CHAPTER 14 SOLVING QUADRATIC EQUATIONSdocuments.routledge-interactive.s3.amazonaws.com/...to_Exerc_Chap… · 1. + 5 = ( ) ( )( ) ( ) −± − − −± = = −+ −− places =

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CHAPTER 14 SOLVING QUADRATIC EQUATIONS

EXERCISE 56 Page 119

1. Solve by factorization: x2 – 16 = 0

Since 2 16 0x − = then (x + 4)(x – 4) = 0 from which, x + 4 = 0 i.e. x = –4 and x – 4 = 0 i.e. x = 4 Hence, if 2 16 0x − = then x = –4 and x = 4 2. Solve by factorization: x2 + 4x – 32 = 0

Since 2 4 32 0x x+ − = then (x – 4)(x + 8) = 0 from which, x – 4 = 0 i.e. x = 4 and x + 8 = 0 i.e. x = –8 Hence, if 2 4 32 0x x+ − = then x = 4 and x = –8 3. Solve by factorization: (x + 2)2 = 16

Since ( )22 16x + = then 2 4 4 16x x+ + = and 2 4 4 16 0x x+ + − = i.e. 2 4 12 0x x+ − = Thus, (x + 6)(x – 2) = 0 from which, x + 6 = 0 i.e. x = –6 and x – 2 = 0 i.e. x = 2 Hence, if ( )22 16x + = then x = –6 and x = 2 4. Solve by factorization: 24 9 0x − =

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Since 24 9 0x − = then (2x + 3)(2x – 3) = 0

from which, 2x + 3 = 0 i.e. 2x = –3 and x = –1.5

and 2x – 3 = 0 i.e. 2x = 3 and x = 1.5

5. Solve by factorization: 23 4 0x x+ = Since 23 4 0x x+ = then x(3x + 4) = 0 from which, x = 0

or 3x + 4 = 0 i.e. 3x = –4 and x = – 43

6. Solve by factorization: 28 32 0x − = Since 28 32 0x − = then (4x – 8)(2x + 4) = 0 from which, 4x – 8 = 0 i.e. 4x = 8 and x = 2

and 2x + 4 = 0 i.e. 2x = –4 and x = –2

7. Solve by factorization: 2 8 16 0x x− + = Since 2 8 16 0x x− + = then (x – 4)(x – 4) = 0 from which, x – 4 = 0 (twice) i.e. x = 4 (twice)

8. Solve by factorization: 2 10 25 0x x+ + = Since 2 10 25 0x x+ + = then (x + 5)(x + 5) = 0 from which, x + 5 = 0 (twice) i.e. x = –5 (twice)

9. Solve by factorization: 2 2 1 0x x− + = Since 2 2 1 0x x− + = then (x – 1)(x – 1) = 0

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from which, x – 1 = 0 (twice) i.e. x = 1 (twice)

10. Solve by factorization: x2 + 5x + 6 = 0 Since x2 + 5x + 6 = 0 then (x + 3)(x + 2) = 0 from which, x + 3 = 0 i.e. x = –3

and x + 2 = 0 i.e. x = –2

11. Solve by factorization: x2 + 10x + 21 = 0 Since x2 + 10x + 21= 0 then (x + 3)(x + 7) = 0 from which, x + 3 = 0 i.e. x = –3

and x + 7 = 0 i.e. x = –7

12. Solve by factorization: x2 – x – 2 = 0 Since x2 – x – 2 = 0 then (x – 2)(x + 1) = 0 from which, x – 2 = 0 i.e. x = 2

and x + 1 = 0 i.e. x = –1

13. Solve by factorization: y2 – y – 12 = 0 Since y2 – y – 12 = 0 then (y – 4)(y + 3) = 0 from which, y – 4 = 0 i.e. y = 4

and y + 3 = 0 i.e. y = –3

14. Solve by factorization: y2 – 9y + 14 = 0 Since y2 – 9y + 14 = 0 then (y – 7)(y – 2) = 0 from which, y – 7 = 0 i.e. y = 7

and y – 2 = 0 i.e. y = 2

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15. Solve by factorization: x2 + 8x + 16 = 0 Since x2 + 8x + 16 = 0 then (x + 4)(x + 4) = 0 from which, x + 4 = 0 (twice) i.e. x = –4 (twice)

