KC Sinha Mathematics Solution Class 10 Chapter 2 Polynomials Exercise 2.2

Q1 A Q1 B Q1 C Q1 D Q1 E Q1 F Q1 G Q1 H Q2 A Q2 B Q2 C Q2 D Q2 E Q2 F Q3 A Q3 B Q3 C Q4 Q5 Q6 Q7 Q8 A Q8 B Q8 C Q9 A Q9 B Q9 C Q9 D Q9 E Q9 F Q10 A Q10 B Q10 C Q10 D Q10 E Q10 F Q11 Q12 A Q12 B Q13 A Q13 B

Exercise 2.1 Exercise 2.2 Exercise 2.3

Exercise 2.2

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Question 1 A 

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

x² – 3
Sol :
Let f (x) = x² – 3
Now, if we recall the identity
(a² – b²) = (a – b)(a + b)
Using this identity, we can write
x² – 3 = (x – √3) (x + √3)
So, the value of x² – 3 is zero when x = √3 or x = – √3
Therefore, the zeroes of x² – 3 are √3 and – √3.
Verification
Now,
Sum of zeroes = α + β = √3 + ( – √3) = 0 or

$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{0}{1}=0$

Product of zeroes = αβ = (√3)( – √3) = – 3 or

$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{1}=-3$

So, the relationship between the zeroes and the coefficients is verified.

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Question 1 B 

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

2x² – 8x + 6
Sol :
Let f(x) = 2x² – 8x + 6
By splitting the middle term, we get
f(x) = 2 x² – (2 + 6)x + 6 [∵ – 8 = – (2 + 6) and 2×6 = 12]
= 2 x² – 2x – 6x + 6
= 2x(x – 1) – 6(x – 1)
= (2x – 6) (x – 1)
On putting f(x) = 0, we get
(2x – 6) (x – 1) = 0
⇒2x – 6 = 0 or x – 1 = 0
⇒x = 3 or x = 1
Thus, the zeroes of the given polynomial 2 x² – 8x + 6 are 1 and 3
Verification
Sum of zeroes = α + β = 3 + 1 = 4 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{2}=4$

Product of zeroes = αβ = (3)(1) = 3 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{6}{2}=3$
So, the relationship between the zeroes and the coefficients is verified.

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Question 1 C 

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

x² – 2x – 8
Sol :
Let f(x) = x² – 2x – 8
By splitting the middle term, we get
f(x) = x² – 4x + 2x – 8 [∵ – 2 = 2 – 4 and 2×4 = 8]
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
On putting f(x) = 0, we get
(x + 2) (x – 4) = 0
⇒ x + 2 = 0 or x – 4 = 0
⇒x = – 2 or x = 4
Thus, the zeroes of the given polynomial x² – 2x – 8 are – 2 and 4
Verification
Sum of zeroes = α + β = – 2 + 4 = 2 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-2)}{1}=2$

Product of zeroes = αβ = ( – 2)(4) = – 8 or

$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-8}{1}=-8$

So, the relationship between the zeroes and the coefficients is verified.

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Question 1 D 

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

3x² + 5x – 2

Sol :
Let f(x) = 3x² + 5x – 2
By splitting the middle term, we get
f(x) = 3x² + (6 – 1)x – 2 [∵ 5 = 6 – 1 and 2×3 = 6]
= 3x² + 6x – x – 2
= 3x(x + 2) – 1(x + 2)
= (3x – 1) (x + 2)
On putting f(x) = 0 , we get
(3x – 1) (x + 2) = 0
⇒ 3x – 1 = 0 or x + 2 = 0
$\Rightarrow \mathrm{x}=\frac{1}{3}$ or x = – 2
Thus, the zeroes of the given polynomial 3x² + 5x – 2 are – 2 and 
Verification

Sum of zeroes

$=\alpha+\beta=\frac{1}{3}+(-2)$

$=\frac{1-6}{3}=-\frac{5}{3}$

  or

$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{5}{3}$

$=\alpha \beta=\left(\frac{1}{3}\right)(-2)=\frac{-2}{3}$

Product of zeroes or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-2}{3}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 1 E 

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

3x² – x – 4

Sol :
Let f(x) = 3x² – x – 4
By splitting the middle term, we get
f(x) = 3x² – (4 – 3)x – 4 [∵ – 1 = 3 – 4 and 4×3 = 12]
= 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (3x – 4) (x + 1)
On putting f(x) = 0, we get
(3x – 4) (x + 1) = 0
⇒ 3x – 4 = 0 or x + 1 = 0
$\Rightarrow x=\frac{4}{3}$ or x = – 1
Thus, the zeroes of the given polynomial 3x² – x – 4 are – 1 and $\frac{4}{3}$

Verification

Sum of zeroes

$=\alpha+\beta=\frac{4}{3}+(-1)$

$=\frac{4-3}{3}=\frac{1}{3}$

 or

$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-1)}{3}=\frac{1}{3}$

Product of zeroes

$=\alpha \beta=\left(\frac{4}{3}\right)(-1)=\frac{-4}{3}$ 

or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-4}{3}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 1 F 

