Class10 NCRT Polynomials Exercise – 2.2 pdf || UP Board

Polynomials
Exercise – 2.2

 

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

(i) x2-2x-8

Solution:

By the factorization method,

x2-4x+2x-8 = 0

x(x-4) +2(x-4) = 0

(x+2) (x-4) = 0

x = -2, 4

So, zeroes α=-2, β=4

Verification,

Polynomials

    -2+4 = -(-2)/1

     2 = 2

Polynomials

   -2×4 = -8/1

    -8 = -8

The relation between the coefficients is true in both the cases.

 

(ii) 4s2-4s+1

Solution:

By the factorization method,

4s2-2s-2s+1 = 0

2s(2s-1) -1(2s-1) = 0

(2s-1) (2s-1) = 0

s = 1/2, 1/2

So, zeroes α= 1/2, β= 1/2

Verification,

sum of zeroes(α+β) =(-b )/a

    ½ + 1/2 = -(-4/4)

     2/2 = 4/4

       1 = 1

product of zeroes (αβ) =(c )/a

   ½ × 1/2 = 1/4

    1/4 = 1/4

The relation between the coefficients is true in both the cases.

 

(iii) 6x2-3-7x

Solution:

By the factorization method,

6x2-9x+2x-3 = 0

3x(2x-3) +1(2x-3) = 0

(3x+1) (2x-3) = 0

x = -1/3, 3/2

So, zeroes α= 3/2, β= -1/3

Verification,

sum of zeroes(α+β) =(-b )/a

    3/2 + -1/3 = -(-7)/6

     (9-2)/6 = 7/6

       7/6 = 7/6

product of zeroes (αβ) =(c )/a

   3/2 × -1/3 = -3/6

    -3/6 = -3/6

The relation between the coefficients is true in both the cases.

 

(iv) 4u2+8u

Solution:

By the factorization method,

4u(u+2) = 0

u = 0, -2

So, zeroes α= 0, β= -2

Verification,

sum of zeroes (α+β) =(-b )/a

    0 + -2 = -8/4

     -2 = -2

product of zeroes (αβ) =(c )/a

  0 × -2 = 0/4

    0 = 0

The relation between the coefficients is true in both the cases.

 

(v) t2-15

Solution:

By the factorization method,

t2 - 15 = 0

t = ±√15

t = √15, -√15

So, zeroes α= √15, β= -√15

Verification,

sum of zeroes (α+β) =(-b )/a

    √15 + -√15 = -(0)/1

     0 = 0

product of zeroes (αβ) =(c )/a

   √15 × -√15 = -15

    -15 = -15

The relation between the coefficients is true in both the cases.

 

(vi) 3x2-x-4

Solution:

By the factorization method,

3x2-4x+3x-4 = 0

x(3x-4) +1(3x-4) = 0

(3x-4) (x+1) = 0

x = 4/3, -1

So, zeroes α= 4/3, β= -1

Verification,

sum of zeroes (α+β) =(-b )/a

    4/3 + -1 = -(-1)/3

     (4-3)/3 = 1/3

       1/3 = 1/3

product of zeroes (αβ) =(c )/a

   4/3 × -1 = -4/3

    -4/3 = -4/3

The relation between the coefficients is true in both the cases.

 

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:

(i) ¼, -1

Solution:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

           sum of zeroes α + β = -b/a = ¼

              -b/a = ¼

by comparison,

            -b = 1, a = 4

                       b = -1, a = 4

          product of zeros αβ = c/a = -1

                 c/a = -1

or                c/4 = -1  (a = 4)

                   c = -4

Now,

            b = -1, a = 4, c = -4

Hence,

                 ax2+bx+c

                                  4x2-1x-4

The quadratic polynomial is 4x2-1x-4.

(ii) √2, 1/3

Solution:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

           sum of zeroes α + β = -b/a = √2

              -b/a = √2

or            -b/a = √2/1

by comparison,

            -b = √2, a = 1

             b = -√2, a = 1

 

          product of zeros αβ = c/a = 1/3

                 c/a = 1/3

or               c/1 = 1/3  (a = 1)

                   c = 1/3

Now,

            b = -√2, a = 1, c = 1/3

Hence,

                 ax2+bx+c = 0     

         

                          3x2-3√2+1 = 0

The quadratic polynomial is 3x2-3√2+1.

 

(iii) 0,

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

           sum of zeroes α + β = -b/a = 0

              -b/a = 0

or            -b/a = 0/1

by comparison,

            b = 0, a = 1

          product of zeros αβ = c/a = 5

                 c/a = 5/1

or               c/1 = √5/1  (a = 1)

                   c = 5

Now,

            b = 0, a = 1,                                     c = √5

Hence,

                                 ax2+bx+c = 0

                                 x2-0x+√5 = 0

                                x2+√5 = 0

The quadratic polynomial is x2+√5.

 

(iv) 1, 1

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

           sum of zeroes α + β = -b/a = 1

              -b/a = 1

or            -b/a = 1/1

by comparison,

            b = -1, a = 1

 

          product of zeros αβ = c/a = 1

                 c/a = 1/1

or               c/1 = 1/1  (a = 1)

                   c = 1

Now,

            b = -1, a = 1,                                     c = 1

Hence,

                                 ax2+bx+c = 0

                                 x2-1x+1 = 0

The quadratic polynomial is x2-1x+1.

 

(v) -1/4, 1/4

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

           sum of zeroes α + β = -b/a = -1/4

              -b/a = -1/4

or            b/a = 1/4

by comparison,

            b = 1, a = 4

 

          product of zeros αβ = c/a = 1/4

                 c/a = 1/4

or               c/4 = 1/4  (a = 4)

                   c = 1

Now,

            b = 1, a = 4,                                     c = 1

Hence,

                                 ax2+bx+c = 0

                                 4x2+1x+1 = 0

The quadratic polynomial is 4x2+1x+1.

 

(vi) 4, 1

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

           sum of zeroes α + β = -b/a = 4

              -b/a = 4

or            b/a = -4/1

by comparison,

            b = -4, a = 1

 

          product of zeros αβ = c/a = 1

                 c/a = 1

or               c/1 = 1/1  (a = 1)

                   c = 1

Now,

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

Hence,

                                 ax2+bx+c = 0

                                 x2-4x+1 = 0

The quadratic polynomial is x2-4x+1.

 

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