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SEBA Class 10 Maths Chapter 2 Polynomials Exercise 2.2 Solutions (New Syllabus 2026-27)
SEBA Class 10 Maths Chapter 2 Exercise 2.2 Solutions – New Course
Welcome to the ultimate guide for SEBA Class 10 Mathematics Chapter 2 (Polynomials) Exercise 2.2. If you are a Class 10 student under the Board of Secondary Education, Assam (SEBA), preparing for your upcoming Board Exams, you’re in the right place.
Below, you will find the complete, step-by-step solutions for all the questions in Ex 2.2, meticulously solved according to the latest SEBA New Course Syllabus for the academic session 2026-27.
EXERCISE 2.2
Q.1 Find the Zeroes of the Following Quadratic Polynomials and Verify the Relationship between the Zeroes and the Coefficients
General Formula for Verification
For a quadratic polynomial:
ax² + bx + c
Sum of zeroes (α + β) = -b/a
Product of zeroes (α × β) = c/a
(i) x² – 2x – 8
To Find the Zeroes
Let,
p(x) = x² – 2x – 8 = 0
x² – 4x + 2x – 8 = 0
x(x – 4) + 2(x – 4) = 0
(x – 4)(x + 2) = 0
x = 4 or x = -2
Zeroes
α = 4, β = -2
Verification
a = 1, b = -2, c = -8
Sum of zeroes
α + β = 4 + (-2) = 2
-b/a = -(-2)/1 = 2
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = 4 × (-2) = -8
c/a = -8/1 = -8
Therefore,
α × β = c/a (Verified)
(ii) 4s² – 4s + 1
To Find the Zeroes
Let,
p(s) = 4s² – 4s + 1 = 0
4s² – 2s – 2s + 1 = 0
2s(2s – 1) – 1(2s – 1) = 0
(2s – 1)(2s – 1) = 0
s = 1/2, 1/2
Zeroes
α = 1/2, β = 1/2
Verification
a = 4, b = -4, c = 1
Sum of zeroes
α + β = 1/2 + 1/2 = 1
-b/a = -(-4)/4 = 1
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = (1/2) × (1/2) = 1/4
c/a = 1/4
Therefore,
α × β = c/a (Verified)
(iii) 6x² – 3 – 7x
To Find the Zeroes
Rearranging the terms,
p(x) = 6x² – 7x – 3 = 0
6x² – 9x + 2x – 3 = 0
3x(2x – 3) + 1(2x – 3) = 0
(2x – 3)(3x + 1) = 0
x = 3/2 or x = -1/3
Zeroes
α = 3/2, β = -1/3
Verification
a = 6, b = -7, c = -3
Sum of zeroes
α + β = 3/2 + (-1/3)
= (9 – 2)/6
= 7/6
-b/a = -(-7)/6
= 7/6
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = (3/2) × (-1/3)
= -3/6
= -1/2
c/a = -3/6
= -1/2
Therefore,
α × β = c/a (Verified)
(iv) 4u² + 8u
To Find the Zeroes
Let,
p(u) = 4u² + 8u = 0
4u(u + 2) = 0
4u = 0 or u + 2 = 0
u = 0 or u = -2
Zeroes
α = 0, β = -2
Verification
a = 4, b = 8, c = 0
Sum of zeroes
α + β = 0 + (-2) = -2
-b/a = -8/4 = -2
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = 0 × (-2) = 0
c/a = 0/4 = 0
Therefore,
α × β = c/a (Verified)
(v) t² – 15
To Find the Zeroes
Let,
p(t) = t² – 15 = 0
t² – (√15)² = 0