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

x² + 7x + 10
Sol :
Let f(x) = x² + 7x + 10
By splitting the middle term, we get
f(x) = x² + 5x + 2x + 10 [∵ 7 = 2 + 5 and 2×5 = 10]
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
On putting f(x) = 0 , we get
(x + 2) (x + 5) = 0
⇒ x + 2 = 0 or x + 5 = 0
⇒x = – 2 or x = – 5
Thus, the zeroes of the given polynomial x² + 7x + 10 are – 2 and – 5
Verification
Sum of zeroes = α + β = – 2 + ( – 5) = – 7 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{7}{1}=-7$

Product of zeroes = αβ = ( – 2)( – 5) = 10 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{10}{1}=10$
So, the relationship between the zeroes and the coefficients is verified.

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Question 1 G 

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

t² – 15

Sol :
Let f(x) = t² – 15
Now, if we recall the identity
(a² – b²) = (a – b)(a + b)
Using this identity, we can write
t² – 15 = (t – √15) (x + √15)
So, the value of t²– 15 is zero when t = √15 or t = – √15
Therefore, the zeroes of t²– 15 are √15 and – √15.
Verification
Now,
Sum of zeroes = α + β = √15 + ( – √15) = 0 or
$=-\frac{\text { Coefficient of } \mathrm{t}}{\text { Coefficient of } \mathrm{t}^{2}}=-\frac{0}{1}=0$

Product of zeroes = αβ = (√15)( – √15) = – 15 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{t}^{2}}=\frac{-15}{1}=-15$
So, the relationship between the zeroes and the coefficients is verified.

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Question 1 H 

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

4 s² – 4s + 1
Sol :
Let f(x) = 4 s² – 4s + 1
By splitting the middle term, we get
f(x) = 4 s² – (2 – 2)s + 1 [∵ – 4 = – (2 + 2) and 2×2 = 4]
= 4 s² – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
On putting f(x) = 0 , we get
(2s – 1) (2s – 1) = 0
⇒ 2s – 1 = 0 or 2s – 1 = 0
$\Rightarrow s=\frac{1}{2}$ or $s=\frac{1}{2}$
Thus, the zeroes of the given polynomial 4 s² – 4s + 1 are $\frac{1}{2}$ and $\frac{1}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{1}{2}+\frac{1}{2}$=1

 or
$=-\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}=-\frac{(-4)}{4}=1$

Product of zeroes $=\alpha \beta=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$   or

$=\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}=\frac{1}{4}$

So, the relationship between the zeroes and the coefficients is verified.

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Question 2 A 

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

8x²– 22x – 21
Sol :
Let f(x) = 8x² – 22x – 21
By splitting the middle term, we get
f(x) = 8x² – 28x + 6x – 21
= 4x(2x – 7) + 3(2x – 7)
= (4x + 3) (2x – 7)
On putting f(x) = 0 , we get
(4x + 3) (2x – 7) = 0
⇒ 4x + 3 = 0 or 2x – 7 = 0

⇒ x = $\frac{-3}{4}$  or x = $\frac{7}{2}$
Thus, the zeroes of the given polynomial 8x² – 22x – 21 are $\frac{-3}{4}$ and $\frac{7}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-3}{4}+\frac{7}{2}=\frac{-3+14}{4}=\frac{11}{4}$       or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-22)}{8}=\frac{11}{4}$
The product of zeroes $=\alpha \beta=\frac{-3}{4} \times \frac{7}{2}=\frac{-21}{8}$            or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-21}{8}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 2 B 

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

2x² – 7x
Sol :
Let f(x) = 2 x² – 7x
In this the constant term is zero.
f(x) = 2 x² – 7x
= x(2x – 7)
On putting f(x) = 0 , we get
x(2x – 7) = 0
⇒ 2x – 7 = 0 or x = 0
 ⇒ X = $\frac{7}{2}$ or x = 0

Thus, the zeroes of the given polynomial 2x² – 7x are 0 and $\frac{7}{2}$
Verification
Sum of zeroes $=\alpha+\beta=0+\frac{7}{2}=\frac{7}{2}$        or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-7)}{2}=\frac{7}{2}$
Product of zeroes $=\alpha \beta=0 \times \frac{7}{2}=0$              or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{0}{2}=0$
So, the relationship between the zeroes and the coefficients is verified.

