Mathematics 10 (Science) - KPK - Gain - Solution

Mathematics 10 (Science) - KPK - Gain - Solution

Solutions of the Textbook of Mathematics for Grade X, published by The Khyber Pakhtunkhwa Textbook Board, Peshawar, Pakistan are given here.

There are thirteen (13) chapters in this book.

Solutions

Chapters 01: Quadratic Equations

CHAPTER - 1

Exercise 1.1 

Chapter 1 

Exp 2: A ball is thrown straight up, from 3m above the ground with a velocity of 14m/s. Gravity pulls it down as 5m per square second. Quadratic Equations 

When does it hit the ground. Quadratic equation in one variable 

Solution: Height start at 3m = 3 or Second degree equation:-An equation of the 

With Velocity 14m/sec = 14t form ax 2 + bx + c = 0 , abc , , R∈ where a ≠ 0 is 

De-acceleration 5m/sec2 = - 5t 2 called quadratic equation. Or 

Add them up and height h at any time t is An eq of degree 2 is called quadratic equation. 

h = 3 + 14 t - 5 t 2 Solution set of the equation:The set of those 

And ball will hit the ground when height is zero values of variable for which the equation is true, is called solutions/roots set of the equation. Roots of the equation: Those values of the variables in which the given equation is true is 

3 + 14 t - 5 t 2 = 0 or 5 t 2 - 14 t - 3 = 0 5 t 2 - 15 t + t 

- 3 =

0 called roots of the equation. Solution of Quadratic equation: We may solve a quadratic equation by any one 

5 t ( t - 1 ) + 1 ( t 

- 3 ) 

= 

0 ( 5 t + 1 )( t 

- 1 ) 

= 

0 of the following three methods: 1. By Factorization 2. By Completing the Square 

Either 5 t+ 1 = 

0 5 t 

= - 1 

or t- 1 = 

0 t 

= 

1 3. By the Quadratic Formula Solution of Quadratic equation By Factorization: 

t = - 15 which is impossible In Quadratic eq ax 2 + bx + c = 0 we find out 

Therefore ball hits the ground after time t = 3 sec product of a(coefficient of x 2) and c (constant 

Solution of Quadratic equation By term) i.e. .ac . Then we find two numbers b 1and b 2such that b 1 . b 2 = ac . and b 1 + b 2 = b . Hence ax 2 + bx + c = ax 2 

+ ( b 1 + b 2 ) x + c then we 

Completing square: 1.Write the equation in standard form. 2.Divide the equation by coefficient of x 2(if other then 1) factorize after grouping. Or Solution of Quadratic equation By Factorization: 

3.Shift the constant to right hand side 4.Add square of half of coefficient ofxboth sides 1.Write the equation in standard form. 

5.Simplify R.H.S to make L.H.S as a perfect square. 2.Factorize the equation. 

6.Take square root on both sides & solve 3.Equate each factor to zero separately. 4.Solve equations for values of given variables. Exp 1i). Solve 2 x 2 + 2 x - 11 = 1 by factorization Solution: we have 2 x 2 + 2 x - 11 = 

1 2 x 2 + 2 x - 11 - 1 = 0 2 x 2 + 2 x - 12 = 0 divided by 2 x 2 + x - 6 = 0 Standard form 

Exp3:Solve x 2 - 8 x + 9 = 0 by completing square Solution: we have x 2 - 8 x + 9 = 

0 x 2 - 8 x = - 9 x 2- ( x )( 8 ) 

= - 

9 

x 2 

- 2 ( x 

) 

⎛ │ ⎝ 8 2 

⎞ │ ⎠ = - 9 x 2 + 3 x - 2 x 

- 6 = 

0 

x 2 

- 2 ( x 

)( 4 ) + ( 4 ) 2 = - 9 +

( 4 

) 2 x ( x + 3 ) - 2 ( x 

+ 3 ) 

= 

0 ( x - 2 )( x 

+ 3 ) 

= 

0 ∴ x- 2 = 

0 x 

= 2 

or x+ 3 = 

0 x= - 

3 

( x- 4 ) 

