) 4, 1
Let the
quadratic polynomial be ax2 + bx
+ c.
Let α and β be two zeroes of above quadratic polynomial.
Then,
sum of zeroes = α + β = 4 =
−(−4)/1 = −b/a
And
product of zeroes = α × Î² = 1 =
1/1 = c/a
Thus,
a = 1 , b = −4 , c = 1
Therefore, the
quadratic polynomial which satisfies the above conditions is x2 − 4x
+ 1.