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Question 2 C 

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

10x² + 3x – 1
Sol :
Let f(x) = 10x² + 3x – 1
By splitting the middle term, we get
f(x) = 10x² – 2x + 5x – 1
= 2x(5x – 1) + 1(5x – 1)
= (2x + 1) (5x – 1)
On putting f(x) = 0, we get
(2x + 1) (5x – 1) = 0
⇒ 2x + 1 = 0 or 5x – 1 = 0

⇒ X = $\frac{-1}{2}$  or  X =$\frac{1}{5}$
Thus, the zeroes of the given polynomial10x² + 3x – 1 are $\frac{-1}{2}$ and $\frac{1}{5}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-1}{2}+\frac{1}{5}=\frac{-5+2}{10}=\frac{-3}{10}$  or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{3}{10}=-\frac{3}{10}$
Product of zeroes $=\alpha \beta=\frac{-1}{2} \times \frac{1}{5}=\frac{-1}{10}$             or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-1}{10}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 2 D 

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

px² + (2q – p²)x – 2pq, p≠0
Sol :
Let f(x) = px² + (2q – p²)x – 2pq
f(x) = px² + 2qx – p² x – 2pq
= x(px + 2q) – p(px + 2q)
= (x – p) (px + 2q)
On putting f(x) = 0, we get
(x – p) (px + 2q) = 0
⇒ x – p = 0 or px + 2q = 0
⇒x = p or X = $\frac{-2 q}{p}$
Thus, the zeroes of the given polynomial px² + (2q – p²)x – 2pq are p and $\frac{-2 q}{p}$
Verification
Sum of zeroes $=\alpha+\beta=p+\frac{(-2 q)}{p}=\frac{p^{2}-2 q}{p}$        or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{\left(-\mathrm{p}^{2}+2 \mathrm{q}\right)}{\mathrm{p}}=\frac{\mathrm{p}^{2}-2 \mathrm{q}}{\mathrm{p}}$
Product of zeroes $=\alpha \beta=p \times \frac{-2 q}{p}=-2 q$         or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-2 \mathrm{pq}}{\mathrm{p}}=-2 \mathrm{q}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 2 E 

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

x² – (2a + b)x + 2ab
Sol :
Let f(x) = x² – (2a + b)x + 2ab
f(x) = x² – 2ax – bx + 2ab
= x(x – 2a) – b(x – 2a)
= (x – 2a) (x – b)
On putting f(x) = 0 , we get
(x – 2a) (x – b) = 0
⇒ x – 2a = 0 or x – b = 0
⇒x = 2a or x = b
Thus, the zeroes of the given polynomial x² – (2a + b)x + 2ab are 2a and b
Verification
Sum of zeroes = α + β = 2a + b      or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-2 \mathrm{a}-\mathrm{b})}{1}=2 \mathrm{a}+\mathrm{b}$
Product of zeroes = αβ = 2a × b = 2ab       or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{2 \mathrm{ab}}{1}=2 \mathrm{ab}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 2 F 

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

r²s²x² + 6rstx + 9t²
Sol :
Let f(x) = r²s²x² + 6rstx + 9t²
Now, if we recall the identity
(a + b)² = a² + b² + 2ab
Using this identity, we can write
r²s²x² + 6rstx + 9t² = (rsx + 3t)²
On putting f(x) = 0 , we get
(rsx + 3t)² = 0

$\Rightarrow X=\frac{-3 t}{r s}, \frac{-3 t}{r s}$
Thus, the zeroes of the given polynomial r²s²x² + 6rstx + 9t² are $\frac{-3 t}{r s}$ and$\frac{-3 t}{r s}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-3 t}{r s}+\frac{-3 t}{r s}=-\frac{6 t}{r s}$       or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{6 \mathrm{rst}}{\mathrm{r}^{2} \mathrm{s}^{2}}=-\frac{6 \mathrm{t}}{\mathrm{rs}}$
Product of zeroes $=\alpha \beta=\frac{-3 \mathrm{t}}{\mathrm{rs}} \times \frac{-3 \mathrm{t}}{\mathrm{rs}}=\frac{9 \mathrm{t}^{2}}{\mathrm{r}^{2} \mathrm{s}^{2}}$          or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{9 \mathrm{t}^{2}}{\mathrm{r}^{2} \mathrm{s}^{2}}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 3 A 

Find the zeroes of the quadratic polynomial 5x² – 8x – 4 and verify the relationship between the zeroes and the coefficients of the polynomial.

Sol :
Let f(x) = 5x² – 8x – 4
By splitting the middle term, we get
f(x) = 5x² – 10x + 2x – 4
= 5x(x – 2) + 2(x – 2)
= (5x + 2) (x – 2)
On putting f(x) = 0 we get
(5x + 2) (x – 2) = 0
⇒ 5x + 2 = 0 or x – 2 = 0

⇒ X = $\frac{-2}{5}$ or x = 2
Thus, the zeroes of the given polynomial5x² – 8x – 4 are $\frac{-2}{5}$ and 2
Verification
Sum of zeroes $=\alpha+\beta=\frac{-2}{5}+2=\frac{-2+10}{5}=\frac{8}{5}$          or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{5}=\frac{8}{5}$
Product of zeroes $=\alpha \beta=\frac{-2}{5} \times 2=\frac{-4}{5}$                          or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-4}{5}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 3 B 

Find the zeroes of the quadratic polynomial 4 x² – 4x – 3 and verify the relationship between the zeroes and the coefficients of the polynomial.