16. Solve by factorization: x2 – 4x + 4 = 0 Since 2 4 4 0x x− + = then (x – 2)(x – 2) = 0 from which, x – 2 = 0 i.e. x = 2 (twice) 17. Solve by factorization: 2 6 9 0x x+ + = Since 2 6 9 0x x+ + = then (x + 3)(x + 3) = 0 from which, x + 3 = 0 i.e. x = –3 (twice) Hence, if 2 4 4 0x x− + = then x = –3 18. Solve by factorization: 2 9 0x − = Since x2 – 9 = 0 then (x – 3)(x + 3) = 0 from which, x – 3 = 0 i.e. x = 3

and x + 3 = 0 i.e. x = –3

19. Solve by factorization: 23 8 4 0x x+ + = Since 3x2 + 8x + 4 = 0 then (3x + 2)(x + 2) = 0

from which, 3x + 2 = 0 i.e. 3x = –2 i.e. x = 23

−

and x + 2 = 0 i.e. x = –2

20. Solve by factorization: 24 12 9 0x x+ + =

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Since 24 12 9 0x x+ + = then (2x + 3)(2x + 3) = 0 from which, 2x + 3 = 0 (twice) i.e. 2x = –3 i.e. x = –1.5 (twice)

21. Solve by factorization: 2 14 016

z − =

Since 2 14 016

z − = then 1 12 24 4

z z − +

= 0

from which, 2z – 14

= 0 i.e. 2x = 14

i.e. x = 18

and 2z + 14

= 0 i.e. 2x = – 14

i.e. x = 18

−

22. Solve by factorization: x2 + 3x – 28 = 0 Since x2 + 3x – 28 = 0 then (x + 7)(x – 4) = 0 from which, x + 7 = 0 i.e. x = –7

and x – 4 = 0 i.e. x = 4

23. Solve by factorization: 2x2 – x – 3 = 0 Since 22 3 0x x− − = then (2x – 3)(x + 1) = 0

from which, 2x – 3 = 0 i.e. 2x = 3 and x = 3 1or 1 or 1.52 2

and x + 1 = 0 i.e. x = –1 Hence, if 22 3 0x x− − = then x = 1.5 and x = –1 24. Solve by factorization: 6x2 – 5x + 1 = 0 Since 26 5 1 0x x− + = then (3x – 1)(2x – 1) = 0

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from which, 3x – 1 = 0 i.e. 3x = 1 and x = 13

and 2x – 1 = 0 i.e. 2x = 1 and x = 12

Hence, if 26 5 1 0x x− + = then x = 13

and x = 12

25. Solve by factorization: 10x2 + 3x – 4 = 0 Since 10x2 + 3x – 4 = 0 then (5x + 4)(2x – 1) = 0

from which, 5x + 4 = 0 i.e. 5x = – 4 and x = 45

−

and 2x – 1 = 0 i.e. 2x = 1 and x = 12

26. Solve by factorization: 21x2 – 25x = 4 Since 221 25 4x x− = then 221 25 4 0x x− − = and (7x + 1)(3x – 4) = 0

from which, 7x + 1 = 0 i.e. 7x = –1 and x = 17

−

and 3x – 4 = 0 i.e. 3x = 4 and x 4 113 3

= =

27. Solve by factorization: 8x2 + 13x – 6 = 0 Since 8x2 + 13x – 6 = 0 then (8x – 3)(x + 2) = 0


and x + 2 = 0 i.e. x = –2 28. Solve by factorization: 5x2 + 13x – 6 = 0 Since 5x2 + 13x – 6 = 0 then (5x – 2)(x + 3) = 0

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and x + 3 = 0 i.e. x = –3 29. Solve by factorization: 6x2 – 5x – 4 = 0 Since 6x2 – 5x – 4 = 0 then (3x – 4)(2x + 1) = 0


or 113

and 2x + 1 = 0 i.e. 2x = –1 and x = 12

−

30. Solve by factorization: 8x2 + 2x – 15 = 0 Since 28 2 15 0x x+ − = then (4x – 5)(2x + 3) = 0


and 2x + 3 = 0 i.e. 2x = –3 and x = 32

−

31. Determine the quadratic equations in x whose roots are 3 and 1 If the roots are 3 and 1 then: (x – 3)( x – 1) = 0 i.e. 2 3 3 0x x x− − + = i.e. 2 4 3 0x x− + = 32. Determine the quadratic equations in x whose roots are 2 and –5 If the roots are 2 and –5 then: (x – 2)(x + 5) = 0 i.e. 2 5 2 10 0x x x+ − − = i.e. 2 3 10 0x x+ − =

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33. Determine the quadratic equations in x whose roots are –1 and –4 If the roots are –1 and –4 then: (x + 1)(x + 4) = 0 i.e. 2 4 4 0x x x+ + + = i.e. 2 5 4 0x x+ + = 34. Determine the quadratic equations in x whose roots are 2.5 and –0.5