2= - 9 + 

16 ( x 

- 4 ) 

2 

= 

7 Taking square root on both sides 

The solution set = { 2, - 3 } Exp 1ii). Solve 12 t 2 t= + 1 by factorization Solution: we have 12 t 2 t= + 1 form 

( - 4 )2 = 

7 

- 4 = ± 

7 

12 t 2 t- - 1 = 0 Standard form 

Solution set = { 4 = + 4 ± 

7 7,4 - 7 } Solution of Quadratic equation By Formula: ( ) ( ) ( )( ) 

1.Write the equation in standard form. 2.Take the values of abc , , by comparing the standard equation. 

Put in the formula 

x 

xx 

12 2 - 4 + 3 - 1 = 

0 4 3 - 1 + 13 - 1 = 

0 

4 + 1 3 - 1 = 

0 

14 

t t t t t t 

t t 

∴ 4 t 

+ 1 = 

0 

3 t- 1 = 

0 

4 t 

= - 

1 

3 t 

=1 t = 

- 

t 

= Solution set 4 1 , 13 

or 

x = - b ± b 2 - 4 

ac and solve 2 

a 1the R.H.S to get roots. 3 = -⎧ ⎨ ⎩ ⎫ ⎬ ⎭ 

2 Chapter 1 

Exercise 1.1 

Derivation of Quadratic Formula 

Exp5: A company is making frames as part of a We have Quadratic equation 

new product they are launching. The frame will ax 2 + bx + c = 0, a ≠ 0 ax 2 + bx = - c or x 2 + b a x = - ac dividing by a be cut of a piece of steel. To keep the weight down, the final area should be 28 cm2. The inside 

of a frame has to be 11cm by 6cm. what should the width x of a metal be? Solution: Let width of 

frame = xcm x 2 + ( x ) ⎛ │ ⎝ b a ⎞ │ ⎠ = - ac x 2 + 2 ( x ) ⎛ │ ⎝ 2b a ⎞ │ ⎠ = - ac x 2 

+ 2 ( x ) 

⎛ │ ⎝ 2 b a b 2 b 2 

c According to condition a a a 

Length = 11 + 2x width = 6 + 2x x b a ⎞ │ ⎠ + ⎛ │ ⎝ 2 ⎞ │ ⎠ = ⎛ │ ⎝ 2 

⎞ │ ⎠ - ⎛ │ ⎝ + 2 ⎞ │ ⎠ 2 = 4 b a 22 

- a c × 4 4 

a a 

So Area covered by frame = ( 11 + 2 x )( 6 + 

2 x ) = 66 + 22 x + 12 x + 

4 

x 

2 

⎛ │ ⎝ x + 2 b a ⎞ │ ⎠ 

2 = b 2 

4 - a4 

2ac = 66 + 34 x + 

4 

x 

2 

Total area = Area of inside rectangle + area of frame 

= 11 × 6 + 

28 ⎛ │ ⎝ x + b a 2 2 

2 

66 28 94 

22 

⎞ │ ⎠= - = + 

=

+ = 

± - So, both area are equal 

= - ± 

- 2 4 

( ) 2 

b a ac x b a b a ac x b a b a 4 2 44 2 2 

4 ac 2 2 

4 x 2+ 34 x 

+ 66 = 

94 4 x 2+ 34 x 

+ 66 - 94 = 

0 

4 x 2 

+ 34 x 

- 28 = 

0 

x = - b ± b a 

- ac required formula 

2 2 x 2 

+ 17 x 

- 14 = 

0 

Or 2 x 2 + 17 x - 14 = 0 Exp4: Solve 3 x 2 - 6 x + 2 = 0 by quadratic formula 

By comparing with ax 2 + bx + c = 0 Solution: We have 3 x 2 - 6 x + 2 = 

0 Here a = 2,b = 17,c = - 14 using By comparing with ax 2 + bx + c = 

0 2 4 Here a = 3,b = - 6,c = 2 using 

2 2 4 2 

x = - b ± b a 

- ac putting the values 

x = - b ± b a 

- ac putting the values 

( 17 ) ( 17 ) 2 4 ( 2 )( 14 

) 2 ( 2 

) ( ) ( ) ( )( ) 

