Mathematics 10 (Science) - KPK - Gain - Solution - PART - 10 - 18 Chapter 1

18 Chapter 1 

Review Exercise 1 

121 - 9 = 

2 16 11 - 3 = 

2 ( 4 

) 8 8= satisfies Therefore x = - 9 is a real root 

S.S = { - }9 Review Exercise 1 

Q1. Fill the correct circle only i). if ( x + 1 )( x - 5 ) = 0 then solutions are O x = 1, - 5 O x = 1,5 O x = - 1, - 5 O x = - 1,5 ii). If x 2 x- - 1 = 0 then x = ................ O 1 2 

- ± 5 

O - 1 ± 2 5 O 1 ± 2 5

O 1 ± 2 5 iii). - 1 ± 5 

2 

in simplified form is 

O 1 ± 24 O 1 ± 6 O 2 ± 6 O Cannot be simplified iv). Apply the quadratic formula to 2 x 2 - x =

3 O a = 2, b = - 1, c = 3 O a = 2, b = 1, c = 3 O a = 1, b = - 1, c = - 3 O a = 1, b = - 1, c = 0 v). if x 2 - 3 x - 4 = 0 then the solution are O x = 4, - 1 O x = - 4,1 O x = 4,1 O x = - 4, - 1 vi). If 2 x 2 - 4 x + 9 = 0 solutions are O x = 2 ± 2 22 

O x = 

- 2 ± 2 22 O x = 2 ± 22 2 O x = - 2 ± 22 2 vii). If x 2 - 1 4 = 0 the solutions are O x = ± 12 O x = ± 14 O x = ± 18 O x = ± 16 1viii). What are the solution of x 2 + 7 x - 18 = 0 O 2 or -9 O -2 or 9 O -2 or -9 O 2 or 9 ix). Roots of x 2 - 8 x + 15 = 0 are O 1 or -7 O 2 or 4 O -2 or 4 O 3 or 5 Q2. Solve 2 w 4 - 5 w 2 + 2 = 0 Solution: we have 2 w 4 - 5 w 2 + 2 = 0 Let y = w 2 ⇒ y 2 = w 4 Given equation becomes 

2 y 2- 5 y 

+ 2 = 

0 

2 y 2 

- 4 y - 1 y 

+ 2 = 

0 2 y ( y - 2 ) - 1 ( y 

- 2 ) 

= 

0 ( 2 y - 1 )( y 

- 2 ) 

= 

0 Either 2 y- 1 = 0 or y- 2 = 

0 

1y = 2 y = 2 Putting back the value of y 

w2 12w 12 == 

± w2 =2 

w 

= ± 

2 

Solution set = ⎧ ⎨ ⎩ ± 21 , ± 2 ⎫ ⎬ ⎭ Q3. Find the constant a and b such that x = 1 and x = - 1 are the solutions to ax 2 + bx + 2 = 0 Sol: since ax 2 + bx + 2 = 0 have solution x = 1 So a

( 1 ) 2 b+ ( 1 ) + 2 = 0 Or a b+ + 2 = 0 ...........................(1) And ax 2 + bx + 2 = 0 have solution x = - 1 So a ( - 1 ) 2 + b ( - 1 ) + 2 = 0 Or a b- + 2 = 0 ...........................(2) Adding eq (1) and eq (2) 

a + b 

+ 2 = 

0 a - b 

+ 2 = 

0 2 a+ 4 = 

0 2 a = - 4 a = - 2 put in eq (1) - 2 + b 

+ 2 = 

0 ⇒ b 

= 

0 Q4. Find all solutions of x such that x 2 + 5 x + 6 and x 2 + 19 x + 32 are equal Sol: according to condition 

x 2 + 19 x + 32 = x 2 

+ 5 x 

+ 

6 x 2 - x 2+ 19 x - 5 x 

= 6 - 

32 13 x 

= - 

26 x 

= - 

2 Q5. Find the solution to 49 x 2 - 316 x + 132 = 0 Solution we have 49 x 2 - 316 x + 132 = 0 49 x 2 - 316 x + 132 = 0 49 x 2 - 294 x - 22 x 

+ 132 = 

0 49 x ( x - 6 ) - 22 ( x 

- 6 ) 

= 

0 ( 49 x - 22 )( x 

- 6 ) 

= 

0 Either 49 x- 22 = 0 x- 6 = 

0 x = 2249 x = 6 Solution set = ⎧ ⎨ ⎩ 6, 22 49 

⎫ ⎬ ⎭