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**Exercise 2.1**

**Choose the correct answer from the given four options in the following questions:****1. If one of the zeroes of the quadratic polynomial ( k–1) x^{2} + k x + 1 is –3, then the value of k is**

According to the question,

-3 is one of the zeros of quadratic polynomial (k-1)x

Substituting -3 in the given polynomial,

(k - 1)(-3)² + k(-3) + 1 = 0

(k - 1)9 + k(-3) + 1 = 0

9k - 9 -3k + 1 = 0

6k - 8 = 0

k = 8/6

Therefore, k =4 /3

Hence, option (A) is the correct answer.

Explanation:

Sum of zeroes, α+ β= -3+4 =1

Product of Zeroes, αβ = -3× 4 = -12

Therefore, the quadratic polynomial becomes,

x²- (sum of zeroes)x+(product of zeroes)

= x²- (α+ β)x+(αβ)

= x² – (1)x + (-12)

= x² – x -12

divide by 2, we get

= x²/2 – x/2 - 12/2

= x²/2 – x/2 -6

Hence,

Explanation:

According to the question,

x² + (a + 1)x + b

Given that, the zeroes of the polynomial = 2 and -3,

When x = 2

2² + (a + 1)(2) + b = 0

4 + 2a + 2 + b = 0

6 + 2a + b = 0

2a + b = -6 —– (1)

When x = -3,

(-3)² + (a + 1)(-3) + b = 0

9 – 3a - 3 + b = 0

6 – 3a + b = 0

-3a + b = -6 —– (2)

Subtracting equation (2) from (1)

2a + b – (-3a + b) = -6-(-6)

2a + b + 3a - b = -6 + 6

5a = 0

a = 0

Substituting the value of ‘a’ in equation (1), we get,

2a + b = -6

2(0) +b = -6

b = -6

Hence,

Explanation:

According to the question,

The zeroes of the polynomials = -2 and 5

We know that the polynomial is of the form,

p(x) = ax

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x

Sum of the zeroes = – b/a

– 2 + 5 = – b/a

3 = – b/a

b = – 3 and a = 1

Product of the zeroes = constant term ÷ coefficient of x

Product of zeroes = c/a

(- 2)5 = c/a

– 10 = c

Substituting the values of a, b and c in the polynomial p(x) = ax

We get, x

Therefore, we can conclude that x can take any value.

Hence,

Explanation:

According to the question,

We have the polynomial,

We know that,

Sum of product of roots of a cubic equation is given by c/a

It is given that one root = 0

Now, let the other roots be α, β

So, we get,

αβ + β(0) + (0)α = c/a

αβ = c/a

Hence the product of other two roots is c/a

Hence,

**Exercise 2.2**

**1. Answer the following and justify:****(i) Can x^{2} – 1 be the quotient on division of x^{6} + 2x^{3} + x – 1 by a polynomial in x of degree 5?**

No, x

Justification:

When a degree 6 polynomial is divided by degree 5 polynomial,

The quotient will be of degree 1.

Assume that (x

According to our assumption,

(degree 6 polynomial) = (x

= (degree 7 polynomial) + r(x) [ Since, (x

= (degree 7 polynomial)

From the above equation, it is clear that, our assumption is contradicted.

x

Hence Proved.

Degree of the polynomial px

Degree of the polynomial ax

Here, degree of px

Therefore, the quotient would be zero,

And the remainder would be the dividend = ax

We know that,

p(x)= g(x) × q(x)+r(x)

According to the question,

q(x) = 0

When q(x)=0, then r(x) is also = 0

So, now when we divide p(x) by g(x),

Then p(x) should be equal to zero

Hence, the relation between the degrees of p (x) and g (x) is the degree p(x)<degree g(x)

In order to divide p(x) by g(x)

We know that,

Degree of p(x) > degree of g(x)

or

Degree of p(x)= degree of g(x)

Therefore, we can say that,

The relation between the degrees of

A Quadratic Equation will have equal roots if it satisfies the condition:

b² – 4ac = 0

Given equation is x² + kx + k = 0

a = 1, b = k, x = k

Substituting in the equation we get,

k² – 4 ( 1 ) ( k ) = 0

k² – 4k = 0

k ( k – 4 ) = 0

k = 0 , k = 4

But in the question, it is given that k is greater than 1.

