EXERCISE-3

Find the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.

Example 1: p(x) = 2x² – 8x + 6.

Solution:

split the middle term ‘–8x’ as a sum of two terms, whose product is

6 × 2x² = 12 x². So, we write

2x² – 8x + 6 = 2x² – 6x – 2x + 6

= 2x(x – 3) – 2(x – 3)

= (2x – 2) (x – 3) = 2(x – 1) (x – 3)

p(x) = 2x² – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0,

i.e., when x = 1 or x = 3. So, the zeroes of 2x² – 8x + 6 are 1 and 3.

the sum of the zeroes = 1 + 3 = 4 =`-((-8)/2)=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = 1 × 3 = 3 =`6/2=(constant   term)/(coefficient  of  x^2)`

Example 2: p(x) = 3x² + 5x – 2.

Solution:

By splitting the middle term we see,

3x² + 5x – 2 = 3x² + 6x – x – 2

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

3x² + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0

i.e., when x =`1/3`or x = –2.

The zeroes of 3x² + 5x – 2 are `1/3` and -2.

Sum of its zeroes =`1/3`+ (-2) =`1/3-2 =–5/3=–(coefficient  of  x)/(coefficient  of  x^2)`

Product of its zeroes =`1/3 (-2) =(-2)/3=(constant   term)/(coefficient  of  x^2)`

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Do this:

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(i) p(x) = x² – x – 6

Solution:

split the middle term ‘–x’ as a sum of two terms, whose product is

–6 × x² = –6 x². So, we write

x² – x – 6 = x² – 3x + 2x – 6

= x(x – 3) + 2(x – 3)= (x – 3) (x + 2)

p(x) = x² – x – 6 is zero when x – 3 = 0 or x + 2 = 0,

i.e., when x = 3 or x = –2. So, the zeroes of x² – x – 6 are –2 and 3.

the sum of the zeroes = –2 + 3 = 1 =`-((-1)/1)=–(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = –2 × 3 = –6 =`(-6)/1=(constant   term)/(coefficient  of  x^2)`

(ii) p(x) = x² – 4x + 3

Solution:

split the middle term ‘–4x’ as a sum of two terms, whose product is

3 × x² = 3 x². So, we write

x² – 4x + 3 = x² – 3x – x + 3

= x(x – 3) – 1(x – 3)= (x – 3) (x – 1)

p(x) = x² – 4x + 3 is zero when x – 1 = 0 or x – 3 = 0,

i.e., when x = 1 or x = 3. So, the zeroes of x² – 4x + 3 are 1 and 3.

the sum of the zeroes = 1 + 3 = 4 =`-((-4)/1)=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = 1 × 3 = 3 =`3/1=(constant   term)/(coefficient  of  x^2)`

(iii) p(x) = x² – 4

Solution:

Let p(x) = 0

So, x² – 4 =x² -2²=0

(x-2)(x+2)=0 (`because` a²-b²=(a+b)(a-b))

(x-2)=0 or (x+2)=0

x=2 or x=-2

∴ The zeroes of P(x) are 2 and-2

the sum of the zeroes = 2 -2 = 0 =`0/1=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = -2 × 2 = -4 =`(-4)/1=(constant   term)/(coefficient  of  x^2)`

(iv) p(x) = x² + 2x + 1

Solution:

Let p(x) = 0

x² + 2x + 1 =x² + 2x + 1² =0

= (x+1)²=0 (`because` a²+2ab+b²=(a+b)²)

=x+1=0 `Rightarrow`x=-1

So, the zero of x² + 2x + 1 are -1 and -1

the sum of the zeroes = -1 -1 = -2 =`-2/1=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = -1 × -1 = 1 =`1/1=(constant   term)/(coefficient  of  x^2)`

Example-3: Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Solution :

We have x² + 7x + 10 = x²+2x+5x+10= (x + 2) (x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0,

i.e., when x = –2 or x = –5.

`therefore` The zeroes of x² + 7x + 10 are –2 and –5.

Now, sum of the zeroes= –2 + (–5) = –2 –5 =– (7) =`-7/1=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = –2 × (–5) = 10 =`10/1=(constant   term)/(coefficient  of  x^2)`

Example-4: Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.

Solution :

Recall the identity a² – b² = (a – b) (a + b). Using it, we can write:

x² – 3 = (x – `sqrt 3` ) (x + `sqrt 3` )

So, the value of x² – 3 is zero when x = `sqrt 3` or x = – `sqrt 3` .

`therefore` The zeroes of x² – 3 are `sqrt 3` and – `sqrt 3` .

Sum of the zeroes = `sqrt 3` + (–`sqrt 3` ) = 0 =-`(coefficient  of  x)/(coefficient  of  x^2)`

Product of zeroes = (`sqrt 3` ) × (–`sqrt 3` ) = – 3 =`(-3)/1 =(constant   term)/(coefficient  of  x^2)`

Example-5: Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.