If the roots are 2.5 and –0.5 then: 5 12 2

x x − +

= 0

i.e. 2 1 5 5 02 2 4

x x x+ − − =

and 2 52 04

x x− − =

or 24 8 5 0x x− − = 35. Determine the quadratic equations in x whose roots are 6 and –6 If the roots are 6 and –6 then: (x – 6)(x + 6) = 0 i.e. 2 6 6 36 0x x x− + − = i.e. 2 36 0x − = 36. Determine the quadratic equations in x whose roots are 2.4 and –0.7 If the roots are 2.4 and –0.7 then: (x – 2.4)(x + 0.7) = 0 i.e. 2 0.7 2.4 1.68 0x x x+ − − = i.e. 2 1.7 1.68 0x x− − =

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1. Solve, correct to 3 decimal places, by completing the square: x2 + 4x + 1 = 0 2 4 1 0x x+ + = i.e. 2 4 1x x+ = − and ( ) ( )2 22 4 2 1 2x x+ + = − + adding to both sides (half the coefficient of x) 2 Hence, ( )22x + = 3 from which, (x + 2) = 3 1.7321= ± Thus, x = –2 + 1.7321 = –0.268 and x = –2 – 1.7321 = –3.732 Hence, if 2 4 1 0x x+ + = then x = –0.268 or –3.732, correct to 3 decimal places 2. Solve, correct to 3 decimal places, by completing the square: 2x2 + 5x – 4 = 0 22 5 4 0x x+ − =

or 2 5 2 02

x x+ − =

i.e. 2 5 22

x x+ =

and 2 2

2 5 5 522 4 4

x x + + = +

adding to both sides (half the coefficient of x) 2

Hence, 25

4x +

= 25 32 25 572

16 16 16 16+ = + =

from which, 54

x + = 57 1.8874616 = ±

Thus, x = 1.88746 – 1.25 = 0.637 and x = –1.88746 – 1.25 = –3.137 Hence, if 22 5 4 0x x+ − = then x = 0.637 or –3.137, correct to 3 decimal places

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3. Solve, correct to 3 decimal places, by completing the square: 3x2 – x – 5 = 0 3x2 – x – 5 = 0

or 2 1 5 03 3

x x− − =

i.e. 2 1 53 3

x x− =

and 2 2

2 1 1 5 13 6 3 6

x x − + − = + −


Hence, 21

6x −

= 5 1 60 1 61

3 36 36 36+

+ = =

from which, 16

x −

= 6136 = ± 1.301708

Thus, x = 16

+ 1.301708 = 1.468

and x = 16

– 1.301708 = –1.135

Hence, if 3x2 – x – 5 = 0 then x = 1.468 or –1.135, correct to 3 decimal places 4. Solve, correct to 3 decimal places, by completing the square: 5x2 – 8x + 2 = 0 25 8 2 0x x− + =

or 2 8 2 05 5

x x− + =

i.e. 2 8 25 5

x x− = −

and 2 2

2 8 4 4 25 5 5 5

x x − + = −


Hence, 24

5x −

= 16 2 16 10 6

25 5 25 25−

− = =

from which, 45

x − = 6 0.489925

= ±

Thus, x = 0.4899 + 0.8 = 1.290

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and x = –0.4899 + 0.8 = 0.310 Hence, if 25 8 2 0x x− + = then x = 1.290 or 0.310, correct to 3 decimal places 5. Solve, correct to 3 decimal places, by completing the square: 4x2 – 11x + 3 = 0 24 11 3 0x x− + =

or 2 11 3 04 4

x x− + =

i.e. 2 11 34 4

x x− = −

and 2 2

2 11 11 11 34 8 8 4

x x − + = −


Hence, 211

8x −

= 121 3 121 48 73

64 4 64 64 64− = − =

from which, 118

x − = 73 1.068064

= ±

Thus, x = 1.0680 + 118

= 2.443

and x = –1.0680 + 118

= 0.307

Hence, if 24 11 3 0x x− + = then x = 2.443 or 0.307, correct to 3 decimal places 6. Solve, correct to 3 decimal places, by completing the square: 2x2 + 5x = 2 2x2 + 5x = 2

i.e. 2 5 12

x x+ =

and 2 2

2 5 5 512 4 4

x x + + = +


Hence, 25

4x +

= 25 16 25 411

16 16 16 16+ = + =

from which, 54

x + = 41 1.6007816 = ±

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Thus, x = 1.60078 – 54

= 0.351

and x = –1.60078 – 54

= –2.851

Hence, if 2x2 + 5x = 2 then x = 0.351 or –2.851, correct to 3 decimal places

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1. Solve, correct to 3 decimal places, using the quadratic formula: 2x2 + 5x – 4 = 0

