Telangana SCERT Class 10 Maths Chapter 3 Polynomials Exercise 3.3 Maths Problems and Solution Here in this Post. Telangana SCERT Class 10 Maths Solution Chapter 3 Polynomials Exercise 3.3
Telangana SCERT Solution Class X (10) Maths Chapter 3 Polynomials Exercise 3.3
Exercise 3.3
(Q1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
=> Solution:
We have.
Let x2 – 2x – 8 = 0
To find the factor of – 8
-8
-4 +2
x2 – 4x + 2x – 8 = 0
x (x – 4) + 2 (x – 4) = 0
(x – 4) (x + 2) = 0
x-4 = 0 or x+2 = 0
x = 4 or x = 2
∴ Zeroes of x2 – 2x – 8 = 4 and – 2
* Verification the relationship between the zeroes and the coefficients:
Sum of the zeroes = -2 +4 = 2 = 2/1
= – (-2/1) =
= – (coefficient of x)/(coefficient of x2)
Product of zeroes = (-2) (4) = -8/1 = -8/1
= -8/1 = Constant term/ coefficient of x2
∴ verified the relationship between the zeroes and the coefficients.
(ii) 4s2 – 4s + 1
We have,
Let 4s2 – 4s + 1 = 0
[1 X coefficient of s2 = 1 X 4 => to find the factor of 4]
+4
-2 -2
(-2-2 = -4 middle term)
4s2 – 2s – 2s + 1 = 0
2s (2s – 1) – 1 (2s – 1) = 0
(2s – 1) (2s – 1) = 0
2s – 1 = 0 or 2s – 1 = 0
2s = 1 or 2s = 1
S = 1/2 or s = 1/2
∴ Zeroes of 4s2 – 4s + 1 = 1/2 and 1/2
* Verification the relationship between the zeroes and the coefficients:
* Sum of the zeroes = 1/2 + 1/2
= 1
[1 is written as +4/+4]
= (-4)/+4 = (coefficient or s/coefficient of s2
* Product of zeroes = 1/2 X 1/2 = 1/4
= Constant term/coefficient of s2
∴ verified the relationship bet∩ the zeroes and the coefficient.
(iii) 6x2 – 3 – 7x
=> Solution:
First we have to write in standard form.
6x2 – 7x – 3
We have,
Let 6x2 – 7x – 3 = 0
To find factor of 3X6 = 18
-18
-9 +2
6x2 – 9x + 2x – 3 = 0
3x (2x – 3) +1 (2x – 3) = 0
(2x – 3) (3x + 1) = 0
(2x – 3) = 0 or 3x + 1 = 0
2x = 3 or 3x = 1
X = 3/2 or x = -1/3
∴ Zeroes of 6x2 – 7x – 3 = 3/2 and -1/3
* Verification the relationship between the zeroes and the coefficient:
Sum of the zeroes = 3/2 + (-1/3)
* Sum of zeroes = -1/3 + 3/2
= -2+9/6
= 7/6
= – (-7)/6 = – (coefficient of x)/coefficient of x2
* Product of the = -1/3 X 3/2 zeroes
= = -3/6
= Constant term/coefficient of x2
∴ verified the relationship between the zeroes and the coefficient.
(iv) 4u2 + 8u
=> Solution:
We have,
Let 4u2 + 8u = 0
Here, there is no constant term.
So, we have take common term.
4u (u + 2) = 0
4u = 0 or u +2 = 0
U = 0 or u = -2
∴ Zeroes of 4u2 + 8u = 0 and -2
* Verification the relationship between the zeroes and the coefficient:
* Sum of the zeroes = 0 + (-2) = 0 – 2 = -2
= -8/4 = – (coefficient of u)/(coefficient of u2)
[We have to write -8/4 because the formula of sum of zeroes is – (coefficient of u)/(coefficient of u2) is 8/4 and -8/4 = -2]
* Product of the = 0 X -2 = 0
= 0/4
= constant term/coefficient of u2
[∵ constant term is zero = 0]
∴ verified the relationship between the zeroes and the coefficient.
(v) t2 – 15
=> Solution:
We have,
Let t2 – 15 = 0
T2 – (√15)2 = 0
[This is a2 – b2 form]
By using formula, a2 – b2 = (a + b) (a – b)
(t + √15) (t – √15) = 0
T + √15 = 0 or t – √15 = 0
T = -√15 or t = √15
∴ zeroes of t2 – 15 = -√15 and √15
* Verification the relationship between the zeroes and the coefficient:
* Sum of the zeroes = -√15 + √15 = 0
= 0/1 = – (coefficient of t)/(coefficient of t2)
[In given polynomial there is no coefficient of t term].