Solution :

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and β. We have

α + β = – 3 =`(-b)/a`

and αβ = 2 =`c/a`

If we take a = 1, then b = 3 and c = 2

So, one quadratic polynomial which fits the given conditions is x² + 3x + 2.

Similarly, we can take 'a' to be any real number. Let us say it is k. This gives `(-b)/k =-3` or b = 3k and `c/k`=2 or c = 2k. Putting the values of a, b and c, we get the polynomial is

kx²+ 3kx + 2k.

Example-6: Find a quadratic polynomial if the zeroes of it are 2 and `(-1)/3` respectively.

Solution :

Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.

Here α = 2, β =`(-1)/3`

Sum of the zeroes = (α + β) = 2 +`((-1)/3) = 5/3`

Product of the zeroes = (αβ) = 2`times (-1)/3 = (-2)/3`

`therefore`The quadratic polynomial ax² + bx + c is

k[x² – (α + β)x + α β], where k is a constant

= k[x² –`5/3`x`-2/3`]

We can put different values of k.

When k = 3, the quadratic polynomial will be 3x² – 5x – 2.

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Try this

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(i) Find a quadratic polynomial with zeroes −2 and `1/3`

Solution :

Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.

Here α = -2, β =`1/3`

Sum of the zeroes = (α + β) = -2 +`1/3 = -5/3`

Product of the zeroes = (αβ) = -2`× 1/3 = (-2)/3`

`therefore`The quadratic polynomial ax² + bx + c is

k[x² – (α + β)x + α β], where k is a constant

= k[x² –`((-5)/3)`x`-2/3`]

We can put different values of k.

When k = 3, the quadratic polynomial will be 3x² + 5x – 2.

(ii) What is the quadratic polynomial whose sum of zeroes is `(-3)/2`and the product of zeroes is −1.

Solution :

Sum of the zeroes = (α + β) = `-3/2`

Product of the zeroes = (αβ) = -1

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –`((-3)/2)`x`-1`]

= k[`(2x^2 +3x -2)/2`]

When k = 2, the quadratic polynomial will be 2x² + 3x – 2.

Example:Verify that 4, – 2, and `1/2` are the zeroes of the cubic polynomial 2x³ – 5x² – 14x + 8 and check the relationship between zeroes and the coefficients.

Solution :

Let us consider p(x) =2x³ – 5x² – 14x + 8

p(4) =2`times`4³ – 5`times`4² – 14`times`4 + 8

=2`times`64 - 5`times`16 - 56 +8 = 136 - 136 = 0

p(-2) =2`times`(-2)³ – 5`times`(-2)² – 14`times`(-2) + 8

=2`times`(-8) - 5`times`4 + 28 +8 = -36 + 36 = 0

p(`1/2`) =2`times (1/2)^3` – 5`times(1/2)^2` – 14`times(1/2)` + 8

=2`times(1/8)` - 5`times(1/4)` - 7 +8 = `1/4-5/4+1 = -1+1` = 0

So,4, – 2, and `1/2` are the zeroes of the cubic polynomial 2x³ – 5x² – 14x + 8

Sum of its zeroes =`4+(-2)+(1/2) = 2+1/2`

`=5/2 = -((-5)/2) = -(coefficient  of  x^2)/(coefficient  of  x^3 )`

Product of its zeroes =`4 × (-2) × 1/2`

`= -4 = (-8)/2=-(constant   term)/(coefficient  of  x^3)`

The sum of the products of the zeroes taken two at a time =`4 × (-2) + (-2) × 1/2 + 1/2 × 4 = -8 -1 + 2 `

`= -9 + 2 = -7 = -14/2= -(coefficient  of  x)/(coefficient  of  x^3)`

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Do This

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Example-7. Verify that 3, –1, –`1/3` are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

Solution :

Comparing the given polynomial with ax³ + bx² + cx + d, we get

a = 3, b = – 5, c = – 11, d = – 3 .

p(3) = 3 × 3³ – 5 × 3² – 11 × 3 – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)³ – 5 × (–1)²– 11 × (–1) – 3 =3 × (–1) – 5 × (1)+ 11 – 3 = – 3 – 5 + 11 – 3 = 0,

p(`-1/3`)=`3×((-1)/3)^3-5×((-1)/3)^2-11×(-1/3)-3`

=`-1/9-5/9+11/3-3= -6/9+2/3=-2/3+2/3=0`

`therefore` 3, –1, and `-1/3` are the zeroes of 3x³ – 5x² – 11x – 3

So, we take α = 3, β = –1 and γ =`-1/3`

Now, α + β + γ = 3 + (–1) +`(-1/3)`=3 –1`-1/3`= 2 `-1/3=5/3 = - ((-5)/3) = -b/a`

αβ + βγ + γα = 3 × (–1) + (–1) ×`((-1)/3) + ((-1)/3)×3 `

`= -3 + 1/3 -1= -4 + 1/3 = - 11/3 = c/a`

α β γ = 3 × (–1) × `((-1)/3)=1 = -((-3)/3)=-d/a`

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Exercise:

