Exercise 2.2 NCERT
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) 6×2–3–7x (iv) 4u2+8u (v) t2–15 (vi) 3×2–x–4
To find the zeroes of quadratic polynomials, we can use the quadratic formula:
For a quadratic polynomial of the form ax^2 + bx + c = 0, the zeroes can be found using the quadratic formula:
x = (-b ± √(b^2 – 4ac)) / (2a)
Let’s apply this formula to find the zeroes of each quadratic polynomial and verify the relationship between the zeroes and the coefficients:
(i) x^2 – 2x – 8: Using the quadratic formula,
we have: x = (2 ± √((-2)^2 – 4(1)(-8))) / (2(1))
x = (2 ± √(4 + 32)) / 2
x = (2 ± √36) / 2
x = (2 ± 6) / 2
x = (2 + 6) / 2 or (2 – 6) / 2
x = 8 / 2 or -4 / 2
x = 4 or -2
The zeroes of the polynomial x^2 – 2x – 8 are 4 and -2.
(ii) 4s^2 – 4s + 1: Using the quadratic formula,
we have: s = (4 ± √((-4)^2 – 4(4)(1))) / (2(4))
s = (4 ± √(16 – 16)) / 8
s = (4 ± √0) / 8 s = (4 ± 0) / 8
s = 4 / 8 or -4 / 8
s = 1/2 or -1/2
The zeroes of the polynomial 4s^2 – 4s + 1 are 1/2 and -1/2.
(iii) 6x^2 – 3 – 7x: To find the zeroes,
we set the polynomial equal to zero: 6x^2 – 7x – 3 = 0
Using factoring, quadratic formula, or any other method,
we find the zeroes as x = 3/2 and x = -1/3.
(iv) 4u^2 + 8u: Setting the polynomial equal to zero: 4u^2 + 8u = 0 4u(u + 2) = 0
From this equation, we can see that the zeroes are u = 0 and u = -2.
(v) t^2 – 15: Setting the polynomial equal to zero: t^2 – 15 = 0
By taking the square root of both sides, we find the zeroes as t = √15 and t = -√15.
(vi) 3x^2 – x – 4: Using the quadratic formula: x = (1 ± √((-1)^2 – 4(3)(-4))) / (2(3))
x = (1 ± √(1 + 48)) / 6
x = (1 ± √49) / 6
x = (1 ± 7) / 6
x = 8 / 6 or -6 / 6
x = 4 / 3 or -1
The zeroes of the polynomial 3x^2 – x – 4 are 4/3 and -1.
To verify the relationship between the zeroes and the coefficients, we can observe that the sum of the zeroes is equal to the negative coefficient of x divided by the coefficient of x^2, and the product of the zeroes is equal to the constant term divided by
the coefficient of x^2.
For example, in the polynomial x^2 – 2x – 8, the sum of the zeroes is (-2)/1 = -2, and the product of the zeroes is (-8)/1 = -8, which matches the relationship.
You can perform similar calculations for the other polynomials to verify the relationship between the zeroes and the coefficients.
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4 , -1
Given the sum of the zeroes (α + β) as 1/4 and the product of the zeroes (α * β) as -1.
Using the relationship between the sum and product of the zeroes and the coefficients of the quadratic polynomial, we can write the equation as:
x^2 – (α + β)x + αβ = 0
Substituting the values, we have:
x^2 – (1/4)x + (-1) = 0
Simplifying, we get:
x^2 – (1/4)x – 1 = 0
Therefore, the quadratic polynomial with the sum of zeroes 1/4 and product of zeroes -1 is x^2 – (1/4)x – 1.
(ii)√2, 1/3
Solution:
Given the sum of the zeroes (α + β) as √2 and the product of the zeroes (α * β) as 1/3.
Using the relationship between the sum and product of the zeroes and the coefficients of the quadratic polynomial, we can write the equation as:
x^2 – (α + β)x + αβ = 0
Substituting the values, we have:
x^2 – (√2)x + (1/3) = 0
Therefore, the quadratic polynomial with the sum of zeroes √2 and 1/3 is x^2 – (√2)x + (1/3).
(iii) 0, √5
Solution:
Given the sum of the zeroes (α + β) as 0 and the product of the zeroes (α * β) as √5.
Using the relationship between the sum and product of the zeroes and the coefficients of the quadratic polynomial, we can write the equation as:
x^2 – (α + β)x + αβ = 0
Substituting the values, we have:
x^2 – (0)x + (√5) = 0
Simplifying, we get:
x^2 + √5 = 0
Therefore, the quadratic polynomial with the sum of zeroes 0 and product of zeroes √5 is x^2 + √5.
(iv) 1, 1
Solution:
Given the sum of the zeroes (α + β) as 1 and the product of the zeroes (α * β) as 1.
Using the relationship between the sum and product of the zeroes and the coefficients of the quadratic polynomial, we can write the equation as:
x^2 – (α + β)x + αβ = 0
Substituting the values, we have:
x^2 – (1)x + (1) = 0
Simplifying, we get:
x^2 – x + 1 = 0
Therefore, the quadratic polynomial with the sum of zeroes 1 and product of zeroes 1 is x^2 – x + 1.
(v) -1/4, 1/4
Solution:
Given the sum of the zeroes (α + β) as -1/4 and the product of the zeroes (α * β) as 1/4.
Using the relationship between the sum and product of the zeroes and the coefficients of the quadratic polynomial, we can write the equation as:
x^2 – (α + β)x + αβ = 0
Substituting the values, we have:
x^2 – (-1/4)x + (1/4) = 0
Multiplying through by 4 to eliminate the fractions, we get:
4x^2 + x + 1 = 0
Therefore, the quadratic polynomial with the sum of zeroes -1/4 and 1/4 is 4x^2 + x + 1.
(vi) 4, 1
solution:
Given the sum of the zeroes (α + β) as 4 and the product of the zeroes (α * β) as 1.
Using the relationship between the sum and product of the zeroes and the coefficients of the quadratic polynomial, we can write the equation as:
x^2 – (α + β)x + αβ = 0
Substituting the values, we have:
x^2 – (4)x + (1) = 0
Simplifying, we get:
x^2 – 4x + 1 = 0
Therefore, the quadratic polynomial with the sum of zeroes 4 and product of zeroes 1 is x^2 – 4x + 1.