Hence the value of k is 4 if the equation has common roots.

Hence if the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.

**Exercise 2.3**

**Find the zeroes of the following polynomials by factorisation method.****1. 4 x^{2} – 3x – 1**

4

Splitting the middle term, we get,

4x

Taking the common factors out, we get,

4x(x - 1) +1(x - 1)

On grouping, we get,

(4x + 1)(x - 1)

So, the zeroes are,

4x + 1 = 0⇒ 4x = -1 ⇒x = (-1/4)

(x-1) = 0 ⇒ x = 1

Therefore, zeroes are (-1/4) and 1

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x

α + β = – b/a

1 – 1/4 = – (- 3)/4 = ¾

Product of the zeroes = constant term ÷ coefficient of x

α β = c/a

1(- 1/4) = – ¼

– 1/4 = – 1/4

3x

Splitting the middle term, we get,

3x

Taking the common factors out, we get,

3x(x + 2) -2(x + 2)

On grouping, we get,

(x + 2)(3x - 2)

So, the zeroes are,

x + 2 =0 ⇒ x = -2

3x - 2 = 0 ⇒ 3x = 2 ⇒ x = 2/3

Therefore, zeroes are (2/3) and -2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x

α + β = – b/a

– 2 + (2/3) = – (4)/3

= – 4/3 = – 4/3

Product of the zeroes = constant term ÷ coefficient of x

α β = c/a

Product of the zeroes = (- 2) (2/3) = – 4/3

5t

Splitting the middle term, we get,

5t

Taking the common factors out, we get,

5t (t + 1) +7(t + 1)

On grouping, we get,

(t + 1)(5t + 7)

So, the zeroes are,

t + 1 = 0 ⇒ y = -1

5t+7=0 ⇒ 5t=-7⇒t=-7/5

Therefore, zeroes are (-7/5) and -1

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x

α + β = – b/a

(- 1) + (- 7/5) = – (12)/5

= – 12/5 = – 12/5

Product of the zeroes = constant term ÷ coefficient of x

α β = c/a

(- 1)(- 7/5) = 7/5

7/5 = 7/5

Taking t common, we get,

t ( t

Splitting the middle term of the equation t

t( t

Taking the common factors out, we get,

t (t (t-5) +3(t-5)

On grouping, we get,

t (t+3)(t-5)

So, the zeroes are,

t=0

t+3=0 ⇒ t= -3

t -5=0 ⇒ t=5

Therefore, zeroes are 0, 5 and -3

Verification:

Sum of the zeroes = – (coefficient of x

α + β + γ = – b/a

(0) + (- 3) + (5) = – (- 2)/1

= 2 = 2

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x

αβ + βγ + αγ = c/a

(0)(- 3) + (- 3) (5) + (0) (5) = – 15/1

= – 15 = – 15

Product of all the zeroes = – (constant term) ÷ coefficient of x

αβγ = – d/a

(0)(- 3)(5) = 0

**Solution:**

2*x*^{2} +(7/2)*x *+3/4

The equation can also be written as,

8x^{2}+14x+3

Splitting the middle term, we get,

8x^{2}+12x+2x+3

Taking the common factors out, we get,

4x (2x+3) +1(2x+3)

On grouping, we get,

(4x+1)(2x+3)

So, the zeroes are,

4x+1=0 ⇒ x = -1/4

2x+3=0 ⇒ x = -3/2

Therefore, zeroes are -1/4 and -3/2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x^{2}

α + β = – b/a

(- 3/2) + (- 1/4) = – (7)/4

= – 7/4 = – 7/4

Product of the zeroes = constant term ÷ coefficient of x^{2}

α β = c/a

(- 3/2)(- 1/4) = (3/4)/2

3/8 = 3/8

**Exercise 2.4**

**1. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.****(i) (–8/3), 4/3****(ii) 21/8, 5/16****(iii) -2√3, -9****(iv) (-3/(2√5)), -½****Solution:**