(t – √15)(t + √15) = 0
t = √15 or t = -√15
Zeroes
α = √15, β = -√15
Verification
a = 1, b = 0, c = -15
Sum of zeroes
α + β = √15 + (-√15) = 0
-b/a = -0/1 = 0
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = (√15) × (-√15) = -15
c/a = -15/1 = -15
Therefore,
α × β = c/a (Verified)
(vi) 3x² – x – 4
To Find the Zeroes
Let,
p(x) = 3x² – x – 4 = 0
3x² – 4x + 3x – 4 = 0
x(3x – 4) + 1(3x – 4) = 0
(3x – 4)(x + 1) = 0
x = 4/3 or x = -1
Zeroes
α = 4/3, β = -1
Verification
a = 3, b = -1, c = -4
Sum of zeroes
α + β = 4/3 + (-1)
= (4 – 3)/3
= 1/3
-b/a = -(-1)/3
= 1/3
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = (4/3) × (-1)
= -4/3
c/a = -4/3
Therefore,
α × β = c/a (Verified)
(vii) x² + 7x + 12
To Find the Zeroes
Let,
p(x) = x² + 7x + 12 = 0
x² + 4x + 3x + 12 = 0
x(x + 4) + 3(x + 4) = 0
(x + 4)(x + 3) = 0
x = -4 or x = -3
Zeroes
α = -4, β = -3
Verification
a = 1, b = 7, c = 12
Sum of zeroes
α + β = (-4) + (-3) = -7
-b/a = -7/1 = -7
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = (-4) × (-3) = 12
c/a = 12/1 = 12
Therefore,
α × β = c/a (Verified)
(viii) x² – 4x + 3
To Find the Zeroes
Let,
p(x) = x² – 4x + 3 = 0
x² – 3x – x + 3 = 0
x(x – 3) – 1(x – 3) = 0
(x – 3)(x – 1) = 0
x = 3 or x = 1
Zeroes
α = 3, β = 1
Verification
a = 1, b = -4, c = 3
Sum of zeroes
α + β = 3 + 1 = 4
-b/a = -(-4)/1 = 4
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = 3 × 1 = 3
c/a = 3/1 = 3
Therefore,
α × β = c/a (Verified)
(ix) x² – 6x – 7
To Find the Zeroes
Let,
p(x) = x² – 6x – 7 = 0
x² – 7x + x – 7 = 0
x(x – 7) + 1(x – 7) = 0
(x – 7)(x + 1) = 0
x = 7 or x = -1
Zeroes
α = 7, β = -1
Verification
a = 1, b = -6, c = -7
Sum of zeroes
α + β = 7 + (-1) = 6
-b/a = -(-6)/1 = 6
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = 7 × (-1) = -7
c/a = -7/1 = -7
Therefore,
α × β = c/a (Verified)
(x) 2x² – 5x – 7
To Find the Zeroes
Let,
p(x) = 2x² – 5x – 7 = 0
2x² – 7x + 2x – 7 = 0
x(2x – 7) + 1(2x – 7) = 0
(2x – 7)(x + 1) = 0
2x – 7 = 0 or x + 1 = 0
x = 7/2 or x = -1
Zeroes
α = 7/2, β = -1
Verification
a = 2, b = -5, c = -7
Sum of zeroes
α + β = 7/2 + (-1)
= (7 – 2)/2
= 5/2
-b/a = -(-5)/2
= 5/2
Therefore,
α + β = -b/a (Verified)
Product of zeroes
α × β = (7/2) × (-1)
= -7/2
c/a = -7/2
Therefore,
α × β = c/a (Verified)
Q.2 Find a Quadratic Polynomial Each with the Given Numbers as the Sum and Product of Its Zeroes Respectively
General Formula Reference
The required quadratic polynomial is given by:
p(x) = k [x² – (Sum of zeroes)x + (Product of zeroes)]
Where k is any real non-zero constant used to clear fractions.
