145-NCERT-10-2-Polynomials - Ex-2.2

NCERT

10th Mathematics

Exercise 2.2

Topic: 2 Polynomials

Click here for ⇨ NCERT-10-2-Polynomials - Ex-2.1

EXERCISE 2.2

Q1. 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

Explanation:

1) Let ax² + bx + c be the quadratic polynomial. Let us denote it as p(x).

2) That is, p(x) = ax² + bx + c.

3) Zeroes of p(x) can be found out by taking p(x) = 0. That is ax² + bx + c = 0.

4) To get zeroes of p(x), we will solve the equation ax² + bx + c = 0.

5) We know that if 𝝰 and 𝛃 be zeroes of the quadratic equation 

ax² + bx + c = 0, then, 

sum of the zeroes (𝝰 + 𝛃) = - (b/a) = - (Coefficient of x)/Coefficient of x²)

product of the zeroes (𝝰 x 𝛃) = (c/a) = (Constant term)/Coefficient of x²)

Solution:

(i) x² – 2x – 8 

1) Quadratic polynomial = x² – 2x – 8                     

here last term is 8 and its sign is minus, factorise 8 in such a way

that their difference will be - 2.

- 8 = (- 4) x (2) (- 4 + 2 = - 2).

= x² – 4x + 2x – 8

= (x² – 4x) + (2x – 8)

= x(x – 4) + 2(x – 4)

= (x – 4)(x + 2)

2) So x² – 2x – 8 = (x – 4)(x + 2)

3) The value of p(x) is 0 when (x – 4) = 0 or (x + 2) = 0.

4) So, p(x) is 0 when x = 4 or x = - 2.

5) Therefore zeroes of x² – 2x – 8 are 4 and - 2. i.e 𝝰 = 4 and 𝛃 = - 2,

a = 1, b = - 2 and c = - 8

6) Now we will verify the relationship between the zeroes and the coefficients.

7) (𝝰 + 𝛃) = - (b/a)

LHS = (𝝰 + 𝛃)

= (4 + (- 2))

= (2) ------------------equation 1

RHS = - (b/a)

= - (- 2)/1

= (2) ------------------equation 2

From equation 1 and equation 2, we have LHS = RHS.

8) Similarly, we will prove (𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

= (4 x (- 2))

= (- 8) ------------------equation 3

RHS = (c/a)

= (- 8)/1

= (- 8) ------------------equation 4

From equation 3 and equation 4, we have LHS = RHS.

9) Hence verified.

(ii) 4s² – 4s + 1

1) Quadratic polynomial = 4s² – 4s + 1

here last term is 1 and its sign is plus, factorise 1 & 4 in such a way that their addition will be - 4. 

4 = (- 2) x (- 2) (- 2 - 2 = - 4).

= 4s² – 2s - 2s + 1

= (4s² – 2s) - (2s - 1)

= 2s(2s – 1) - (2s - 1)

= (2s – 1)(2s – 1)

2) So 4s² – 4s + 1 = (2s – 1)(2s – 1)

3) The value of p(s) is 0 when (2s – 1) = 0 or (2s – 1) = 0.

4) So, p(s) is 0 when s = 4 or s = - 2.

5) Therefore zeroes of 4s² – 4s + 1 are 1/2 and 1/2. i.e 𝝰 = 1/2 and 𝛃 = 1/2,

a = 4, b = - 4 and c = 1

6) Now we will verify the relationship between the zeroes and the coefficients.

7) (𝝰 + 𝛃) = - (b/a)

LHS = (𝝰 + 𝛃)

= [(1/2) + (1/2)]

= (1) ------------------equation 1

RHS = - (b/a)

= - (- 4)/4

= (1) ------------------equation 2

From equation 1 and equation 2, we have LHS = RHS.

8) Similarly, we will prove (𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

= [(1/2) x (1/2)]

= (1/4) ------------------equation 3

RHS = (c/a)

= (1)/4

= (1/4) ------------------equation 4

From equation 3 and equation 4, we have LHS = RHS.

9) Hence verified.

(iii) 6x² – 3 – 7x

1) Quadratic polynomial = 6x² – 7x - 3

here last term is 3 and its sign is minus, factorise 6 & 3 in such a way that their difference will be - 7. 