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1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² – 2x – 8

Solution:

split the middle term ‘–2x’ as a sum of two terms, whose product is

-8 × x² = –8 x². So, we write

x² – 2x –8 = x² – 4x +2x –8

= x(x – 4) +2 (x – 4)= (x – 4) (x +2)

p(x) = x² – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0,

i.e., when x = 4 or x = –2. So, the zeroes of x² – 2x – 8 are 4 and –2.

the sum of the zeroes = 4 + (–2) = 4 – 2 =2 =`-((-2)/1)=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = 4 × –2 = – 8 =`(–8)/1=(constant   term)/(coefficient  of  x^2)`

(ii) 4s² – 4s + 1

Solution:

split the middle term ‘–4s’ as a sum of two terms, whose product is

1 × 4s² = 4s². So, we write

4s² – 4s + 1 = 4s² – 2s – 2s + 1

= 2s(2s – 1) -1 (2s – 1)= (2s – 1) (2s – 1)

p(s) = 4s² – 4s + 1 is zero when 2s – 1 = 0

i.e., when s = `1/2`. So, the zeroes of 4s² – 4s + 1 are `1/2` and `1/2`.

the sum of the zeroes = `1/2 + 1/2` =1=`-((-4)/4)=-(coefficient  of  s)/(coefficient  of  s^2)`

Product of the zeroes =`1/2 × 1/2 = 1/4=(constant   term)/(coefficient  of  s^2)`

(iii) 6x² – 3 – 7x

Solution:

p(x)=6x² – 7x – 3

split the middle term ‘–7x’ as a sum of two terms, whose product is

-3 × 6x² = –18 x². So, we write

6x² – 7x –3 = 6x² – 9x +2x –3

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

p(x) = 6x² – 7x – 3 =0 when 2x – 3 = 0 or 3x + 1 = 0,

i.e., when x = `3/2 `or x = `–1/3`. So, the zeroes of 6x² – 7x – 3 are `3/2 ` and `–1/3`.

the sum of the zeroes = `3/2 + (–1/3) = (9 – 2)/6 =7/6 =-((-7)/6)=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = `3/2 × –1/3 = (-3)/6 =(constant   term)/(coefficient  of  x^2)`

(iv) 4u²+ 8u

Solution:

If p(u)=4u²+ 8u = 0

4u(u + 2) = 0

4u= 0 or u+2 = 0

u=0 or u= –2

So, the zeroes of p(u)=4u²+ 8u are 0 and –2.

the sum of the zeroes = 0 –2 = –2 = `-8/4= -(coefficient  of  u)/(coefficient  of  u^2)`

Product of the zeroes = 0 × – 2 = `0/4 =(constant   term)/(coefficient  of  u^2)`

(v) t² – 15

Solution :

Recall the identity a² – b² = (a – b) (a + b). Using it, we can write:

t² – 15 = (t – `sqrt 15` ) (t + `sqrt 15` )

So, the value of t² – 15 is zero when t = `sqrt 15` or t = – `sqrt 15` .

`therefore` The zeroes of t² – 15 are `sqrt 15` and – `sqrt 15` .

Sum of the zeroes = `sqrt 15` + (–`sqrt 15` ) = 0=`0/1` =-`(coefficient  of  t)/(coefficient  of  t^2)`

Product of zeroes = (`sqrt 15` ) × (–`sqrt 15` ) = – 15 =`(-15)/1 =(constant   term)/(coefficient  of  t^2)`

(vi) 3x² – x – 4

Solution:

split the middle term ‘–x’ as a sum of two terms, whose product is

-4 × 3x² = –12 x². So, we write

3x² – x –4 = 3x² – 4x +3x –4

= x(3x – 4) +1 (3x – 4)= (3x – 4) (x +1)

p(x) = 3x² – x – 4 is zero when 3x – 4 = 0 or x + 1 = 0,

i.e., when x = `4/3` or x = –1. So, the zeroes of 3x² – x – 4 are `4/3` and –1.

the sum of the zeroes = `4/3` + (–1) = `(4 – 3)/3 =1/3 =-((-1)/3)=-(coefficient  of  x)/(coefficient  of  x^2)`

Product of the zeroes = `4/3× (–1) =(–4)/3=(constant   term)/(coefficient  of  x^2)`

2. Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

(i)`1/4`, – 1

Solution :

Sum of the zeroes = (α + β) = `1/4`

Product of the zeroes = (αβ) = –1

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –`1/4`x–1]

= k[`(4x^2 – x –4)/4`]

When k = 4, the quadratic polynomial will be 4x² – x – 4.