17 289 112 ( ) 

4 17 401 

4 

( ) 

x= - ± - - x= 

- - 6 ± - 6 2 - 4 3 2 2 3 

x= - ± + x= 

6 ± 36 - 24 6 

x= 

6 ± 6 

x 

= - ± ∴ x = - 17 - 4 12 401 

or x = 

- 17 + 4 401 x = - 9.3 approximate x = 0.8 approximate 

x= 

6 ± 6 

4 × 3 

x= 6 ± 6 2 3 

= 

2 3 ± 6 

As distance can not be negative so width of frame = 0.8 cm 3 Exercise 1.1 

Q1. Solve x 2+ 5x + 4 = 0 by factorization x= 3 ± 3 Sol: Since a=+1, b=+5, c=+4 

( ) ( ) ( )( ) 3 = 3 3 ± 

3 3 

x 

= 1 ± 

3 3 The solution set 1 3 3 ,1 3 

3 x 2+ 5x + 4 = 

0 R.W x 2+ 4x + x + 4 = 0 x x + 4 + 1 x + 4 = 

0 

+ a.c Sign of b 

x + 1 x + 4 = 

0 

× Result 

+ 4 + × + = ⎧ │ ⎨ │ ⎩ + - ⎫ │ ⎬ │ ⎭ 

4 × 1 Either x + 1 = 0 Or x + 4 =0 

x = - 1 x = - 4 Solution Set = { - 1, - 

4 } 

3 Chapter 1 

Exercise 1.1 

Q1ii). Solve ( x - 3 )2 = 4 by factorization 

3x 2 - 5x = x 2 

- 13x + 

42 

Sol: Since ( x - 3 )2 = 4 x 26x 9 4 

3x 2 - x 2 

- 5x + 13x - 42 = 

0 

- + = 

2x 2 + 8x - 42 = 0 Divided by 2 

x 2- 6x + 9 - 4 = 

0 x 2- 6x + 5 = 

0 

x 2+ 4x - 21 = 0 So a = + 1, b = + 4, c = - 21 

So a = + 1, b = - 6, c = + 5 

( ) ( ) ( ( )( ) ( ) 

x 2+ 4x - 21 = 

0 

R.W. - 21 

x 2- 6x + 5 = 

0 

R.W. + 5 

x 2+ 7x - 3x - 21 = 

0 

x 2- 5x - 1x + 5 = 

0 

x x + 7 - 3 x + 7 = 

0 ) 

( )( ) 

+ × - - × - 

x x - 5 - 1 x - 5 = 

0 

5 × 1 

x - 3 x + 7 = 

0 

5 × 1 6 × 2 7 × 3 x - 1 x - 5 = 

0 

Either x + 7 = 0 Or x - 3 =0 Either x - 1 = 0 Or x - 5 =0 

Solution x Set = 1 = { 1,5 

} x= 5 

x = - 7 Solution Set = { 3, 7- 

} x= 3 

Q2i).Solve x 2+ 6x - 40 = 0 by completing square Q1iii). Solve x 2+ 3x - 10 = 0 by factorization 

Sol: Since x 2+ 6x - 40 = 0 Sol: Since a = + 1, b = + 3, c = - 10 

x 2+ 3X - 10 = 

0 

R.W. - 10 

x 2+ 5x - 2x - 10 = 

0 

x 2 

+ 6x = 

40 ( x ) 2+ ( x )( 6 ) 

= 

40 

x ( x + 5 ) - 2 ( x + 5 ) 

= 

0 ( )( ) 

+ × - 4 × 1 5 × 2 x - 2 x + 5 = 

0 Either x – 2 = 0 Or x + 5 =0 

x =2 x= - 5 

( x ) 2+ 2 ( x ) 

⎛ │ ⎝ 2 

6 ⎞ │ ⎠ = 40 ( x ) 2+ 2 ( x )( 3 ) 

= 

40 ( x ) 2 + 2 ( x )( 3 ) + ( 3 ) 2 = 40 + 

( 3 ) 

