Ncert Solutions For Class 10 Maths Ex 2.2

Ncert Solutions For Class 10 Maths Chapter 2 Ex 2.2

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

(i) 6x² + 11x + 5 = 0

Soln: –

6x2 + 11x + 5 = 6x2 + 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation 6x2 +11x +5 are { −1, 56 }

Now, Sum of zeroes of this given polynomial equation = −1+( 56 ) = 116

But, the Sum of zeroes of any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

= 116

And Product of these zeroes will be = 1×56 = 56

But, the Product of zeroes of any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 56

Hence the relationship is verified.

 

(ii) 4s2 – 4s + 1

Sol:

     4s2 – 4s + 1 = 4s2 – 2s 2s + 1

= 2s (2s 1) 1(2s 1)

= (2s 1) (2s 1)

∴ zeroes of the given polynomial are: {12,12}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = coeff.ofscoeff.ofs2

= 44 = 1

And the Product of these zeroes will be = 12×12
=14

But, Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofs2

= 14.

Hence, the relationship is verified.

 

(iii) 6x2 – 3 – 7x

Sol:

     6x2 – 7x – 3 = 6x2 – 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (13,32)

∴ sum of these zeroes will be =  13+32
=76

But, The Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

= 76

And Product of these zeroes will be = 13×32=12
Also, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 36 = 12

Hence, the relationship is verified.

 

(iv) 4u2 + 8u

Sol:

     4u2 + 8u = 4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0

Hence, the zeroes of the above polynomial equation will be (0, −2)

∴ Sum of these zeroes will be = −2

But, the Sum of the zeroes in any quadratic polynomial equation is given by = coeff.ofucoeff.ofu2

 = 84 = −2

And product of these zeroes will be = 0 × −2 = 0

But, the product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofu2                                                                          = 04 = 0

Hence, the relationship is verified.

 

(v) t2 – 15

Sol:

     t2 – 15 = (t+ 15) (t − 15)

Therefore, zeroes of the given polynomial are: – {15, −15}

∴ sum of these zeroes will be = 15 −  15 = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

                                                                = 01 = 0   

And the product of these zeroes will be = (15) × (−15)  = −15

But, the product of zeroes in any quadratic polynomial equation is given by

= constanttermcoeff.oft2                                                                     = 151  = −15

Hence, the relationship is verified.

 

(vi) 3x2 – x – 4

Sol:

 3x2 − x − 4   = 3x2 – 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – {−1, 43 }

∴ sum of these zeroes will be = −1 +43 = 13

But, the Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2

= (1)3 = 13

And the Product of these zeroes will be = {−1 × 43 }

= 43

But, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 43

Hence, the relationship is verified.

 

Q2. Form a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i). 26 , −3

Sol.

Given,

α + β = 26

αβ = −3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ = 0

Thus, the required quadratic equation will be:

x2 – (26)x −3 = 0

6x2 − 2x – 18 = 0.

 

(ii). 3 , 43

Sol.

Given,

α + β = 3

αβ = 43

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x2 – (3)x + 43 =0

3x2 33x + 4 = 0.

 

(iii).  0, 7

Sol.

Given,

α + β = 0

αβ = 7

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x2 – (0)x + 7 = 0

x2 + 7 = 0.

 

(iv).  −2, −2

Sol.

Given,

α + β = −2

αβ = −2

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x2 – (α+β)x +αβ=0

∴ The required quadratic polynomial will be:

x2 – (−2)x −2 = 0

x2 + 2x – 2 = 0.

 

(v). 72, 39

Sol.

Given,

α + β = 72

αβ = 39

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:

x2 – (72)x + 39 = 0

18x2 + 63x + 6 = 0.

 

(vi). 6, 0

Sol.

Given,

α + β = 6

αβ = 0

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:-

x2 – 6x + 0 = 0

x2 – 6x = 0.

 

                                  EXTRA QUESTIONS

Q.1 Find a quadratic polynomial whose zeroes are: –2+12, 212.

Sol.

Given: –

α + β = 2+12 + 212

= 4.

αβ = (2+12)(212)=412=72

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ = 0

Thus, the required quadratic polynomial equation will be :-

x2 – (4)x −72 = 0

2x2−8x+7 = 0.

 

Q.2 If α and β are the roots of a quadratic polynomial ax2+bx+c, then find the value of  α2 + β2.

Sol.

From the equation, (α+β=ba)

And,               α×β=ca

(α+β)2=α2+β2+(2αβ)

α2+β2=(α+β)22αβ

α2+β2=(ba)22ca

α2+β2=b2a22ca

 ∴α2+β2=b22aca2.

Similarly, we can find out the values of (α3 + β3) and (α3 – β3).

 

Q3. If α and β are zeroes of a quadratic polynomial x2+4x+3, form the polynomial whose zeroes are  1+αβand1+βα.

Sol.

Since α and β are zeroes of a quadratic polynomial x2+4x+3,

α+β= -4, αβ = 3

Given: –  α1 = 1+αβ

β1 = 1βα

Now, sum of zeroes = 1+αβ+1+βα

= 2+αβ+βα

= 2αβ+α2+β2αβ

= (α+β)2αβ

On putting values of α+β and αβ from above we get:-

Sum of zeroes = α1 + β1 = 423

= 163

Now, Product of zeroes = (1+αβ)(1+βα)

=1+βα+αβ+αββα

=2αβ+β2+α2αβ

= (α+β)2αβ

On putting values of α+β and αβ we get:

Product of zeroes = α1 × β1 = 423

= 163

Thus the required quadratic polynomial equation will be:-

x2 – (α1 + β1)x + α1β1 = 0

x2 – (163)x + 163 = 0

3x2 – 16x +16=0.

 

Q4. If α and β are zeroes of a quadratic polynomial p(x) = rx2+4x+4, Find the values of “r” if: – α2 + β2 = 24.

Sol.

From the given polynomial p(x),

α + β = 4r, and αβ = 4r………. (1)

Since,(α + β)2 = α2 + β2 + 2αβ

Given,   α2 + β2 = 24 and from equation (1).

Therefore,  (4r)2=24+2×4r

16r2=24+2×4r

16 = 24 r2 + 8r

3r2 + r – 2 = 0

3r2 + 3r – 2r -2 = 0

3r(r+1) – 2(r+1) = 0

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

∴  r = 23   or   r = -1

 

                                 DIVISION ALOGORITHM

If suppose p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that: –

p(x)=g(x) × q(x) + r(x), where r(x)=0 or  degree of r(x) < degree of g(x) …….(1)

This is known as the division algorithm for polynomials.

And since, Dividend = Divisor × Quotient + Remainder

On comparing it with equation (1) we can conclude that:

Dividend = p(x).

Divisor = g(x).

Quotient = q(x).

Remainder = r(x).