Polynomials ⇒⇒ 2.2

Question 1 (i)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients : ${x^2 -2x - 8 }$

Solution :

To find the zeroes of the polynomial P(x ) = ${x^2 -2x - 8 }$,Simplify the quadratic polynomial by factorisation:

$${x^2 - 2x - 8 }$$

$$⇒ {x^2 -4x + 2x - 8 }$$

$$⇒ {x(x -4 )+ 2(x - 4)}$$

$$⇒ {(x - 4 )(x + 2)}$$

Setting each factor to zero, we find the zeroes:

⇒ x + 2 = 0 or x - 4 = 0

⇒ x = -2 or x = 4.

The zeroes of the given quadratic polynomial are -2 and 4 which means : ${\alpha = -2 }$ and ${\beta = 4 }$.

Compare with Standard form of the Quadratic Polynomials : $ ax^2 + bx + c $

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

Verification of the Relationship

Sum of Zeroes

$${\alpha } + {\beta } = {− b \over a}$$

$${ -2}+ {4 } = {− (-2) \over 1}$$

$$ 2 = 2 $$

The calculated sum and the formulaic sum are equal, which verifies the relationship.

Product of Zeroes

$${\alpha } × {\beta } = {c \over a}$$

$${ -2} × {4 } = { (-8) \over 1}$$

$$ -8 = -8 $$

The calculated product and the formulaic product are equal, which also verifies the relationship.

Hence, for zero of the polynomial = -2 and 4 .

Question 1 (ii)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients : ${4s^2 -4s + 1 }$

Solution :

To find the zeroes of the polynomial P(x ) = ${4s^2 -4s + 1 }$, Simplify the quadratic polynomial by factorisation:

$$⇒{{4s^2 -4s + 1 } }$$

This quadratic polynomial is a perfect square trinomial, which can be factored as follows:

$$⇒{4s^2 -2s - 2s + 1 }$$

$$⇒{2s(2s - 1 )- (2s - 1)}$$

$$⇒{(2s - 1)(2s - 1)}$$

Setting each factor to zero, we find the zeroes:

⇒ 2s - 1 = 0 or 2s - 1 = 0

⇒ s = ${1 \over 2}$ or s = ${1 \over 2}$.

The zeroes of the given quadratic polynomial are ${1 \over 2}$ and ${1 \over 2}$ which means : ${\alpha = {1 \over 2} }$ and ${\beta = {1 \over 2} }$.

Compare with Standard form of the Quadratic Polynomials : $ ax^2 + bx + c $

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

Verification of the Relationship

Sum of Zeroes

$${\alpha } + {\beta } = {− b \over a}$$

$${1 \over 2}+ {1 \over 2} = {− (-4) \over 4}$$

$$ 1 = 1 $$

The calculated sum and the formulaic sum are equal, which verifies the relationship.

Product of Zeroes

$${\alpha } × {\beta } = {c \over a}$$

$${1 \over 2} × {1 \over 2} = { 1 \over 4}$$

$$ { 1 \over 4} = { 1 \over 4} $$

The calculated product and the formulaic product are equal, which also verifies the relationship.

Hence, for zero of the polynomial = ${1 \over 2}$ and ${1 \over 2}$.

Question 1 (iii)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients : ${6x^2 -3 - 7x }$

Solution :

To find the zeroes of the polynomial P(x ) = ${6x^2 -3 - 7x }$ , Simplify the quadratic polynomial by factorisation:

$$⇒{6x^2 -3 - 7x } $$

we rewrite the polynomial in standard form

$$⇒{6x^2 - 7x -3 }$$

$$⇒{6x^2 -9x + 2x - 3 }$$

$$⇒{3x(2x -3 )+ 1(2x - 3)}$$

$$⇒{(2x - 3)(3x + 1)}$$

Setting each factor to zero, we find the zeroes:

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ x = ${3 \over 2} $ or x = ${-1 \over 3}$.

The zeroes of the given quadratic polynomial are ${3 \over 2} $ and ${-1 \over 3}$ which means : ${\alpha }= {3 \over 2} $ and ${\beta }= {-1 \over 3}$.

Compare with Standard form of the Quadratic Polynomials : $ ax^2 + bx + c $

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

Verification of the Relationship

Sum of Zeroes

$${\alpha } + {\beta } = {− b \over a}$$

$$⇒{3 \over 2}+ {-1 \over 3} = {− (-7) \over 6}$$

$$⇒ {7 \over 6} = {7 \over 6} $$

The calculated sum and the formulaic sum are equal, which verifies the relationship.