2 

Solution Set = { 2, - 5 } Q1iv).Solve 6x 2 - 13x + 5 = 0 by factorization Sol: Since a = + 6, b = - 13, c = + 5 

( x + 3 ) 

2= 40 + 

9 ( x + 3 ) 

2 

= 

49 

6x 2- 13x + 5 = 

0 

R.W. + 30 

Taking square root on both sides x6 2 

- 10x - 3x + 5 = 

0 2x ( 3x - 5 ) - 1 ( 3x - 5 ) 

= 

0 ( )( ) 

- × - 12 × 1 

( x 3 )2 49 11 × 2 2x - 1 3x - 5 = 

0 

10 × 3 Either 2x – 1 = 0 Or 3x - 5 =0 

x 3 7 x 3 7 

2x =1 3x= 5 x = 12 x = 53 Solution Set 1 2 ,53 

+ = ± 

+ = ± = - ± Either or X = -3 + 7 x = - 3 – 7 X = 4 Solution set { 4, - 

10 } x = -10 = ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ Q1v) Solve 3 ( x 2 - 1 ) = 4 ( x + 1 ) by factorization 

Q2ii).Solve x 2- 10x + 11 = 0 by completing square Sol: Since x 2- 10x + 11 = 0 x 2 

- 10x = - 

11 Sol: Since 3 ( x 2 - 1 ) = 4 ( x + 

1 ) ( x ) 2- ( x )( 10 ) 

= - 

11 

3x 2- 3 = 4x + 

4 

3x 2 

- 4x - 4 - 3 = 

0 

( x ) 2- 2 ( x ) 

⎛ │ ⎝ 10 2 ( ) 2 

( )( ) 

⎞ │ ⎠ = - 11 3x 2 - 4x - 7 = 

0 x - 2 x 5 = - 

11 So a = + 3, b = - 4, c = - 7 

( ) - ( )( ) + ( ) = - + 

( ) ( - ) 

= - + ( 3x 2- 4x - 7 = 

0 

3x 2 

- 7x + 3x - 7 = 

0 x ( 3x - 7 ) + 1 ( 3x - 7 ) 

= 

0 x + 1 )( 3x - 7 ) 

= 

0 

R.W. - 21 

x 2 2 x 5 5 2 11 5 

2 

- × + 5 × 1 6 × 2 7 × 3 

x 5 211 25 ( x - 5 )2 

= 

14 Taking square root on both sides Either x + 1 = 0 Or 3x - 7 =0 

x = - 1 3x= 7 

( x 5 )2 14 

Solution Set = ⎧ -⎨ ⎩ 1, 7 3 

x 5 14 

x 5 14 ⎫ ⎬ ⎭- = ± 

- = ± 

= ± Either X = 5 + Solution x = 73 1vi Solve x ( 3x - 5 ) = ( x - 6 )( x - 7 ) by factorization Sol: Since x ( 3x - 5 ) = ( x - 6 )( x - 

7 ) 14 set { or 5 + 14,5 x = - 

5 - 14 } 14 

3x 2 - 5x = x 2 - 7x - 6x + 42 

4 Chapter 1 

Exercise 1.1 

Q2iii).Solve 4x 2 + 12x = 0 by completing square Sol: Since 4x 2 + 12x = 0 Divided by 4 x 2+ 3x = 0 ( ) + ( )( ) = 

x 2 - 2 3 x = - 81 5 ( ) - ( ) ⎛ │ ⎝ ⎞ │ ⎠ = - ( ) - ( ) ⎛ │ ⎝ × ⎞ │ ⎠ = - ( ) - ( ) 

⎛ │ ⎝ ⎞ │ ⎠ = - ( ) - ( )⎛ │ ⎝ x 2x 2 3 ⎞ │ ⎠ + ⎛ │ ⎝  ⎞ │ 

⎠ = - + ⎛ │ ⎝ ⎞ │ ⎠ ⎛ │ ⎝ - ⎞ │ ⎠ = - + × ⎛ │ ⎝ 5 81 x 22 x 2 - ⎞ │ ⎠ x 2 x 3 0 2 5 ( x ) 2 + 2