(i) Sum = 1/4, Product = -1
Sum of zeroes (α + β) = 1/4
Product of zeroes (α × β) = -1
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = k [x² – (1/4)x + (-1)]
p(x) = k [(4x² – x – 4) / 4]
Taking k = 4,
p(x) = 4x² – x – 4
(ii) Sum = √2, Product = 1/3
Sum of zeroes (α + β) = √2
Product of zeroes (α × β) = 1/3
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = k [x² – (√2)x + 1/3]
p(x) = k [(3x² – 3√2x + 1) / 3]
Taking k = 3,
p(x) = 3x² – 3√2x + 1
(iii) Sum = 0, Product = √5
Sum of zeroes (α + β) = 0
Product of zeroes (α × β) = √5
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = 1 [x² – 0x + √5]
p(x) = x² + √5
(iv) Sum = 1, Product = 1
Sum of zeroes (α + β) = 1
Product of zeroes (α × β) = 1
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = 1 [x² – 1x + 1]
p(x) = x² – x + 1
(v) Sum = -1/4, Product = 1/4
Sum of zeroes (α + β) = -1/4
Product of zeroes (α × β) = 1/4
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = k [x² – (-1/4)x + 1/4]
p(x) = k [x² + 1/4x + 1/4]
p(x) = k [(4x² + x + 1) / 4]
Taking k = 4,
p(x) = 4x² + x + 1
(vi) Sum = 4, Product = 1
Sum of zeroes (α + β) = 4
Product of zeroes (α × β) = 1
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = 1 [x² – 4x + 1]
p(x) = x² – 4x + 1
Final Answers of Q.2
(i) 4x² – x – 4
(ii) 3x² – 3√2x + 1
(iii) x² + √5
(iv) x² – x + 1
(v) 4x² + x + 1
(vi) x² – 4x + 1
Q.3 Find the Quadratic Polynomials Whose Zeroes Are
General Formula Reference
The required quadratic polynomial is given by:
p(x) = k [x² – (Sum of zeroes)x + (Product of zeroes)]
Where k is any real non-zero constant used to clear fractions.
(i) Zeroes = -4 and 3/2
Let the zeroes be:
α = -4 and β = 3/2
Sum of zeroes (α + β)
= -4 + 3/2
= (-8 + 3)/2
= -5/2
Product of zeroes (α × β)
= -4 × 3/2
= -6
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = k [x² – (-5/2)x + (-6)]
p(x) = k [(2x² + 5x – 12) / 2]
Taking k = 2,
p(x) = 2x² + 5x – 12
(ii) Zeroes = 5 and 2
Let the zeroes be:
α = 5 and β = 2
Sum of zeroes (α + β)
= 5 + 2
= 7
Product of zeroes (α × β)
= 5 × 2
= 10
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = 1 [x² – 7x + 10]
p(x) = x² – 7x + 10
(iii) Zeroes = 1/3 and -1
Let the zeroes be:
α = 1/3 and β = -1
Sum of zeroes (α + β)
= 1/3 + (-1)
= (1 – 3)/3
= -2/3
Product of zeroes (α × β)
= 1/3 × (-1)
= -1/3
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = k [x² – (-2/3)x + (-1/3)]
p(x) = k [(3x² + 2x – 1) / 3]
Taking k = 3,
p(x) = 3x² + 2x – 1
(iv) Zeroes = 3/2 and -2
Let the zeroes be:
α = 3/2 and β = -2
Sum of zeroes (α + β)
= 3/2 + (-2)
= (3 – 4)/2
= -1/2
Product of zeroes (α × β)
= 3/2 × (-2)
= -3
Required Polynomial
p(x) = k [x² – (α + β)x + (α × β)]
p(x) = k [x² – (-1/2)x + (-3)]
p(x) = k [(2x² + x – 6) / 2]
Taking k = 2,
p(x) = 2x² + x – 6
Final Answers of Q.3
(i) 2x² + 5x – 12
(ii) x² – 7x + 10
(iii) 3x² + 2x – 1
(iv) 2x² + x – 6
Q.4 If α and β are the Two Zeroes of the Polynomial x² – p(x + 1) + c Such That (α + 1)(β + 1) = 0, Then Find the Value of c
Options
(a) 1
(b) 2
(c) -1
(d) -2
Given Polynomial
x² – p(x + 1) + c
Rearranging into Standard Form ax² + bx + c
x² – px – p + c
x² – px + (c – p)
Comparing with ax² + bx + c, we get:
a = 1
b = -p
Constant term = c – p
Using Relationships of Zeroes
Sum of zeroes:
α + β = -b/a
= -(-p)/1
= p
Product of zeroes:
αβ = (Constant term)/a
= (c – p)/1
= c – p
Given Condition
(α + 1)(β + 1) = 0
αβ + α + β + 1 = 0
αβ + (α + β) + 1 = 0
Substituting the Values
(c – p) + p + 1 = 0
c – p + p + 1 = 0
c + 1 = 0
c = -1
Correct Option (c) -1
Q.5 If a – b, a and a + b are the Zeroes of the Polynomial p(x) = x³ – 3x² – 6x + 8, Then Find the Values of a and b
Statements
(i) a = 1
(ii) b = ±3
(iii) a = -1
(iv) b = a
Options
(a) (i) and (iv)
(b) (i) and (ii)
(c) (iii) and (ii)
(d) (iii) and (iv)
Given Cubic Polynomial
p(x) = x³ – 3x² – 6x + 8
Comparing with the standard form Ax³ + Bx² + Cx + D, we get:
A = 1
B = -3
C = -6
D = 8
The given zeroes are:
α = a – b
β = a
γ = a + b
Step 1: Using Sum of Zeroes
α + β + γ = -B/A
(a – b) + a + (a + b) = -(-3)/1
3a = 3
a = 1
Therefore, Statement (i) is correct.