* Product of zeroes = (-√15) X (√15)
= -15/1 = constant term/coefficient of t2
∴ verified the relationship between the zeroes and the coefficient.
(vi) 3x2 – x – 4
=> Solution:
We have,
Let 3x2 – x – 4 = 0
To find the factor of 4X3 = 12
[Constant term X coefficient of x2]
-12
-4 +3
3x2 – 4x + 3x – 4 = 0
1x (3x – 4) +1 (3x – 4) = 0
(3x – 4) = 0 (x+1) = 0
3x = 4 or x = -1
X = 4/3 or x = -1
∴ Zeroes of 3x2 – x – 4 = 4/3 and -1
* Verification the relation between the zeroes and the coefficient:
* Sum of the zeroes = 4/3 – 1 = 4-3/3 = 1/3
= -(-1)/3 = – (coefficient of x)/(coefficient of x2)
* Product of the zeroes = -1 X 4/3
= -4/3
= constant term/coefficient of x2
∴ verified the relationship between the zeroes and the coefficient.
(Q2) Find the quadratic polynomial in each case with the given numbers as the sum and product of its zeroes respectively.
(i)
(ii)
(iii) 0, √5
= Solution:
Sum of the zeroes = (α + β) = 0
And product of the zeroes = (αβ) = √5
∴ required quadratic polynomial
= k [x2 – (α+ β) x + αβ] (k = constant)
= k [x2 – (0) x + √5]
= k [x2 + √5]
If k = 1 then Q.P = 1 [x2 + √5]
= x2 + √5
∴ the required Q.P. is x2 + √5
(iv) 1, 1
=> Solution:
Sum of the zeroes = (α + β) = 1 and product of the zeroes = (αβ) = 1
∴ Required quadratic polynomial
= k [x2 – (α+β) x +αβ] (k = constant)
Substitute (α+β) = 1 & (αβ) = 1
K [x2 – x + 1]
If k = 1 then Q.P. = 1 [x2 – x + 1]
∴ the required Q.P. is x2 – x + 1
(v)
(vi) 4, 1
=> Solution:
Sum of the zeroes = (α+β) = 4
And product of the zeroes = (αβ) = 1
∴ Required quadratic polynomial
= k [x2 – (α+β) x + (αβ)] (k = constant)
K [x2 – 4x + 1]
If k = 1 then Q.P. = 1 [x2 – 4x + 1]
= x2 – 4x + 1
∴ the required Q.P. is x2 – 4x + 1.
(Q3) Find that quadratic polynomial for the zeroes α, β given in each case.
(i) 2, 1
=> Given that α = 2 and β= -1
α + β= 2+ (-1) = 2-1 = 1 and
α.β = 2 X (-1) = -2
∴ The required quadratic polynomial
= k [x2 – (α + β)x + αβ]
Substitute the value of α+β & αβ.
= k [x2 – (1)x + (-2)] (k= constant)
= k [x2 – x – 2]
If k = 1 then Q.P.
= 1 [x2 – x =- 2]
= x2 – x – 2
∴ The required Q.P. is x2 – x – 2 where zeroes are 2 and -1
(ii) √3, -√3
=> Solution:
Given that α = √3 and β = √3
α+β = √3 + (-√3) = √3 -√3 = 0
and α.β = (√3) X (-√3) = -3
∴ Require quadratic polynomial =
K [x2 – (α+β)x + (αβ)] (k = constant)
= k [x2 – (0) x + (-3)]
= k [x2 – 0 – 3]
= k [x2 – 3]
If k = 1 then Q.P = 1 [x2 – 3]
= x2 – 3
∴ The required Q.P. is x2 – 3 Where zeroes are √3 & -√3
(iii) 1/4, -1
= Solution:
Given that α = 1/4 and β = -1
α+ β = 1/4 + (-1)
= 1/4 -1 = 1-4/4
= -3/4
and αβ = 1/4 X (-1)
= -1/4
∴ The required quadratic polynomial
= K [x2 – (α+β) x + (αβ)] (k = constant)
= k [x2 – (-3/4) x + (-1/4)]
= k [x2 + 3/4 x -1/4]
= k [4x2 +3x – 1/4]
(iv)
(Q4)
Here is your solution of Telangana SCERT Class 10 Math Chapter 3 Polynomials Exercise 3.3
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