# 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$\frac{-5}{6}$ }

Now, Sum of zeroes of this given polynomial equation = −1+( 56$\frac{-5}{6}$ ) = 116$\frac{-11}{6}$

But, the Sum of zeroes of any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2$\frac{-coeff.\;\;of\;\; x}{coeff.\;\;of\;\;x^{2}}$

= 116$\frac{-11}{6}$

And Product of these zeroes will be = 1×56$-1\times\frac{-5}{6}$ = 56$\frac{5}{6}$

But, the Product of zeroes of any quadratic polynomial equation is given by = constanttermcoeff.ofx2$\frac{constant\;\; term}{coeff.\;\;of\;\;x^{2}}$

= 56$\frac{5}{6}$

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$\frac{1}{2},\frac{1}{2}$}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = coeff.ofscoeff.ofs2$\frac{-coeff.\;\;of\;\; s}{coeff.\;\;of\;\;s^{2}}$

= 44$\frac{-4}{4}$ = 1

And the Product of these zeroes will be = 12×12$\frac{1}{2}\times \frac{1}{2}$
=14$\frac{1}{4}$

But, Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofs2$\frac{constant\;\; term}{coeff.\;\;of\;\;s^{2}}$

= 14$\frac{1}{4}$.

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$\frac{-1}{3},\frac{3}{2}$)

∴ sum of these zeroes will be =  13+32$\frac{-1}{3}+\frac{3}{2}$
=76$\frac{7}{6}$

But, The Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2$\frac{-coeff.\;\;of\;\; x}{coeff.\;\;of\;\;x^{2}}$

= 76$\frac{7}{6}$

And Product of these zeroes will be = 13×32=12$\frac{-1}{3}\times \frac{3}{2}=\frac{-1}{2}$
Also, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2$\frac{constant\;\; term}{coeff.\;\;of\;\;x^{2}}$

= 36$\frac{-3}{6}$ = 12$\frac{-1}{2}$

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$\frac{-coeff.\;\;of\;\; u}{coeff.\;\;of\;\;u^{2}}$

= 84$\frac{−8}{4}$ = −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$\frac{constant\;\; term}{coeff.\;\;of\;\;u^{2}}$                                                                          = 04$\frac{-0}{4}$ = 0

Hence, the relationship is verified.

(v) t2 – 15

Sol:

t2 – 15 = (t+ 15$\sqrt{15}$) (t − 15$\sqrt{15}$)

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

∴ sum of these zeroes will be = 15$\sqrt{15}$ −  15$\sqrt{15}$ = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2$\frac{-coeff.\;\;of\;\; x}{coeff.\;\;of\;\;x^{2}}$

= 01$\frac{−0}{1}$ = 0

And the product of these zeroes will be = (15$\sqrt{15}$) × (−15)$\sqrt{15})$  = −15

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

= constanttermcoeff.oft2$\frac{constant\;\; term}{coeff.\;\;of\;\;t^{2}}$                                                                     = 151$\frac{−15}{1}$  = −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$\frac{4}{3}$ }

∴ sum of these zeroes will be = −1 +43$\frac{4}{3}$ = 13$\frac{1}{3}$

But, the Sum of zeroes in any quadratic polynomial equation is given by = coeff.ofxcoeff.ofx2$\frac{-coeff.\;\;of\;\; x}{coeff.\;\;of\;\;x^{2}}$

= (1)3$\frac{−(−1)}{3}$ = 13$\frac{1}{3}$

And the Product of these zeroes will be = {−1 × 43$\frac{4}{3}$ }

= 43$\frac{-4}{3}$

But, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2$\frac{constant\;\; term}{coeff.\;\;of\;\;x^{2}}$

= 43$\frac{-4}{3}$

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$\frac{2}{6}$ , −3

Sol.

Given,

α + β = 26$\frac{2}{6}$

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$\frac{2}{6}$)x −3 = 0

6x2 − 2x – 18 = 0.

(ii). 3$\sqrt{3}$ , 43$\frac{4}{3}$

Sol.

Given,

α + β = 3$\sqrt{3}$

αβ = 43$\frac{4}{3}$

∴ 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 – (3$\sqrt{3}$)x + 43$\frac{4}{3}$ =0

3x2 33x$3\sqrt{3}x$ + 4 = 0.

(iii).  0, 7$\sqrt{7}$

Sol.

