NCERT Solutions for Class 10 Maths Ex 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.