class 10 maths chapter 2 exercise 2.2
Quadratic Polynomials – Middle Term Splitting Method
Question 1: Find the zeroes and verify relationships
(i) x² – 2x – 8
Step 1: Factorize using middle term splitting
x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2)(x – 4)
Step 2: Find zeroes
(x + 2)(x – 4) = 0 ⇒ x = -2 or x = 4
Verification:
Sum of zeroes = -2 + 4 = 2
Compare with -b/a = -(-2)/1 = 2 ✓
Product of zeroes = (-2) × 4 = -8
Compare with c/a = -8/1 = -8 ✓
Zeroes: -2, 4
(ii) 4s² – 4s + 1
Step 1: Factorize
4s² – 4s + 1 = 4s² – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
Step 2: Find zeroes
(2s – 1)² = 0 ⇒ s = ½ (double root)
Verification:
Sum of zeroes = ½ + ½ = 1
Compare with -b/a = -(-4)/4 = 1 ✓
Product of zeroes = ½ × ½ = ¼
Compare with c/a = 1/4 = ¼ ✓
Zeroes: ½ (double root)
(iii) 6x² – 7x – 3 [Rearranged from original]
Step 1: Factorize
6x² – 7x – 3 = 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
Step 2: Find zeroes
(3x + 1)(2x – 3) = 0 ⇒ x = -⅓ or x = 3/2
Verification:
Sum of zeroes = -⅓ + 3/2 = 7/6
Compare with -b/a = -(-7)/6 = 7/6 ✓
Product of zeroes = (-⅓) × (3/2) = -½
Compare with c/a = -3/6 = -½ ✓
Zeroes: -⅓, 3/2
(iv) 4u² + 8u
Step 1: Factorize
4u² + 8u = 4u(u + 2)
Step 2: Find zeroes
4u(u + 2) = 0 ⇒ u = 0 or u = -2
Verification:
Sum of zeroes = 0 + (-2) = -2
Compare with -b/a = -8/4 = -2 ✓
Product of zeroes = 0 × (-2) = 0
Compare with c/a = 0/4 = 0 ✓
Zeroes: 0, -2
(v) t² – 15
Step 1: Factorize (difference of squares)
t² – 15 = (t – √15)(t + √15)
Step 2: Find zeroes
(t – √15)(t + √15) = 0 ⇒ t = √15 or t = -√15
Verification:
Sum of zeroes = √15 + (-√15) = 0
Compare with -b/a = -0/1 = 0 ✓
Product of zeroes = √15 × (-√15) = -15
Compare with c/a = -15/1 = -15 ✓
Zeroes: √15, -√15
(vi) 3x² – x – 4
Step 1: Factorize
3x² – x – 4 = 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (3x – 4)(x + 1)
Step 2: Find zeroes
(3x – 4)(x + 1) = 0 ⇒ x = 4/3 or x = -1
Verification:
Sum of zeroes = 4/3 + (-1) = 1/3
Compare with -b/a = -(-1)/3 = 1/3 ✓
Product of zeroes = (4/3) × (-1) = -4/3
Compare with c/a = -4/3 = -4/3 ✓
Zeroes: 4/3, -1
Question 2: Find quadratic polynomials
(i) Sum = ¼, Product = -1
Quadratic polynomial formula: x² – (sum)x + product
= x² – (¼)x + (-1)
= x² – ¼x – 1
To eliminate fraction, multiply by 4:
4x² – x – 4
Polynomial: 4x² – x – 4
Question 2: Find quadratic polynomials (Complete Solutions)
(ii) Sum = √2, Product = ⅓
Using standard form: x² – (sum)x + product
= x² – √2x + ⅓
To eliminate fraction, multiply by 3:
3x² – 3√2x + 1
Verification:
For polynomial 3x² – 3√2x + 1:
Sum of roots = -(-3√2)/3 = √2 ✓
Product of roots = 1/3 ✓
Polynomial: 3x² – 3√2x + 1
(iii) Sum = 0, Product = √5
Standard form: x² – (sum)x + product
= x² – 0x + √5
= x² + √5
Verification:
For polynomial x² + √5:
Sum of roots = -0/1 = 0 ✓
Product of roots = √5/1 = √5 ✓
Polynomial: x² + √5
(iv) Sum = 1, Product = 1
Standard form: x² – (sum)x + product
= x² – 1x + 1
= x² – x + 1
Verification:
For polynomial x² – x + 1:
Sum of roots = -(-1)/1 = 1 ✓
Product of roots = 1/1 = 1 ✓
Polynomial: x² – x + 1
(v) Sum = -¼, Product = ¼
Standard form: x² – (sum)x + product
= x² – (-¼)x + ¼
= x² + ¼x + ¼
To eliminate fractions, multiply by 4:
4x² + x + 1
Verification:
For polynomial 4x² + x + 1:
Sum of roots = -1/4 = -¼ ✓
Product of roots = 1/4 = ¼ ✓
Polynomial: 4x² + x + 1
(vi) Sum = 4, Product = 1
Standard form: x² – (sum)x + product
= x² – 4x + 1
Verification:
For polynomial x² – 4x + 1:
Sum of roots = -(-4)/1 = 4 ✓
Product of roots = 1/1 = 1 ✓
Polynomial: x² – 4x + 1
Key Concept Summary
Standard Form of Quadratic Polynomial:
x² – (sum of zeroes)x + (product of zeroes)
Steps to Find Polynomial:
- Write standard form with given sum and product
- Simplify the expression
- Eliminate fractions by multiplying with LCD if needed
- Verify using -b/a for sum and c/a for product
Note: The polynomial k[x² – (sum)x + product] where k is any non-zero constant will satisfy the conditions.
Additional RD Sharma Questions
1. Find zeroes of x² + 5x + 6 using middle term splitting
x² + 5x + 6 = x² + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 2)(x + 3)
Zeroes: -2, -3
2. Form quadratic polynomial with zeroes 3 and -2
Sum = 1, Product = -6 ⇒ x² – x – 6
3. Find k if sum of zeroes of 2x² – kx + 3 is 4
Sum = k/2 = 4 ⇒ k = 8
4. Find quadratic polynomial whose zeroes are squares of x² – 5x + 6
Original zeroes: 2, 3 ⇒ New zeroes: 4, 9
Sum = 13, Product = 36 ⇒ x² – 13x + 36
5. If one zero of x² – 6x + k is double the other, find k
Let zeroes be α, 2α ⇒ Sum = 3α = 6 ⇒ α = 2
Product = 2α² = k ⇒ k = 8