Step 2: Using Product of Zeroes
α × β × γ = -D/A
(a – b) × a × (a + b) = -8/1
a(a² – b²) = -8
Substituting the value of a = 1:
1(1² – b²) = -8
1 – b² = -8
-b² = -8 – 1
-b² = -9
b² = 9
b = ±3
Therefore, Statement (ii) is correct.
Conclusion
The correct statements are (i) and (ii).
Correct Option(b) (i) and (ii)
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Q.6 Assertion and Reason
Assertion (A)
If α and β are the zeroes of the polynomial x² − 6x + p such that
(α + β)² − 2αβ = 40,
then the value of p is -1.
Reason (R)
The sum and the product of the zeroes of the quadratic polynomial ax² + bx + c are -b/a and c/a respectively.
Options
(a) Both Assertion (A) and Reason (R) are true and R is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but R is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Step 1: Check Reason (R)
For a quadratic polynomial ax² + bx + c:
Sum of zeroes = -b/a
Product of zeroes = c/a
Thus, Reason (R) is TRUE.
Step 2: Check Assertion (A)
Given polynomial:
x² – 6x + p
Comparing with the standard form, we have:
a = 1
b = -6
c = p
Sum of zeroes:
α + β = -(-6)/1
= 6
Product of zeroes:
αβ = p/1
= p
Using the Given Condition
(α + β)² – 2αβ = 40
Substituting the values:
(6)² – 2(p) = 40
36 – 2p = 40
-2p = 40 – 36
-2p = 4
p = 4/(-2)
p = -2
Since the calculated value of p is -2 but the assertion states it is -1, Assertion (A) is FALSE.
Conclusion
Assertion (A) is false but Reason (R) is true.
Correct Option (d) Assertion (A) is false but Reason (R) is true.
Q.7 Match the Polynomial in Column I with the Sum and Product of Its Zeroes in Column II
Column I
(A) x² – 7x + 12
(B) x² + 7x + 12
(C) x² – 7x – 12
Column II
(P) 7, -12
(Q) 7, 12
(R) -7, 12
Options
(a) A → Q, B → P, C → R
(b) A → Q, B → R, C → P
(c) A → P, B → Q, C → R
(d) A → P, B → R, C → Q
Formula Reference
Sum of zeroes = -(Coefficient of x) / (Coefficient of x²)
Product of zeroes = Constant term / (Coefficient of x²)
Step 1: Solve Each Polynomial
(A) x² – 7x + 12
Sum of zeroes = -(-7)/1 = 7
Product of zeroes = 12/1 = 12
Match: (Q)
(B) x² + 7x + 12
Sum of zeroes = -7/1 = -7
Product of zeroes = 12/1 = 12
Match: (R)
(C) x² – 7x – 12
Sum of zeroes = -(-7)/1 = 7
Product of zeroes = -12/1 = -12
Match: (P)
Step 2: Final Matching
A → Q
B → R
C → P
Correct Option (b)
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Q.8 Assertion and Reason
Statement (i)
The graph of a linear polynomial is a straight line.
Statement (ii)
A polynomial cannot have a variable in the denominator.