Sol :
Let f(x) = 4 x² – 4x – 3
By splitting the middle term, we get
f(x) = 4 x² – 6x + 2x – 3
= 2x(2x – 3) + 1(2x – 3)
= (2x + 1) (2x – 3)
On putting f(x) = 0, we get
(2x + 1) (2x – 3) = 0
⇒ 2x + 1 = 0 or 2x – 3 = 0
⇒ X = $\frac{-1}{2}$  or  X = $\frac{3}{2}$
Thus, the zeroes of the given polynomial4x² – 4x – 3 are $\frac{-1}{2}$ and $\frac{3}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-1}{2}+\frac{3}{2}=\frac{-1+3}{2}=1$                or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-4)}{4}=1$
Product of zeroes $=\alpha \beta=\frac{-1}{2} \times \frac{3}{2}=\frac{-3}{4}$            or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{4}$
So, the relationship between the zeroes and the coefficients is verified.

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Question 3 C 

Find the zeroes of the quadratic polynomial √3 x² – 8x + 4√3 .

Sol :
Let f(x) = √3 x² – 8x + 4√3
By splitting the middle term, we get
(x) = √3 x² – 6x – 2x + 4√3
= √3 x(x – 2√3) – 2(x – 2√3)
= (√3 x – 2) (x – 2√3)
On putting f (x) = 0, we get
(√3 x – 2) (x – 2√3) = 0
⇒ √3 x – 2 = 0 or x – 2√(3 = 0)
$\Rightarrow \mathrm{X}=\frac{2}{\sqrt{3}}$ or X =  $2 \sqrt{3}$
Thus, the zeroes of the given polynomial√3 x² – 8x + 4√3 are $\frac{2}{\sqrt{3}}$ and $2 \sqrt{3}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{2}{\sqrt{3}}+2 \sqrt{3}=\frac{2+6}{\sqrt{3}}=\frac{8}{\sqrt{3}}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{\sqrt{3}}=\frac{8}{\sqrt{3}}$
Product of zeroes $=\alpha \beta=\frac{2}{\sqrt{3}} \times 2 \sqrt{3}=4$                    or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{4 \sqrt{3}}{\sqrt{3}}=4$
So, the relationship between the zeroes and the coefficients is verified.

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Question 4 

If α and β be the zeroes of the polynomial 2x² + 3x – 6, find the values of

(i) α² + β² (ii) α² + β² + αβ
(iii) α²β + αβ² (iv)$\frac{1}{\alpha}+\frac{1}{\beta}$
(v)$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ (vi) α – β
(vii) α³ + β³ (viii)$\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}$
Sol :
Let the quadratic polynomial be 2 x² + 3x – 6, and its zeroes are α and β.
We have
$\alpha+\beta=\frac{-b}{2}$ and $\alpha \beta=\frac{c}{a}$
Here, a = 2 , b = 3 and c = – 6
$\alpha+\beta=\frac{-b}{a}=\frac{-3}{2}$ ….(1)
$\alpha \beta=\frac{c}{a}=\frac{-6}{2}=-3$ ….(2)
(i) α² + β²
We have to find the value of α² + β²
Now, if we recall the identity
(a + b)² = a² + b² + 2ab
Using the identity, we get
(α + β)² = α² + β² + 2αβ
$\Rightarrow\left(\frac{-3}{2}\right)^{2}=\alpha^{2}+\beta^{2}+2(-3)$   {from eqⁿ (1) &(2)} $\Rightarrow \frac{9}{4}=\alpha^{2}+\beta^{2}-6$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{9}{4}+6$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{9+24}{4}=\frac{33}{4}$

(ii) α² + β² + αβ
$\alpha^{2}+\beta^{2}=\frac{33}{4}$ { from part (i)}
and, we have αβ = – 3
So, $\alpha^{2}+\beta^{2}+\alpha \beta=\frac{33}{4}+(-3)$
= $\frac{33-12}{4}$
= $=\frac{21}{4}$

(iii) α²β + α β²
Firstly, take $\alpha \beta$ common, we get
αβ(α + β)
and we already know the value of $\alpha \beta$ and $\alpha+\beta$.
So, α²β + α β² = αβ(α + β)
$=(-3)\left(\frac{-3}{2}\right)$ {from eqⁿ (1) and (2)}
$=\frac{9}{2}$

(iv) $\frac{1}{\alpha}+\frac{1}{\beta}$

Let’s take the LCM first then we get,
$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}$
$=\frac{\left(\frac{-3}{2}\right)}{-3}$
$=\frac{1}{2}$

(v) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
Let’s take the LCM first then we get,
$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\beta \alpha}$
$=\frac{\left(\frac{33}{4}\right)}{-3}$    {from part(i) and eqⁿ (2)}
$=\frac{-11}{4}$