Product of Zeroes

$${\alpha } × {\beta } = {c \over a}$$

$$⇒{3 \over 2} × {-1 \over 3} = {− 3 \over 6}$$

$$⇒{− 3 \over 6} = {− 3 \over 6} $$

$$⇒{− 1 \over 2} = {− 1 \over 2} $$

The calculated product and the formulaic product are equal, which also verifies the relationship.

Hence, for zero of the polynomial = ${3 \over 2} $ and ${-1 \over 3}$.

Question 1 (iv)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients : ${4u^2 + 8u }$

Solution :

To find the zeroes of the polynomial P(x ) = ${4u^2 + 8u }$ , Simplify the quadratic polynomial by factorisation:

$$⇒{4u^2 + 8u} $$

$$⇒{4u(u + 2 )}$$

Setting each factor to zero, we find the zeroes:

⇒ 4u = 0 or u + 2 = 0

⇒ u = 0 or u = -2.

The zeroes of the given quadratic polynomial are 0 and -2, which means : ${\alpha }$ = 0 and ${\beta }$ = -2 .

Compare with Standard form of the Quadratic Polynomials : $ ax^2 + bx + c $

here a = 4 , b= 8 and c = 0

Verification of the Relationship

Sum of Zeroes

$${\alpha } + {\beta } = {− b \over a}$$

$$⇒ 0 + {(-2 )} = {− (8) \over 4}$$

$$⇒ -2 = -2 $$

The calculated sum and the formulaic sum are equal, which verifies the relationship.

Product of Zeroes

$${\alpha } × {\beta } = {c \over a}$$

$$⇒ 0 × {(-2 )} = {0 \over 4}$$

$$⇒ 0 = 0 $$

The calculated product and the formulaic product are equal, which also verifies the relationship.

Hence, for zero of the polynomial = 0 and -2.

Question 1 (v)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients : ${t^2 -15 }$

Solution :

To find the zeroes of the polynomial P(x ) = ${t^2 -15 }$ , Simplify the quadratic polynomial by factorisation:

$$⇒{t^2 -15} $$

Taking the square root of both sides, we get two possible values for t

$$⇒{(t - \sqrt{15} )}{(t + \sqrt{15} ) }$$

Setting each factor to zero, we find the zeroes:

⇒ $ {t - \sqrt {15}} $ = 0, or $ {t + \sqrt {15}} $ = 0

⇒ t = $ \sqrt{15}$ ,or t = $ -\sqrt{15}$.

The zeroes of the given quadratic polynomial are $ { \sqrt {15}} $ and $ {- \sqrt {15}} $ which means : $${\alpha }= \sqrt{15}$$ $${\beta }= - \sqrt{15}$$.

Compare with Standard form of the Quadratic Polynomials : $ ax^2 + bx + c $

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

Verification of the Relationship

Sum of Zeroes

$${\alpha } + {\beta } = {− b \over a}$$

$$⇒ \sqrt{15}- {\sqrt{15} } = {0 \over 1}$$

$$⇒ 0 = 0 $$

The calculated sum and the formulaic sum are equal, which verifies the relationship.

Product of Zeroes

$${\alpha } × {\beta } = {c \over a}$$

$$⇒ \sqrt{15} × - {\sqrt{15} } = {-15 \over 1}$$

$$⇒ -15 = -15 $$

The calculated product and the formulaic product are equal, which also verifies the relationship.

Hence, for zero of the polynomial = $ {- \sqrt {15}} $ and $ { \sqrt {15}} $.

Question 1 (vi)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients : ${3x^2 - x - 4 }$

Solution :

To find the zeroes of the polynomial P(x ) = ${3x^2 - x - 4 }$ , Simplify the quadratic polynomial by factorisation:

$$⇒{3x^2 - x - 4 } $$

$$⇒{3x^2 + 3x- 4x - 4}$$

$$⇒{3x( x + 1)- 4(x + 1)}$$

$$⇒ {(3x - 4 )(x + 1)}$$

Setting each factor to zero, we find the zeroes:

⇒ 3x − 4 = 0 or x + 1 = 0

⇒ x = ${4 \over 3} $ or x = $-1$.

The zeroes of the given quadratic polynomial are ${4 \over 3} $ and -1 which means : ${\alpha }= {4 \over 3} $ and ${\beta }= -1 $.

Compare with Standard form of the Quadratic Polynomials : $ ax^2 + bx + c $

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

Verification of the Relationship

Sum of Zeroes

$${\alpha } + {\beta } = {− b \over a}$$

$$⇒{4 \over 3}- {1 \over 1} = {− (-1) \over 3}$$

$$⇒ {1 \over 3 } = {1 \over 3 }$$

The calculated sum and the formulaic sum are equal, which verifies the relationship.