Options
(a) Both (i) and (ii) are true
(b) Both (i) and (ii) are false
(c) (i) is true but (ii) is false
(d) (i) is false but (ii) is true
Step 1: Check Statement (i)
A linear polynomial is of the form:
y = ax + b
Its graph is always a straight line.
Therefore, Statement (i) is TRUE.
Step 2: Check Statement (ii)
A polynomial is defined such that:
All exponents of variables must be non-negative integers.
If a variable appears in the denominator, for example:
1/x = x⁻¹
The exponent becomes negative, so it is NOT a polynomial.
Therefore, Statement (ii) is TRUE.
Conclusion
Both (i) and (ii) are true.
Correct Option (a)
Q.9 If p² = 32/50, then the value of p is
Options
(i) An integer
(ii) A rational number
(iii) An irrational number
(iv) A real number
Choose the correct option:
(a) Both (ii) and (iv) are true
(b) Both (i) and (iv) are true
(c) (i) is true but (ii) is false
(d) (ii) is false but (iii) is true
Given Equation
p² = 32/50
Simplifying the fraction:
p² = 16/25
Taking Square Root
p = ±√(16/25)
p = ±4/5
Analysis of Value of p
p = ±4/5 is in the form a/b where b ≠ 0
So, p is:
(ii) A rational number
Also, every rational number is a real number
So, p is also:
(iv) A real number
Conclusion
Statements (ii) and (iv) are true
Correct Option (a) Both (ii) and (iv) are true
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Q.10 If One of the Zeroes of the Polynomial x³ + ax² + bx + c is -1, Then the Product of the Other Two Zeroes is
Options
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1
Given Cubic Polynomial
p(x) = x³ + ax² + bx + c
Let the zeroes be:
α, β, γ
Given one zero is:
α = -1
Step 1: Product of Zeroes Formula
α × β × γ = -c/1
(-1) × β × γ = -c
-βγ = -c
βγ = c (Equation 1)
Step 2: Use Condition α = -1
Since -1 is a zero:
p(-1) = 0
(-1)³ + a(-1)² + b(-1) + c = 0
-1 + a – b + c = 0
c = b – a + 1 (Equation 2)
Step 3: Substitute Value of c
From Equation 1 and Equation 2:
βγ = c
βγ = b – a + 1
Conclusion
Product of the other two zeroes = b – a + 1
Correct Option (a) b – a + 1
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Q.11 α and β are the Zeroes of the Quadratic Polynomial x² – 6x + a. If 3α + 2β = 20 Then Find the Value of a
Given Polynomial
x² – 6x + a
Comparing with standard form px² + qx + r:
p = 1, q = -6, r = a
Step 1: Using Sum of Zeroes
α + β = -q/p
α + β = -(-6)/1
α + β = 6
α = 6 – β (Equation 1)
Step 2: Using Given Condition
3α + 2β = 20 (Equation 2)
Substitute α = 6 – β:
3(6 – β) + 2β = 20
18 – 3β + 2β = 20
18 – β = 20
-β = 2
β = -2
Step 3: Find α
α = 6 – (-2)
α = 8
Step 4: Find Value of a
Product of zeroes:
α × β = a
8 × (-2) = a
a = -16
Correct Option (d) -16
Q.12 If α and β are the Zeroes of the Polynomial 2x² – 5x + 7, Then Find another Polynomial Whose Zeroes are 2α + 3β and 3α + 2β
Given Polynomial
2x² – 5x + 7
Comparing with standard form ax² + bx + c:
a = 2
b = -5
c = 7
Sum and Product of Zeroes
Sum of zeroes:
α + β = -b/a
= -(-5)/2
= 5/2
Product of zeroes:
αβ = c/a
= 7/2
New Zeroes