(vi) $\alpha-\beta$
Now, recall the identity
(a – b)² = a² + b² – 2ab
Using the identity , we get
(α – β)² = α² + β² – 2 αβ
$\Rightarrow(\alpha-\beta)^{2}=\left(\frac{33}{4}\right)-2(-3)$ {from part(i) and eqⁿ (2)}
$\Rightarrow(\alpha-\beta)^{2}=\frac{33}{4}+6$
$\Rightarrow(\alpha-\beta)^{2}=\frac{33+24}{4}=\frac{57}{4}$
$\Rightarrow(\alpha-\beta)=\pm \frac{\sqrt{57}}{2}$

(vii) $\alpha^{3}+\beta^{3}$
Now, recall the identity
(a + b)³ = a³ + b³ + 3a² b + 3ab²
Using the identity, we get
⇒(α + β)³ = α³ + β³ + 3α² β + 3αβ²
$\Rightarrow\left(\frac{-3}{2}\right)^{3}=\alpha^{3}+\beta^{3}+3\left(\alpha^{2} \beta+\alpha \beta^{2}\right)$
$\Rightarrow \frac{-27}{8}=\alpha^{3}+\beta^{3}+3 \times \frac{9}{2}$
$\Rightarrow \alpha^{3}+\beta^{3}=\frac{-27-108}{8}$
$\Rightarrow \alpha^{3}+\beta^{3}=-\frac{135}{4}$

(viii) $\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}$
Let’s take the LCM first then we get,
$\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\beta \alpha}$
$=\frac{\left(\frac{-135}{8}\right)}{-3}$ {from part(vii) and eqⁿ (2)}
$=\frac{45}{8}$

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Question 5 

If α and β be the zeroes of the polynomial ax² + bx + c, find the values of

(i) $\mathrm{a}^{2}+\beta^{2}$
(ii) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
(ii) $\alpha^{3}+\beta^{3}$

Sol :
Let the quadratic poynomial be ax² + bx + c , and its zeroes be α and β.
We have
α + β = $\frac{-b}{a}$ and α β = $\frac{c}{a}$
(i) $\mathrm{a}^{2}+\beta^{2}$
We have to find the value of $\alpha^{2}+\beta^{2}$
Now, if we recall the identity
(a + b)² = a² + b² + 2ab
Using the identity, we get $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$
$\Rightarrow\left(\frac{-b}{a}\right)^{2}=\alpha^{2}+\beta^{2}+2 \times \frac{c}{a}$ {from eqⁿ (1) & (2)}
$\Rightarrow \frac{b^{2}}{a^{2}}=\alpha^{2}+\beta^{2}+\frac{2 c}{a}$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a}$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{b^{2}-2 c a}{a^{2}}$

(ii) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
Let’s take the LCM first then we get,
$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\beta \alpha}$
$=\frac{\left(\frac{b^{2}-2 c a}{a^{2}}\right)}{\frac{c}{a}}$ {$\because \alpha \beta=\frac{c}{a}$}

$=\frac{b^{2}-2 c a}{c a}$

(iii) $\alpha^{3}+\beta^{3}$
Now, recall the identity
(a + b)³ = a³ + b³ + 3a² b + 3ab²
Using the identity, we get
⇒(α + β)³ = α³ + β³ + 3α² β + 3αβ²
$\Rightarrow\left(\frac{-b}{a}\right)^{3}=\alpha^{3}+\beta^{3}+3\left(\alpha^{2} \beta+\alpha \beta^{2}\right)$
$\Rightarrow\left(\frac{-b^{3}}{a^{3}}\right)=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta)$
$\Rightarrow \alpha^{3}+\beta^{3}=\left(\frac{-b^{3}}{a^{3}}\right)+3 \times \frac{c}{a} \times\left(\frac{-b}{a}\right)$
$\Rightarrow \alpha^{3}+\beta^{3}=\frac{3 a b c-b^{3}}{a^{3}}$

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Question 6 

If α, β are the zeroes of the quadratic polynomial x² + kx = 12, such that α – β = 1, find the value of k.

Sol :
The given quadratic polynomial is x² + kx = 12 and $\alpha{-} \beta$ = 1
If we rearrange the polynomial then we get
p(x) = x² + kx – 12
We have,
$\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta=\frac{c}{a}$
So, $\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}=\frac{-\mathrm{k}}{1}=-\mathrm{k}$ …(1)
$\alpha \beta=\frac{c}{a}=\frac{-12}{1}=-12$ …(2)
Now, if we recall the identities
(a + b)² = a² + b² + 2ab
Using the identity, we get
(α + β)² = α² + β² + 2 αβ
( – k)² = α² + β² + 2( – 12)
⇒ α² + β² = k² + 24 …(3)
Again, using the identity
(a – b)² = a² + b² – 2ab
Using the identity, we get
(α – β)² = α² + β² – 2 αβ
(1)² = α² + β² – 2( – 12) {∵ (α – β) = 1}
⇒ α² + β² = 1 – 24
⇒ α² + β² = – 23 …(4)
From eqⁿ (3) and (4), we get
k² + 24 = – 23
⇒ k² = – 23 – 24
⇒ k² = – 47
Now the square can never be negative, so the value of k is imaginary.