6 x (- 3) = (2) x (- 3 x 3) (2 - 9 = - 7).

= 6x² + 2x - 9x - 3

= (6x² + 2x) - (9x + 3)

= 2x(3x + 1) - 3(3x + 1)

= (3x + 1)(2x – 3)

2) So 6x² – 7x - 3 = (3x + 1)(2x – 3)

3) The value of p(x) is 0 when (3x + 1) = 0 or (2x – 3) = 0.

4) So, p(x) is 0 when x = -1/3 or x = 3/2.

5) Therefore zeroes of 6x² – 7x - 3 are -1/3 and 3/2. i.e 𝝰 = -1/3 and 𝛃 = 3/2,

a = 6, b = - 7 and c = -3

6) Now we will verify the relationship between the zeroes and the coefficients.

7) (𝝰 + 𝛃) = - (b/a)

LHS = (𝝰 + 𝛃)

= [(-1/3) + (3/2)]

= (7/6) ------------------equation 1

RHS = - (b/a)

= - (- 7)/6

= (7/6) ------------------equation 2

From equation 1 and equation 2, we have LHS = RHS.

8) Similarly, we will prove (𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

= [(-1/3) x (3/2)]

= (-1/2) ------------------equation 3

RHS = (c/a)

= (-3)/6

= (-1/2) ------------------equation 4

From equation 3 and equation 4, we have LHS = RHS.

9) Hence verified.

(iv) 4u² + 8u

1) Quadratic polynomial = 4u² + 8u

here we can take 4u common directly.

= 4u(u + 2)

2) So 4u² + 8u = 4u(u + 2)

3) The value of p(u) is 0 when u = 0 or (u + 2) = 0.

4) So, p(u) is 0 when u = 0 or u = - 2.

5) Therefore zeroes of 4u² + 8u are 0 and - 2. i.e 𝝰 = 0 and 𝛃 = - 2,

a = 4, b = 8 and c = 0

6) Now we will verify the relationship between the zeroes and the coefficients.

7) (𝝰 + 𝛃) = - (b/a)

LHS = (𝝰 + 𝛃)

= [(0) + (- 2)]

= (- 2) ------------------equation 1

RHS = - (b/a)

= - (8)/4

= (- 2) ------------------equation 2

From equation 1 and equation 2, we have LHS = RHS.

8) Similarly, we will prove (𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

= [(0) x (- 2)]

= (0) ------------------equation 3

RHS = (c/a)

= (0)/4

= (0) ------------------equation 4

From equation 3 and equation 4, we have LHS = RHS.

9) Hence verified.

(v) t² – 15

1) Quadratic polynomial = t² - 15

here, use the formula

(a² - b²) = (a - b) (a + b) directly.

= (t + √15) (t - √15)

2) So t² - 15 = (t + √15) (t - √15)

3) The value of p(t) is 0 when (t - √15) = 0 or (t + √15) = 0.

4) So, p(t) is 0 when t = √15 or t = -√15.

5) Therefore zeroes of t² - 15 are √15 and -√15. i.e 𝝰 = √15 and 𝛃 = -√15,

a = 1, b = 0 and c = -15

6) Now we will verify the relationship between the zeroes and the coefficients.

7) (𝝰 + 𝛃) = - (b/a)

LHS = (𝝰 + 𝛃)

= [(√15) + (-√15)]

= (0) ------------------equation 1

RHS = - (b/a)

= - (0)/1

= (0) ------------------equation 2

From equation 1 and equation 2, we have LHS = RHS.

8) Similarly, we will prove (𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

= [(√15) x (-√15)]

= (-15) ------------------equation 3

RHS = (c/a)

= (-15)/1

= (-15) ------------------equation 4

From equation 3 and equation 4, we have LHS = RHS.

9) Hence verified.

(vi) 3x² – x – 4

1) Quadratic polynomial = 3x² – x – 4                     

here last term is 4 and its sign is minus, factorise 3 x 4 in such a way

that their difference will be - 1.

3 x (- 4) = (3) x (- 4) (3 - 4 = - 1).