( x )⎛ │ ⎝ 2 

3 ⎞ │ ⎠ = 0 3 ( x ) 2 

+ 2 ( x )⎛ │ ⎝ = - + = 81 

x 2 2 x 1 3 5 3 2 ⎞ │ ⎠ 81 

x 2 2 x 1 3 + ⎛ │ ⎝ ⎞ │ ⎠ = + ⎛ │ ⎝ 1 2 5 1 2 

3 81 3 x 1 23 ⎞ │ ⎠ ⎛ │ ⎝ 5 1 9 81 9 9 

x 1 2 3 + ⎞ │ ⎠ 5 9 4 81 81 

⎛ │ ⎝ = ⎛ │ ⎝ x - 3 1 ⎞ │ ⎠2 = ± 4 81 Taking square root 

x - 3 1 = ± 2 9 x = 3 1 ± 2 9 = 3 9 ± 2 Either or 

=- = ⎞ │ ⎠ 

2 2 

2 2 

3 2 0 3 2 

x 3 2 3 2 Taking square root on both sides 

⎛ │ ⎝ x + 2 3 ⎞ │ ⎠ 2 = ± ⎛ │ ⎝ ⎞ │ ⎠ 

+ = ± = - ±3 2 2 x 3 2 3 2 x 3 3 2 2 Either or 

x = - + 

x = 

3 2 0 

3 2 

x 3 2 Solution set = { 0, x 3- } 2 

6 3 2 3 Q2ivSolve 5x 2 - 10x - 840 = 0 completing square Sol: Since 5x 2 - 10x - 840 = 0 Divided by 5 

( ) ( )( ) ( ) ( ) 

( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 

= - - 

= - = - 

x 3 9 2 x =3 9+ 2 x 1 9 

x = 

5 9 

x 2- 2x - 168 = 

0 

Solution set = ⎧ ⎨ ⎩ 9 1 ,59 

x 2 

- 2x = 

168 

⎫ ⎬ ⎭ Q2vi). Solve( x - 1 )( x + 3 ) = 5 ( x + 2 ) - 3 by 

x 2- x 2 = 

168 x 2- 2 x ⎛ │ ⎝ 2completing Sol: Since ( square 

x - 1 )( x + 3 ) = 5 ( x + 2 ) - 3 x 23x x 3 5x 10 3 x 23x x 5x 10 3 3 0 

2 2 2 

x 23x 10 0 

22 

2 2 

⎞ │ ⎠ = 168 x - 2 x 1 = 

168 

x - 2 x 1 + 1 = 168 + 

1 

x - 1 = 168 + 

1 

x - 1 = 

169 Taking square root on both sides 

( x 1 )2 169 x 1 13 x 1 13 

+ - - = + - + - - - + - = - - = ( ) 2 

- ( )( ) 

= 

( ) 2 

- ( ) ⎛ │ ⎝ x x 

3 10 

x 2 x 

3 2 2 2 

2 

⎞ │ ⎠ + ⎛ │ ⎝ ⎞ │ ⎠ = + ⎛ │ ⎝ ⎞ │ ⎠ 

⎛ │ ⎝ x - ⎞ │ ⎠ = + = + ⎛ │ ⎝ x- 3 2 ⎞ │ ⎠ 2 = 49 

4 Taking square root on bs 

3 2 49 2 4 3 7 2 2 3 7 3 7 2 2 2 

3 2 10 3 2 

3 2 10 9 4 40 9 - = ± 

4 

- = ± = ± Either or X = 1 + 13 x = 1 - 13 X = 14 Solution set x= ={ -12 - 12,14 } Q2v Solve 9x 2 - 6x + 5 9 = 0 completing square Sol: Since 9x 2 - 6x + 5 9 = 0 Divided by 9 

- = - × 

x xx⎛ │ ⎝ - ⎞ │ ⎠ = - = ±

= ± = ± Either x = 3 - 2 7or x = 3 + 2 7x 2 6 9 x 5 9 9 x = - 2 4 = - 2 x = 10 2 = 5 Solution set = { - 2,5 } 

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