Mathematics 10 (Science) - KPK - Gain - Solution - PART - 29 - 42 Chapter 2

42 Chapter 2 

Review Exercise 2 

Q10ii). Solve the system 

Activity 1: Find the error in x 2 + x + 11 = 0 for the real solution 

Sol: Since 

Sol: Given x 2 x+ + 11 = 0 by comparing 

a = 1, b = 1, c = 11 ( ) ( ) 

7x 2 - 4 = 

5y 

2 

3x 2 + 2 = 

4y 

2 

7x 2 - 4 = 

5y 2 

( 1 

) 

3x 2 + 2 = 

4y 2 

( 2 

) 

Discriminant = b 2 - 4 ac putting the values x eq 1 by 4 ⇒ 28x 2 - 16 = 

20y 

2 

= 1 2- 

4 ( 1 )( 11 ) x eq 2 by 5 ⇒ ± 15x 2 ± 10 = ± 

20y 2 

Subtract 

= 1 - 44 13x 2 

- 26 = 

0 

= - 43 < 0 roots are imaginary 13x 2 

= 

26 

But according to question roots should be real x 2= 2 Putting 5y 2= 7 the ( 2 ) value - 

4 

of x2 in equation (1) 

So take small change x 2 x+ - 11 = 0 to get positive discriminant Discriminant = b 2 - 4 ac putting the values 5y 2= 14 - 

4 

= 1 2- 4 ( 1 )( - 

11 ) 5y 2 

=10 

= 1 + 44 y 2= 10 5= 2 Thus 

x 2 = 2 y 2 = 2 Taking square root on both sides 

= 45 > 0 So Roots are real Using Quadratic formula x= 

- 1 ± 1 2 - 4 ( 1 )( - 11 

) 2 ( 1 

) x = ± 2 y = ± 2 Solution set 

= { ( ± 2, ± 2 ) } x= - 1 ± 2 Q11. Area of a rectangle is 48 cm2 if length and width are each increased by 4cm, area of the larger rectangle is 120cm2 Find length and width of the original rectangle. Sol: Let width of the rectangle = x Length of the rectangle = y Area of rectangle = 48cm2 

xy =48 

y = 48 xAfter increasing Width of new rectangle = x + 4 Length of new rectangle = y + 4 Area of ( new rectangle )( ) 

=120cm2 

( ) ( ) 

Putting the value of xy and y 48 4 4 48 16 120 

4 192 48 16 120 0 

45 = 

- 1 ± 2 

9 × 5 

x 

= 

- 1 ± 2 

3 5 

Solution set = ⎧ │ ⎨ │ ⎩ - 1 - 3 2 5 , - 1 + 2 

3 5 ⎫ │ ⎬ │ ⎭ 

Activity 2: Divide x 5 + x 2 + 5 x + 7 by x+ 2 and find quotient and Remainder. Verify your answer by using long division. Sol: Given P ( x ) = x 5 + 0 x 4 + 0 x 3 + x 2 + 5 x + 7 And divisor x+ 2 = 0 or x = - 2 Then by synthetic division -2 1 0 0 1 5 7 -2 4 -8 14 -38 

x + 4 y + 4 = 

120 

1 -2 4 -7 19 -31 Q(x)= x 4 - 2 x 3 + 4 x 2 - 7 x + 19 and R = - 31 x y + 4 + 4 y + 4 = 

120 

By using Long division 

xy + 4x + 4y + 16 = 

120 

x 4 - 2 x 3 + 4 x 2 - 7 x + 19 x+ 2 x 5 + 0 x 4 + 0 x 3 + x 2 + 5 x + 7 + + ⎛ │ ⎝ ± x 5 ± 

2 x 4 - 2 x 4 + 

0 x 3 2 x 4 4 x 3 x 6 0 x 6 

⎞ │ ⎠ + = + + + - = 

4 x + 192 x - 56 = 0 × by 4x 22 

x x 

x x 

x + 48 - 14 x 

= 

0 x - 8 x - 6 x 

+ 48 = 

0 x ( x - 8 ) - 6 ( x 

- 8 ) 

= 

0 ( x - 6 )( x 

- 8 ) 

= 

0 Either - = 

= 

Or 

x - 8 = 

0 x = 

8 When width x = 6 cm So length = 48 6 = 

8 cm 

3 2 4x x + 3 2 4 8 x x ± ± 

2 7 5 x x - + 

2 7 14 x x 

19 7 x+ 19 38 x ± ± 

33 R = -