= 3x² + 3x - 4x – 4

= (3x² + 3x) - (4x + 4)

= 3x(x + 1) - 4(x + 1)

= (3x – 4)(x + 1)

2) So 3x² – x – 4 = (3x – 4)(x + 1)

3) The value of p(x) is 0 when (3x – 4) = 0 or (x + 1) = 0.

4) So, p(x) is 0 when x = 4/3 or x = - 1.

5) Therefore zeroes of 3x² – x – 4 are 4/3 and - 1. i.e 𝝰 = 4/3 and 𝛃 = - 1,

a = 3, b = - 1 and c = - 4

6) Now we will verify the relationship between the zeroes and the coefficients.

7) (𝝰 + 𝛃) = - (b/a)

LHS = (𝝰 + 𝛃)

= (4/3 + (- 1))

= (1/3) ------------------equation 1

RHS = - (b/a)

= - (- 1)/3

= (1/3) ------------------equation 2

From equation 1 and equation 2, we have LHS = RHS.

8) Similarly, we will prove (𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

= (4/3 x (- 1))

= (- 4/3) ------------------equation 3

RHS = (c/a)

= (- 4)/3

= (- 4/3) ------------------equation 4

From equation 3 and equation 4, we have LHS = RHS.

9) Hence verified.

Q2. 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

Explanation:

1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x).

2) So p(x) = (x - 𝝰)(x - 𝛃)

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃) 

Solution:

(i) 1/4, -1

1) Let 𝝰 and 𝛃 be the zeroes of p(x)

2) Here sum of zeroes is 1/4 and the product of zeroes is -1.

3) i.e. (𝝰 + 𝛃) = 1/4 and (𝝰 x 𝛃) = -1, so,

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃)

= x² – (1/4) x + (-1)

= 4x² – x + (-4)

p(x) = 4x² – x - 4

4) Quadratic polynomial is 4x² – x - 4.

(ii) √2, 1/3

1) Let 𝝰 and 𝛃 be the zeroes of p(x)

2) Here sum of zeroes is √2 and the product of zeroes is 1/3.

3) i.e. (𝝰 + 𝛃) = √2 and (𝝰 x 𝛃) = 1/3, so,

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃)

= x² – (√2) x + (1/3)

= x² – √2x + (1/3)

= 3x² – 3√2x + 1 

p(x) = 3x² – 3√2x + 1 

4) Quadratic polynomial is 3x² – 3√2x + 1.

(iii) 0, √5

1) Let 𝝰 and 𝛃 be the zeroes of p(x)

2) Here sum of zeroes is 0 and the product of zeroes is √5.

3) i.e. (𝝰 + 𝛃) = 0 and (𝝰 x 𝛃) = √5, so,

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃)

= x² – (0) x + (√5)

= x² – 0x + √5

p(x) = x² + √5

4) Quadratic polynomial is x² + √5.

(iv) 1, 1

1) Let 𝝰 and 𝛃 be the zeroes of p(x)

2) Here sum of zeroes is 1 and the product of zeroes is 1.

3) i.e. (𝝰 + 𝛃) = 1 and (𝝰 x 𝛃) = 1, so,

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃)

= x² – (1) x + (1)

= x² – x + 1

p(x) = x² – x + 1

4) Quadratic polynomial is x² – x + 1.

(v) -1/4, 1/4

1) Let 𝝰 and 𝛃 be the zeroes of p(x)

2) Here sum of zeroes is -1/4 and the product of zeroes is 1/4.

3) i.e. (𝝰 + 𝛃) = -1/4 and (𝝰 x 𝛃) = 1/4, so,

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃)

= x² – (-1/4) x + (1/4)

= 4x² + x + 1

p(x) = 4x² + x + 1

4) Quadratic polynomial is 4x² + x + 1.

(vi) 4, 1

1) Let 𝝰 and 𝛃 be the zeroes of p(x)

2) Here sum of zeroes is 4 and the product of zeroes is 1.

3) i.e. (𝝰 + 𝛃) = 4 and (𝝰 x 𝛃) = 1, so,

p(x) = x² – (𝝰 + 𝛃) x + (𝝰 x 𝛃)

= x² – (4) x + (1)

= x² – 4x + 1

p(x) = x² – 4x + 1

4) Quadratic polynomial is x² – 4x + 1.

Click here for ⇨ NCERT-10-2-Polynomials - Ex-2.3

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