(ii)`sqrt2`,`1/3`

Solution :

Sum of the zeroes = (α + β) = `sqrt2`

Product of the zeroes = (αβ) = `1/3`

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –`sqrt2`x + `1/3`]

= k[`(3x^2 – sqrt2 x + 1)/3`]

When k = 3, the quadratic polynomial will be 3x² – `sqrt2`x + 1.

(iii) 0, `sqrt5`

Solution :

Sum of the zeroes = (α + β) = 0

Product of the zeroes = (αβ) = `sqrt5`

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –0x + `sqrt5`]

When k = 1, the quadratic polynomial will be x² + `sqrt5`.

(iv) 1, 1

Solution :

Sum of the zeroes = (α + β) = 1

Product of the zeroes = (αβ) = 1

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –1x + 1]

When k = 1, the quadratic polynomial will be x² – x + 1.

(v) –`1/4`,`1/4`

Solution :

Sum of the zeroes = (α + β) = `-1/4`

Product of the zeroes = (αβ) = `1/4`

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –`(-1/4)`x + `1/4`]

= k[`(4x^2 + x + 1)/4`]

When k = 4, the quadratic polynomial will be 4x² + x + 1.

(vi) 4, 1

Solution :

Sum of the zeroes = (α + β) = 4

Product of the zeroes = (αβ) = 1

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –4x + 1]

When k = 1, the quadratic polynomial will be x² – 4x + 1.

3. Find the quadratic polynomial, for the zeroes α, β given in each case.

(i) 2, –1

Solution :

Sum of the zeroes = (α + β) = 2 –1 = 1

Product of the zeroes = (αβ) = 2 (–1)= –2

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² – 1x + (–2)]

= k[x² – 1x – 2]

When k = 1, the quadratic polynomial will be x² – x –2.

(ii) `sqrt3` , – `sqrt3`

Solution :

Sum of the zeroes = (α + β) = `sqrt3 – sqrt3` = 0

Product of the zeroes = (αβ) = `sqrt3 (–sqrt3)`= –3

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² – 0x + (–3)]

= k[x² – 3]

When k = 1, the quadratic polynomial will be x² – 3.

(iii)`1/4`, – 1

Solution :

Sum of the zeroes = (α + β) = `1/4 - 1 = -3/4`

Product of the zeroes = (αβ) = `1/4(-1)= - 1/4`

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² –`(-3/4)`x + `(-1/4)`]

= k[`(4x^2 + 3x - 1)/4`]

When k = 4, the quadratic polynomial will be 4x² +3x – 1.

(iv)`1/2`,`3/2`

Solution :

Sum of the zeroes = (α + β) = `1/2 + 3/2 = 4/2`=2

Product of the zeroes = (αβ) = `(1/2)( 3/2)=3/4`

`therefore`The quadratic polynomial =k[x² – (α + β)x + α β]

= k[x² – 2x + `3/4`]

= k[`(4x^2 – 8x + 3)/4`]

When k = 4, the quadratic polynomial will be 4x² – 8x + 3.

4. Verify that 1, –1 and –3 are the zeroes of the cubic polynomial x³+ 3x² – x – 3 and check the relationship between zeroes and the coefficients.

Solution :

Let p(x)=x³+ 3x² – x – 3

p(1) =1³+ 3(1)² – 1 – 3

= 1 + 3 -1 -3 = 0

So, 1 is the zero of p(x).

p(-1)= (-1)³+ 3(-1)² – (-1) – 3

= -1 +3 +1 -3 =0

So, -1 is also the zero of p(x).

p(-3)=(-3)³+ 3(-3)² – (-3) – 3

= -27 + 3(9) + 3 -3 = -27 + 27+ 3 -3 =0

So, -3 is also the zero of p(x).

∴ 1, –1 and –3 are the zeroes of the cubic polynomial x³+ 3x² – x – 3

Now checking the relationship between zeroes and the coefficients

So, we take α = 1, β = –1 and γ =-3

Now, α + β + γ = 1 + (–1) +(-3)=1 –1–3= =–3 = `-3/1 = -(coefficient  of  x^2)/(coefficient  of  x^3 )`

αβ + βγ + γα = 1 × (–1) + (–1) ×(–3) + (–3)×1

= –1 + 3 –3= –4 + 3 = – 1 =` (-1)/1= (coefficient  of  x)/(coefficient  of  x^3)`

α β γ = 1 × (–1) × (–3)= 3 = `-((-3)/1)=-(constant   term)/(coefficient  of  x^3)`

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