(i) Sum of the zeroes = – 8/3

Product of the zeroes = 4/3

P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x^{2} – (-8x)/3 + 4/3

P(x)= 3x^{2} + 8x + 4

Using splitting the middle term method,

3x^{2} + 8x + 4 = 0

3x^{2} + (6x + 2x) + 4 = 0

3x^{2} + 6x + 2x + 4 = 0

3x(x + 2) + 2(x + 2) = 0

(x + 2)(3x + 2) = 0

⇒ x = -2, -2/3

(ii) Sum of the zeroes = 21/8

Product of the zeroes = 5/16

P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x^{2} – 21x/8 + 5/16

P(x)= 16x^{2} – 42x + 5

Using splitting the middle term method,

16x^{2} – 42x + 5 = 0

16x^{2} – (2x + 40x) + 5 = 0

16x^{2} – 2x – 40x + 5 = 0

2x (8x – 1) – 5(8x – 1) = 0

(8x – 1)(2x – 5) = 0

⇒ x = 1/8, 5/2

(iii) Sum of the zeroes = – 2√3

Product of the zeroes = – 9

P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)

Then, P(x) = x^{2} – (-2√3x) – 9

Using splitting the middle term method,

x^{2} + 2√3x – 9 = 0

x^{2} + (3√3x – √3x) – 9 = 0

x(x + 3√3) – √3(x + 3√3) = 0

(x – √3)(x + 3√3) = 0

⇒ x = √3, -3√3

(iv) Sum of the zeroes = -3/2√5x

Product of the zeroes = – ½

P(x) = x^{2} – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x^{2} -(-3/2√5x) – ½

P(x)= 2√5x^{2} + 3x – √5

Using splitting the middle term method,

2√5x^{2} + 3x – √5 = 0

2√5x^{2} + (5x – 2x) – √5 = 0

2√5x^{2} – 5x + 2x – √5 = 0

√5x (2x + √5) – (2x + √5) = 0

(2x + √5)(√5x – 1) = 0

⇒ x = 1/√5, -√5/2**2. Given that the zeroes of the cubic polynomial x ^{3} – 6x^{2} + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.**

Given that a, a + b, a + 2b are roots of given polynomial x³ - 6x² + 3x + 10

Sum of the roots ⇒ a + 2b + a + a + b = -coefficient of x²/ coefficient of x³

⇒ 3a+3b = -(-6)/1 = 6

⇒ 3(a+b) = 6

⇒ a+b = 2 ——— (1) b = 2-a

Product of roots ⇒ (a+2b)(a+b)a = -constant/coefficient of x³

⇒ (a+b+b)(a+b)a = -10/1

Substituting the value of a+b=2 in it

⇒ (2+b)(2)a = -10

⇒ (2+b)2a = -10

⇒ (2+2-a)2a = -10

⇒ (4-a)2a = -10

⇒ 4a-a² = -5

⇒ a²-4a-5 = 0

⇒ a²-5a+a-5 = 0

⇒ (a-5)(a+1) = 0

a-5 = 0 or a+1 = 0

a = 5 a = -1

a = 5, -1 in (1) a+b = 2

When a = 5, 5+b=2 ⇒ b=-3

a = -1, -1+b=2 ⇒ b= 3

∴ If a=5 then b= -3

or

If a= -1 then b=3

Given, √2 is one of the zero of the cubic polynomial.

Then, (x-√2) is one of the factor of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.

So, by dividing p(x) by x-√2

6x³+√2x²-10x-4√2= (x-√2) (6x² +7√2x + 4)

By splitting the middle term,

We get,

(x-√2) (6x² + 4√2x + 3√2x + 4)

= (x-√2) [ 2x(3x+2√2) + √2(3x+2√2)]

= (x-√2) (2x+√2) (3x+2√2)

To get the zeroes of p(x),

Substitute p(x)= 0

(x-√2) (2x+√2) (3x+2√2)= 0

x= √2 , x= -√2/2 ,x= -2√2/3

Hence, the other two zeroes of p(x) are -√2/2 and -2√2/3

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