Class 10 - Mathematics
Polynomials - Exercise 2.2
Top Block 1
Exercise 2.2
Question : 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
Answer :
(i) x2 – 2x – 8
= x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)
The value of x2 – 2x – 8 is zero if x + 2 = 0 or x – 4 = 0
So, x = -2, 4
Therefore zeroes of x2 – 2x – 8 are -2 and 4
Now, Sum of zeroes = -2 + 4 = 2 = -(-2)/1 = -(Coefficient of x)/ (Coefficient of x2)
Product of zeroes = (-2) * 4 = -8 = -8/1 = Constant term/ (Coefficient of x2)
(ii) 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
The value of 4s2 – 4s + 1 is zero if 2s – 1 = 0
So, s = 1/2, 1/2
Therefore zeroes of 4s2 – 4s + 1 are 1/2 and 1/2
Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Coefficient of x)/ (Coefficient of x2)
Product of zeroes = 1/2 * 1/2 = 1/4 = Constant term/ (Coefficient of x2)
(iii) 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3)(3x + 1)
The value of 6x2 – 7x – 2 is zero if 2x – 3 = 0 or 3x + 1 = 0
So, x = -1/3, 3/2
Therefore zeroes of 6x2 – 7x – 3 are -1/3 and 3/2
Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(-7)/6 = -(Coefficient of x)/ (Coefficient of x2)
Product of zeroes = (-1/3) * 3/2 = -1/2 = -3/6 = Constant term/ (Coefficient of x2)
(iv) 4u2 + 8u
= 4u(u + 2)
The value of 4u2 + 8u is zero if 4u = 0 or u + 2 = 0
⇒ u = 0, -2
Therefore, the zeroes of 4u2 + 8u are 0 and -2.
Now, Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of x)/ (Coefficient of x2)
Product of zeroes = 0 * (-2) = 0 = 0/4 = Constant term/ (Coefficient of x2)
(v) t2 – 15
= t2 – (√15)2
= (t – √15)(t + √15)
The value of t2 – 15 is zero if t – √15 or t + √15 = 0
⇒ t = √15, -√15
Therefore, the zeroes of t2 – 15 are √15 and -√15.
Now, Sum of zeroes = -√15 + √15 = 0 = -(0)/1 = -(Coefficient of x)/ (Coefficient of x2)
Product of zeroes = (-√15) * √15 = -15 = -15/1 = Constant term/ (Coefficient of x2)
(vi) 3x2 – x – 4
= 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
The value of 6x2 – 7x – 2 is zero if 3x – 4 = 0 or x + 1 = 0
So, x = 4/3, -1
Therefore zeroes of 3x2 – x – 4 are 4/3 and -1
Sum of zeroes = 4/3 + (-1) = (4 – 3)/3 = 1/3 = -(-1)/3 = -(Coefficient of x)/ (Coefficient of x2)
Product of zeroes = 4/3 * 1 = -4/3 = Constant term/ (Coefficient of x2)
Mddle block 1
Question : 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, 1/3 (iii) 0, √5 (iv) 1, 1 (v) -1/4, 1/4 (vi) 4, 1
Answer :
(i) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.
Given, α + β = 1/4 = -b/a
and α * β = -1 = -4/4 = c/a
On comparing, we get
a = 4, b = -1, c = -4
Hence, the required quadratic polynomial is 4x2 – x – 4.
(ii) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.
Given, α + β = √2= 3√2/3 = -b/a
and α * β = 1/3 = c/a
On comparing, we get
a = 3, b = -3√2, c = 1
Hence, the required quadratic polynomial is 3x2 – 3√2x + 1.
(iii) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.
Given, α + β = 0 = 0/1 = -b/a
and α * β = √5 = √5/1 = c/a
On comparing, we get
a = 1, b = 0, c = √5
Hence, the required quadratic polynomial is x2 – 0*x + √5 = x2 + √5.
(iv) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.
Given, α + β = 1 = 1/1 = -b/a
and α * β = 1 = 1/1 = c/a
On comparing, we get
a = 1, b = -1, c = 1
Hence, the required quadratic polynomial is x2 – x + 1.
(v) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.
Given, α + β = -1/4 = -b/a
and α * β = 1/4 = c/a
On comparing, we get
a = 4, b = 1, c = 1
Hence, the required quadratic polynomial is 4x2 + x + 1.
(vi) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.
Given, α + β = 4 = 4/1 = -b/a
and α * β = 1 = 1/1 = c/a
On comparing, we get
a = 1, b = -4, c = 1
Hence, the required quadratic polynomial is x2 – 4x + 1.