S₁ = 2α + 3β
S₂ = 3α + 2β
Step 1: Sum of New Zeroes
S₁ + S₂ = (2α + 3β) + (3α + 2β)
= 5α + 5β
= 5(α + β)
Substituting value:
= 5 × (5/2)
= 25/2
Step 2: Product of New Zeroes
S₁ × S₂ = (2α + 3β)(3α + 2β)
= 6α² + 4αβ + 9αβ + 6β²
= 6(α² + β²) + 13αβ
Now use identity:
α² + β² = (α + β)² – 2αβ
So,
S₁ × S₂ = 6[(α + β)² – 2αβ] + 13αβ
= 6(α + β)² – 12αβ + 13αβ
= 6(α + β)² + αβ
Substituting values:
= 6(5/2)² + 7/2
= 6(25/4) + 7/2
= 150/4 + 7/2
= 75/2 + 7/2
= 82/2
= 41
Step 3: Form the New Polynomial
Required polynomial:
p(x) = k [x² – (Sum)x + (Product)]
p(x) = k [x² – (25/2)x + 41]
p(x) = k [(2x² – 25x + 82) / 2]
Taking k = 2:
p(x) = 2x² – 25x + 82
Final Answer
2x² – 25x + 82
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Q.13 If α and β are the zeroes of the polynomial ax² + bx + c then find the values of the following: (i) α² + β² (ii) α² + αβ + β² (iii) α²β + αβ² (iv) α – β (v) α² + β² – αβ
Given Polynomial
ax² + bx + c
Sum and Product of Zeroes
α + β = -b/a
αβ = c/a
(i) α² + β²
Using identity:
α² + β² = (α + β)² – 2αβ
Substituting values:
= (-b/a)² – 2(c/a)
= b²/a² – 2c/a
= (b² – 2ac) / a²
(ii) α² + αβ + β²
Rearranging:
= (α² + β²) + αβ
Substituting:
= (b² – 2ac)/a² + c/a
= (b² – 2ac + ac) / a²
= (b² – ac) / a²
(iii) α²β + αβ²
Taking common:
= αβ(α + β)
Substituting values:
= (c/a)(-b/a)
= -bc / a²
(iv) α – β
Using identity:
(α – β)² = (α + β)² – 4αβ
So,
α – β = ±√[(α + β)² – 4αβ]
Substituting values:
= ±√[(-b/a)² – 4(c/a)]
= ±√[b²/a² – 4c/a]
= ±√[(b² – 4ac) / a²]
= ±√(b² – 4ac) / a
(v) α² + β² – αβ
Rearranging:
= (α² + β²) – αβ
Substituting:
= (b² – 2ac)/a² – c/a
= (b² – 2ac – ac) / a²
= (b² – 3ac) / a²
Final Answers Summary
(i) (b² – 2ac) / a²
(ii) (b² – ac) / a²
(iii) -bc / a²
(iv) ±√(b² – 4ac) / a
(v) (b² – 3ac) / a²
Q.14 If the sum and product of the two zeroes of the polynomial kx² + 2x + 3k are equal, then find the value of k.
Given Polynomial
kx² + 2x + 3k
Comparing with standard form ax² + bx + c:
a = k
b = 2
c = 3k
Step 1: Sum and Product of Zeroes
Sum of zeroes (α + β) = -b/a
= -2/k
Product of zeroes (α × β) = c/a
= 3k/k
= 3
Step 2: Given Condition
Sum of zeroes = Product of zeroes
-2/k = 3
Step 3: Solve for k
-2 = 3k
k = -2/3
Final Answer
k = -2/3
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Q.15 If α and β are the zeroes of the quadratic polynomial f(x) = x² – 5x + 4, then find the value of 1/α + 1/β – 2αβ
Given Polynomial
f(x) = x² − 5x + 4
Comparing with standard form ax² + bx + c:
a = 1
b = −5
c = 4
Step 1: Sum and Product of Zeroes
α + β = −b/a
= −(-5)/1
= 5
αβ = c/a
= 4
Step 2: Given Expression
1/α + 1/β − 2αβ
First simplify:
1/α + 1/β = (α + β) / αβ
So expression becomes:
(α + β)/αβ − 2αβ
Step 3: Substitute Values
= 5/4 − 2(4)
= 5/4 − 8
= (5 − 32)/4
= −27/4
Final Answer
−27/4
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Q.16 Find the cubic polynomial with the sum of the zeroes, sum of the product of its zeroes taken two at a time and product of its zeroes are 2, -7, -14 respectively.