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Question 7 

If the sum of squares of the zeroes of the quadratic polynomial x² – 8x + k is 40, find k.

Sol :
Given : p(x) = x² – 8x + k
α² + β² = 40
We have,
$\alpha+\beta=\frac{-b}{a}$
and $\alpha \beta=\frac{c}{a}$
So, $\alpha+\beta=\frac{-b}{a}=\frac{-(-8)}{1}=8$ …(1)
$\alpha \beta=\frac{c}{a}=\frac{k}{1}=k$ …(2)
Now, if we recall the identities
(a + b)² = a² + b² + 2ab
Using the identity, we get
(α + β)² = α² + β² + 2 αβ
(8)² = 40+ 2(k)
⇒2k = 64 – 40
$\Rightarrow \mathrm{k}=\frac{24}{2}$= 12

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Question 8 A 

If one zero of the polynomial (α² + 9) x² + 13x + 6α is reciprocal of the other, find the value of a.

Sol :
Let one zero of the given polynomial is $\alpha$
According to the given condition,
The other zero of the polynomial is $=\frac{1}{a}$
We have,
Product of zeroes, $\alpha \beta=\frac{c}{a}=\frac{6 \alpha}{\alpha^{2}+9}$
$\Rightarrow \alpha \times \frac{1}{\alpha}=\frac{6 \alpha}{\alpha^{2}+9}$
⇒ α² – 6α + 9 = 0
⇒ α² – 3α – 3α + 9 = 0
⇒ α(α – 3) – 3(α – 3) = 0
⇒(α – 3)(α – 3) = 0
⇒(α – 3) = 0 & (α – 3) = 0
⇒ α = 3, 3

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Question 8 B 

If the product of zeroes of the polynomial α² – 6x – 6 is 4, find the value of a.

Sol :
Given Product of zeroes, α β = 4
p(x) = α x² – 6x – 6
to find: value of α
We know,
Product of zeroes , $\alpha \beta=\frac{c}{a}=\frac{-6}{\alpha}$
$\Rightarrow 4=\frac{-6}{\alpha}$
$\Rightarrow \alpha=\frac{-6}{4}=\frac{-3}{2}$

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Question 8 C 

If (x + a) is a factor 2x² + 2ax + 5x + 10, find a.

Sol :
Given x + a is a factor of $2 x^{2}+2 a x+5 x+10$
So,g(x) = x + a
x + a = 0
⇒x = – a
Putting the value x = – a in the given polynomial, we get
2( – a)² + 2a( – a) + 5( – a) + 10 = 0
2a² – 2a² – 5a + 10 = 0
– 5a + 10 = 0
$a=\frac{-10}{-5}$
a = 2

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Question 9 A 

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

1,1
Sol :
Given: Sum of zeroes = α + β = 1
Product of zeroes = αβ = 1
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (1)x + 1
= x² – x + 1

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Question 9 B 

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

0,3
Sol :
Given: Sum of zeroes = α + β = 0
Product of zeroes = αβ = 3
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (0)x + 3
= x² + 3

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Question 9 C 

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

$\frac{1}{4},-1$
Sol :
Given: Sum of zeroes = α + β = 1/4
Product of zeroes = αβ = – 1
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= $X^{2}-\left(\frac{1}{4}\right) x+(-1)$
= $x^{2}-\frac{x}{4}-1$
= $\frac{4 x^{2}-x-4}{4}$
We can consider 4x² – x – 4 as required quadratic polynomial because it will also satisfy the given conditions.

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Question 9 D 

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

4,1
Sol :
Given: Sum of zeroes = $\alpha+\beta$ = 4
Product of zeroes = $\alpha \beta$ = 1
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (4)x + 1
= x² – 4x + 1

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Question 9 E 

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

$\frac{10}{3},-1$
Sol :
Given: Sum of zeroes $=\alpha+\beta=\frac{10}{3}$
Product of zeroes = αβ = – 1
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= $x^{2}-\left(\frac{10}{3}\right) X+(-1)$
= $X^{2}-\frac{10 x}{3}-1$
= $\frac{3 x^{2}-10 x-3}{3}$
We can consider 3x² – 10x – 3 as required quadratic polynomial because it will also satisfy the given conditions.

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Question 9 F 

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

$-\frac{1}{2},-\frac{1}{2}$
Sol :
Given: Sum of zeroes $=\alpha+\beta=-\frac{1}{2}$
Product of zeroes $=\alpha \beta=-\frac{1}{2}$
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= $X^{2}-\left(-\frac{1}{2}\right) X+\left(-\frac{1}{2}\right)$
= $X^{2}+\frac{x}{2}-\frac{1}{2}$
= $\frac{2 x^{2}+x-1}{2}$
We can consider 2x² + x – 1 as required quadratic polynomial because it will also satisfy the given conditions.