Product of Zeroes

$${\alpha } × {\beta } = {c \over a}$$

$$⇒ {4 \over 3} × {-1 \over 1} = {− 4 \over 3}$$

$$⇒ {-4 \over 3 }= {-4 \over 3 }$$

The calculated product and the formulaic product are equal, which also verifies the relationship.

Hence, for zero of the polynomial = ${4 \over 3} $ and -1.

Question 2 (i)

Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1/4 , -1

Solution :

In this question the value of Sum of zeroes and Product of zeroes is given.

$${\alpha } + {\beta } = {1 \over 4}$$

$${\alpha } × {\beta } = {-1}$$

To find a quadratic polynomial with a given sum and product of zeroes, we can use the formula :

$ x^2 $- (Sum of zeroes)x + (Product of zeroes)

Substituting these values into the formula:

$$(x^2)-({\alpha } + {\beta })x + {\alpha } × {\beta } $$

$$⇒ x^2-{1 \over 4}x - 1 $$

$$⇒ 4x^2 - x - 4 $$

A quadratic polynomial with the given sum and product of zeroes is ${4x^2 - x - 4} $.

Question 2 (ii)

Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. √2, 1/3

Solution :

In this question the value of Sum of zeroes and Product of zeroes is given.

$${\alpha } + {\beta } = {\sqrt 2 }$$

$${\alpha } × {\beta } = {1 \over 3} $$

To find a quadratic polynomial with a given sum and product of zeroes, we can use the formula :

$ x^2 $- (Sum of zeroes)x + (Product of zeroes)

Substituting these values into the formula:

$$(x^2)-({\alpha } + {\beta })x + {\alpha } × {\beta } $$

$$⇒ x^2- {\sqrt 2 }x + {1 \over 3} $$

$$⇒ 3x^2 - {3\sqrt 2 }x + 1 $$

A quadratic polynomial with the given sum and product of zeroes is ${ 3x^2 - {3\sqrt 2 }x + 1 } $.

Question 2 (iii)

Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 0,√5

Solution :

In this question the value of Sum of zeroes and Product of zeroes is given.

$${\alpha } + {\beta } = 0 $$

$${\alpha } × {\beta } = {\sqrt 5 } $$

To find a quadratic polynomial with a given sum and product of zeroes, we can use the formula :

$ x^2 $- (Sum of zeroes)x + (Product of zeroes)

Substituting these values into the formula:

$$(x^2)-({\alpha } + {\beta })x + {\alpha } × {\beta } $$

$$⇒ x^2-{0}x + {\sqrt 5 } $$

$$⇒ x^2 + {\sqrt 5 } $$

A quadratic polynomial with the given sum and product of zeroes is ${x^2 + {\sqrt 5 } } $.

Question 2 (iv)

Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1, 1

Solution :

In this question the value of Sum of zeroes and Product of zeroes is given.

$${\alpha } + {\beta } = 1 $$

$${\alpha } × {\beta } = 1 $$

To find a quadratic polynomial with a given sum and product of zeroes, we can use the formula :

$ x^2 $- (Sum of zeroes)x + (Product of zeroes)

Substituting these values into the formula:

$$(x^2)-({\alpha } + {\beta })x + {\alpha } × {\beta } $$

$$⇒ x^2- {1 }x + {1} $$

$$⇒ x^2 - x + 1 $$

A quadratic polynomial with the given sum and product of zeroes is ${ x^2 - x + 1 } $.

Question 2 (v)

Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. -1/4, 1/4

Solution :

In this question the value of Sum of zeroes and Product of zeroes is given.

$${\alpha } + {\beta } = {-1 \over 4} $$

$${\alpha } × {\beta } = { 1 \over 4} $$

To find a quadratic polynomial with a given sum and product of zeroes, we can use the formula :

$ x^2 $- (Sum of zeroes)x + (Product of zeroes)

Substituting these values into the formula:

$$(x^2)-({\alpha } + {\beta })x + {\alpha } × {\beta } $$

$$⇒ x^2 +{1 \over 4}x + {1 \over 4} $$

$$ 4x^2 + x + 1 $$

A quadratic polynomial with the given sum and product of zeroes is ${ 4x^2 + x + 1 } $.

Question 2 (v)

Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 4, 1

Solution :

In this question the value of Sum of zeroes and Product of zeroes is given.

$${\alpha } + {\beta } = 4 $$

$${\alpha } × {\beta } = 1 $$

To find a quadratic polynomial with a given sum and product of zeroes, we can use the formula :

$ x^2 $- (Sum of zeroes)x + (Product of zeroes)

Substituting these values into the formula:

$$(x^2)-({\alpha } + {\beta })x + {\alpha } × {\beta } $$

$$⇒ x^2-{4}x + {1 } $$

A quadratic polynomial with the given sum and product of zeroes is ${ x^2 - 4x + 1} $.