Given Information
Let zeroes be α, β, γ
Sum of zeroes:
α + β + γ = 2
Sum of products of zeroes taken two at a time:
αβ + βγ + γα = -7
Product of zeroes:
αβγ = -14
Standard Formula
p(x) = k [x³ − (α + β + γ)x² + (αβ + βγ + γα)x − (αβγ)]
Step 1: Substitute Values
p(x) = k [x³ − (2)x² + (-7)x − (-14)]
p(x) = k [x³ − 2x² − 7x + 14]
Step 2: Take k = 1
Required polynomial:
x³ − 2x² − 7x + 14
Final Answer
x³ − 2x² − 7x + 14
Q.17 The adjacent figure showing a mathematical shape of a bridge with hanging wires. Answer the following questions.
(i) Name the shape of the hanging wire? (A) Linear (B) Spiral (C) Parabola (D) Ellipse
(ii) What will be the expression of the polynomial representing the figure? (a) y = ax + b (b) y = ax² + bx + c (c) y = ax³ + bx² + cx + d (d) y = ax⁴ + cx + d
(iii) Zeros of the polynomial can be expressed graphically. Number of zeroes of the polynomial is equal to number of points where the graph of polynomial (a) Intersects x-axis (b) Intersects y-axis (c) Intersects y-axis and x-axis (d) None of the above
(iv) The representation of hanging wire on the bridge whose sum of the zeroes is -3 and product of zeroes is 5 is (a) x³ − 3x − 5 (b) x² − 3x + 5 (c) x² + 3x − 5 (d) x² + 3x + 5
(v) Graph of a quadratic polynomial is (a) Straight line (b) Circle (c) Parabola (d) Any curve
(i) Name the Shape of the Hanging Wire
Options:
(A) Linear
(B) Spiral
(C) Parabola
(D) Ellipse
Explanation:
The hanging wires of a suspension bridge form a U-shaped curve.
This shape is mathematically called a parabola.
Correct Option:
(C) Parabola
(ii) Expression of the Polynomial Representing the Figure
Options:
(a) y = ax + b
(b) y = ax² + bx + c
(c) y = ax³ + bx² + cx + d
(d) y = ax⁴ + cx + d
Explanation:
Since the shape is a parabola, it represents a quadratic polynomial.
The standard form of a quadratic polynomial is:
y = ax² + bx + c
Correct Option:
(b) y = ax² + bx + c
(iii) Zeroes of the Polynomial (Graphical Meaning)
Options:
(a) Intersects x-axis
(b) Intersects y-axis
(c) Intersects y-axis and x-axis
(d) None of the above
Explanation:
The zeroes of a polynomial are the x-coordinates of the points where the graph touches or cuts the x-axis.
Correct Option:
(a) Intersects x-axis
(iv) Polynomial with Sum of Zeroes = -3 and Product = 5
Formula:
p(x) = x² – (Sum of zeroes)x + (Product of zeroes)
Calculation:
p(x) = x² – (-3)x + 5
p(x) = x² + 3x + 5
Correct Option:
(d) x² + 3x + 5
(v) Graph of a Quadratic Polynomial
Options:
(a) Straight line
(b) Circle
(c) Parabola
(d) Any curve
Explanation:
The graph of any quadratic polynomial (degree 2) is always a parabola.
It may open upward or downward depending on the coefficient of x².
Correct Option:
(c) Parabola
Final Summary
(i) Parabola
(ii) y = ax² + bx + c
(iii) Intersects x-axis
(iv) x² + 3x + 5
(v) Parabola
Wrapping Up: Mastered SEBA Class 10 Maths Ex 2.2?
We hope this step-by-step solution for SEBA Class 10 Maths Chapter 2.2 helped clear all your doubts. Practicing these problems regularly will help you score maximum marks in your General Mathematics paper.
Frequently Asked Questions (FAQs)
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Yes, these answers are fully updated according to the revised SEBA Class 10 Maths textbook for the 2026-27 batch. - Where can I find the rest of the Chapter 2 solutions?
You can access the solutions for Exercise 2.1 and Exercise 2.3 right here on our website.
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