Given,

α + β = 0

αβ = 7$\sqrt{7}$

∴ 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$\sqrt{7}$ = 0

x2 + 7$\sqrt{7}$ = 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$\frac{-7}{2}$, 39$\frac{3}{9}$

Sol.

Given,

α + β = 72$\frac{−7}{2}$

αβ = 39$\frac{3}{9}$

∴ 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$\frac{−7}{2}$)x + 39$\frac{3}{9}$ = 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$2+\frac{1}{\sqrt{2}}$, 212$2-\frac{1}{\sqrt{2}}$.

Sol.

Given: –

α + β = 2+12$2+\frac{1}{\sqrt{2}}$ + 212$2-\frac{1}{\sqrt{2}}$

= 4.

αβ = (2+12)(212)=412=72$(2+\frac{1}{\sqrt{2}})(2-\frac{1}{\sqrt{2}}) =4-\frac{1}{2} =\frac{7}{2}$

∴ 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

Thus, the required quadratic polynomial equation will be :-

x2 – (4)x −72$\frac{7}{2}$ = 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)$(\alpha +\beta =\frac{-b}{a})$

And,               α×β=ca$\alpha \times \beta =\frac{c}{a}$

(α+β)2=α2+β2+(2αβ)$(\alpha +\beta )^{2}= \alpha ^{2}+\beta ^{2}+(2\alpha \beta )$

α2+β2=(α+β)22αβ$\alpha ^{2}+\beta ^{2}=(\alpha +\beta )^{2}- 2\alpha \beta$

α2+β2=(ba)22ca$\alpha ^{2}+\beta ^{2}=(\frac{-b}{a} )^{2}- 2\frac{c}{a}$

α2+β2=b2a22ca$\alpha ^{2}+\beta ^{2}= \frac{b^{2}}{a^{2}}- \frac{2c}{a}$

∴α2+β2=b22aca2$\alpha ^{2}+\beta ^{2}= \frac{b^{2}-2ac}{a^{2}}$.

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$1+\frac{\alpha }{\beta } \;\;and \;\;1+\frac{\beta }{\alpha }$.

Sol.

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

α+β= -4, αβ = 3

Given: –  α1 = 1+αβ$1+\frac{\alpha }{\beta }$

β1 = 1βα$1-\frac{\beta }{\alpha }$

Now, sum of zeroes = 1+αβ+1+βα$1+\frac{\alpha }{\beta } +1+\frac{\beta }{\alpha }$

= 2+αβ+βα$2+\frac{\alpha}{\beta }+\frac{\beta }{\alpha }$

= 2αβ+α2+β2αβ$\frac{2\alpha \beta +\alpha ^{2}+\beta ^{2}}{\alpha \beta }$

= (α+β)2αβ$\frac{(\alpha+\beta)^{2}}{\alpha \beta }$

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

Sum of zeroes = α1 + β1 = 423$\frac{-4^{2}}{3}$

= 163$\frac{16}{3}$

Now, Product of zeroes = (1+αβ)(1+βα)$(1+\frac{\alpha }{\beta })(1+\frac{\beta }\alpha)$

=1+βα+αβ+αββα$1+\frac{\beta }{\alpha }+\frac{\alpha }{\beta }+\frac{\alpha \beta }{\beta \alpha }$

=2αβ+β2+α2αβ$\frac{2\alpha\beta+\beta^{2}+\alpha^{2}}{\alpha\beta}$

= (α+β)2αβ$\frac{(\alpha+\beta )^{2}}{\alpha \beta }$

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

Product of zeroes = α1 × β1 = 423$\frac{-4^{2}}{3}$

= 163$\frac{16}{3}$

Thus the required quadratic polynomial equation will be:-

x2 – (α1 + β1)x + α1β1 = 0

x2 – (163$\frac{16}{3}$)x + 163$\frac{16}{3}$ = 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$\frac{-4}{r}$, and αβ = 4r$\frac{-4}{r}$………. (1)

Since,(α + β)2 = α2 + β2 + 2αβ

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

Therefore,  (4r)2=24+2×4r$(\frac{-4}{r})^{2}= 24+\frac{2×4}{r}$

16r2=24+2×4r$\frac{16}{r^{2}}= 24+\frac{2\times4}{r}$

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$\frac{2}{3}$   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).