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Question 10 A 

Find quadratic polynomial whose zeroes are :

3, – 3
Sol :
Let the zeroes of the quadratic polynomial be
α = 3 , β = – 3
Then, α + β = 3 + ( – 3) = 0
αβ = 3 × ( – 3) = – 9
Sum of zeroes = α + β = 0
Product of zeroes = αβ = – 9
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (0)x + ( – 9)
= x² – 9

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Question 10 B 

Find quadratic polynomial whose zeroes are :

$\frac{2+\sqrt{5}}{2}, \frac{2-\sqrt{5}}{2}$

Sol :
Let the zeroes of the quadratic polynomial be
$\alpha=\frac{2+\sqrt{5}}{2}, \beta=\frac{2-\sqrt{5}}{2}$
Then, $\alpha+\beta=\frac{2+\sqrt{5}}{2}+\frac{2-\sqrt{5}}{2}=\frac{2+\sqrt{5}+2-\sqrt{5}}{2}=2$

$\alpha \beta=\frac{2+\sqrt{5}}{2} \times \frac{2-\sqrt{5}}{2}$

$=\frac{(2+\sqrt{5})(2-\sqrt{5})}{4}=\frac{4-5}{4}$

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

Sum of zeroes = α + β = 2
Product of zeroes = $\alpha \beta$ = $-\frac{1}{4}$
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= $X^{2}-(2) X+\left(-\frac{1}{4}\right)$
= $X^{2}-2 X-\frac{1}{4}$
= $\frac{4 x^{2}-8 x-1}{2}$
We can consider 4x² – 8x – 1 as required quadratic polynomial because it will also satisfy the given conditions.

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Question 10 C 

Find quadratic polynomial whose zeroes are :

$3+\sqrt{7}, 3-\sqrt{7}$
Sol :
Let the zeroes of the quadratic polynomial be
α = 3 + √7, β = 3 – √7
Then, α + β = 3 + √7 + 3 – √7 = 6
αβ = (3 + √(7)) × (3 – √7) = 9 – 7 = 2
Sum of zeroes = α + β = 6
Product of zeroes = αβ = 2
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (6)x + 2
= x² – 6x + 2

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Question 10 D 

Find quadratic polynomial whose zeroes are :

$1+2 \sqrt{3}, 1-2 \sqrt{3}$

Sol :
Let the zeroes of the quadratic polynomial be
α = 1 + 2√3, β = 1 – 2√3
Then, α + β = 1 + 2√3 + 1 – 2√3 = 2
αβ = (1 + 2√3) × (1 – 2√3) = 1 – 12 = – 11
Sum of zeroes = α + β = 2
Product of zeroes = αβ = – 11
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (2)x + ( – 11)
= x² – 2x – 11

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Question 10 E 

Find quadratic polynomial whose zeroes are :

$\frac{2-\sqrt{3}}{3}, \frac{2+\sqrt{3}}{3}$

Sol :
Let the zeroes of the quadratic polynomial be
$\alpha=\frac{2-\sqrt{3}}{3}, \beta=\frac{2+\sqrt{3}}{3}$
Then, $\alpha+\beta=\frac{2-\sqrt{3}}{3}+\frac{2+\sqrt{3}}{3}=\frac{2-\sqrt{3}+2+\sqrt{3}}{3}=\frac{4}{3}$
$\alpha \beta=\frac{2-\sqrt{3}}{3} \times \frac{2+\sqrt{3}}{3}=\frac{(2-\sqrt{3})(2+\sqrt{3})}{9}=\frac{4-3}{9}=\frac{1}{9}$
Sum of zeroes $=\alpha+\beta=\frac{4}{3}$
Product of zeroes $=\alpha \beta=\frac{1}{9}$
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= $X^{2}-\left(\frac{4}{3}\right) X+\left(\frac{1}{9}\right)$
= $X^{2}-\frac{4}{3} X+\frac{1}{9}$
= $\frac{9 x^{2}-12 x+1}{2}$
We can consider 9x² – 12x + 1 as required quadratic polynomial because it will also satisfy the given conditions.

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Question 10 F 

Find quadratic polynomial whose zeroes are :

$\sqrt{2}, 2 \sqrt{2}$
Sol :
Let the zeroes of the quadratic polynomial be
α = √2, β = 2√2
Then, α + β = √2 + 2√2 = √2 (1 + 2) = 3√2
αβ = √2× 2√2 = 4
Sum of zeroes = α + β = 3√2
Product of zeroes = αβ = 4
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (3√2)x + 4
= x² – 3√2 x + 4

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Question 11 

Find the quadratic polynomial whose zeroes are square of the zeroes of the polynomial x² – x – 1.

Sol :
Let the zeroes of the polynomial x² – x – 1 be α and β
We have,
$\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}$
and $\alpha \beta=\frac{c}{a}$
So,
$\alpha+\beta=\frac{-b}{a}=-\frac{-1}{1}=1$
$\alpha \beta=\frac{c}{a}=\frac{-1}{1}=-1$
Now, according to the given condition,
α² β² = ( – 1)² = 1
& (α + β)² = α² + β² + 2 α β
 α² + β² = (α + β)² – 2αβ
 α² + β² = (1)² – 2( – 1)
 α² + β² = 3
So, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (3)x + 1
= x² – 3x + 1

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Question 12 A 

If α and β be the zeroes of the polynomial x² + 10x + 30, then find the quadratic polynomial whose zeroes are α + 2β and 2α + β.

Sol :
Given : p(x) = x² + 10x + 30
So, Sum of zeroes $=\alpha+\beta=\frac{-b}{a}=\frac{-10}{1}=-10$ …(1)
Product of zeroes $=\alpha \beta=\frac{c}{a}=\frac{30}{1}=30$ …(2)
Now,
Let the zeroes of the quadratic polynomial be
α^' = α + 2β , β' = 2α + β
Then, α’ + β’ = α + 2β + 2α + β = 3α + 3β = 3(α + β)
α'β’ = (α + 2β) ×(2α + β) = 2α² + 2β² + 5αβ
Sum of zeroes = 3(α + β)
Product of zeroes = 2α² + 2β² + 5αβ
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (3(α + β))x + 2α² + 2β² + 5αβ
= x² – 3( – 10)x + 2 (α² + β²) + 5(30) {from eqⁿ (1) & (2)}
= x² + 30x + 2(α² + β² + 2αβ – 2αβ) + 150
= x² + 30x + 2 (α + β)² – 4αβ + 150
= x² + 30x + 2( – 10)² – 4(30) + 150
= x² + 30x + 200 – 120 + 150
= x² + 30x + 230
So, the required quadratic polynomial is x² + 30x + 230

---

Question 12 B 

If α and β be the zeroes of the polynomial x² + 4x + 3, find the quadratic polynomial whose zeroes are $1+\frac{\alpha}{\beta}$ and $1+\frac{\beta}{\alpha}$.

Sol :
Given : p(x) = x² + 4x + 3
So, Sum of zeroes $=\alpha+\beta=\frac{-b}{a}=\frac{-4}{1}=-4$ …(1)
Product of zeroes $=\alpha \beta=\frac{c}{a}=\frac{3}{1}=3$ …(2)
Now,
Let the zeroes of the quadratic polynomial be
$\alpha^{\prime}=1+\frac{\alpha}{\beta}, \beta^{\prime}=1+\frac{\beta}{\alpha}$
Then,

$\alpha^{\prime}+\beta^{\prime}=1+\frac{\alpha}{\beta}+1+\frac{\beta}{\alpha}$

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

$\alpha^{\prime} \beta^{\prime}=\left(1+\frac{\alpha}{\beta}\right) \times\left(1+\frac{\beta}{\alpha}\right)$

$=1+\frac{\beta}{\alpha}+\frac{\alpha}{\beta}+1=\frac{2 \alpha \beta+\alpha^{2}+\beta^{2}}{\alpha \beta}$

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

Sum of zeroes $=\frac{(\alpha+\beta)^{2}}{\alpha \beta}$

Product of zeroes $=\frac{(\alpha+\beta)^{2}}{\alpha \beta}$

Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
$=\mathrm{x}^{2}-\frac{(\alpha+\beta)^{2}}{\alpha \beta} \mathrm{x}+\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
$=x^{2}-\frac{(4)^{2}}{3} x+\frac{(4)^{2}}{3}$ {from eqⁿ (1) & (2)}
$=x^{2}-\frac{16}{3} x+\frac{16}{3}$
$=\frac{3 x^{2}-16 x+16}{3}$
So, the required quadratic polynomial is 3x² – 16x + 16

---

Question 13 A 

Find a quadratic polynomial whose zeroes are 1 and – 3. Verify the relation between the coefficients and zeroes of the polynomial.

Sol :
Let the zeroes of the quadratic polynomial be
α = 1 , β = – 3
Then, α + β = 1 + ( – 3) = – 2
αβ = 1 × ( – 3) = – 3
Sum of zeroes = α + β = – 2
Product of zeroes = αβ = – 3
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – ( – 2)x + ( – 3)
= x² + 2x – 3
Verification
Sum of zeroes = α + β = 1 + ( – 3) = – 2 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(2)}{1}=-2$

Product of zeroes = αβ = (1)( – 3) = – 3 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{1}=-3$

So, the relationship between the zeroes and the coefficients is verified.

---

Question 13 B 

Find the quadratic polynomial sum of whose zeroes in 8 and their product is 12. Hence find the zeroes of the polynomial.

Sol :
Given: Sum of zeroes = α + β = 8
Product of zeroes = αβ = 12
Then, the quadratic polynomial
= x² – (sum of zeroes)x + product of zeroes
= x² – (8)x + 12
= x² – 8x + 12
Now, we have
$\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta=\frac{c}{a}$
So,
$\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}=-\frac{-8}{1}=8$
$\alpha \beta=\frac{c}{a}=\frac{12}{1}=12$

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S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1