If 22 5 4 0x x+ − = then x = ( ) ( )( )

( )

25 5 4 2 4 5 572 2 4

− ± − − − ± =

= 5 574

− + or 5 574

− −

= 0.637 or –3.137, correct to 3 decimal

places

2. Solve, correct to 3 decimal places, using the quadratic formula: 5.76x2 + 2.86x – 1.35 = 0

If 5.76x2 + 2.86x – 1.35 = 0 then x = ( ) ( )( )

( )

22.86 2.86 4 5.76 1.35 2.86 39.28362 5.76 11.52

− ± − − − ± =

= 2.86 39.283611.52

− + or 2.86 39.283611.52

− −

= 0.296 or –0.792, correct to 3 decimal places

3. Solve, correct to 3 decimal places, using the quadratic formula: 2x2 – 7x + 4 = 0

If 22 7 4 0x x− + = then x = ( ) ( )( )

( )

27 7 4 2 4 7 172 2 4

− − ± − − ± =

= 7 174

+ or 7 174

−

= 2.781 or 0.719, correct to 3

decimal places

4. Solve, correct to 3 decimal places, using the quadratic formula: 4x + 5 = 3x

34 5xx

+ = i.e. 24 5 3x x+ = and 24 5 3 0x x+ − =

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Hence, x = ( ) ( )( )

( )

25 5 4 4 3 5 732 4 8

− ± − − − ± =

= 5 738

− + or 5 738

− −


5. Solve, correct to 3 decimal places, using the quadratic formula: (2x + 1) = 53x −

( ) 52 13

xx

+ =−

i.e. ( )( )2 1 3 5x x+ − = and 22 6 3 5x x x− + − =

i.e. 22 5 8 0x x− − = from which, x = ( ) ( )( )

( )

25 5 4 2 8 5 892 2 4

− − ± − − − ± =

= 5 894

+ or 5 894

−

= 3.608 or –1.108, correct to 3 decimal places.

6. Solve, correct to 3 decimal places, using the quadratic formula: 23 5 1 0x x− + =

If 23 5 1 0x x− + = then x = ( ) ( )( )

( )

25 5 4 3 1 5 132 3 6

− − ± − − ± =

= 5 136

+ or 5 136

−

= 1.434 or 0.232, correct to 3 decimal places

7. Solve, correct to 3 decimal places, using the quadratic formula: 24 6 8 0x x+ − =

If 24 6 8 0x x+ − = then x = ( ) ( )( )

( )

26 6 4 4 8 6 1642 4 8

− ± − − − ± =

= 6 1648

− + or 6 1648

− −

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8. Solve, correct to 3 decimal places, using the quadratic formula: 25.6 11.2 1 0x x− − =

If 25.6 11.2 1 0x x− − = then x = ( ) ( )( )

( )

211.2 11.2 4 5.6 1 11.2 147.842 5.6 11.2

− − ± − − − ± =

= 11.2 147.8411.2+ or 11.2 147.84

11.2−


9. Solve, correct to 3 decimal places, using the quadratic formula: 3x(x + 2) + 2x(x – 4) = 8

If 3x(x + 2) + 2x(x – 4) = 8 then 2 23 6 2 8 8x x x x+ + − = i.e. 25 2 8x x− − = 0

Hence, x = ( ) ( )( )

( )

22 2 4 5 8 2 1642 5 10

− − ± − − − ± =

= 2 16410

+ or 2 16410

−


10. Solve, correct to 3 decimal places, using the quadratic formula: 24 (2 5) 14x x x− + =

If 24 (2 5) 14x x x− + = then 2 24 2 5 14x x x− − = i.e. 22 5 14 0x x− − =

Hence, x = ( ) ( )( )

( )

25 5 4 2 14 5 1372 2 4

− − ± − − − ± =

= 5 1374

+ or 5 1374

−


11. Solve, correct to 3 decimal places, using the quadratic formula: 5 2 63 2x x+ =

− −

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If 5 2 63 2x x+ =

− − then 5 2( 3)( 2) ( 3)( 2) 6( 3)( 2)

3 2x x x x x x

x x− − + − − = − −

− −

i.e. 5(x – 2) + 2(x – 3) = 6( 2 2 3 6x x x− − + )

i.e. 5x – 10 + 2x – 6 = 26 30 36x x− +

i.e. 0 = 26 30 36x x− + – 5x + 10 – 2x + 6

i.e. 26 37 52x x− + = 0

Hence, x = ( ) ( )( )

( )

237 37 4 6 52 37 1212 6 12

− − ± − − ± =

= 37 1112+ or 37 11

12−

= 4 or 2.167, correct to 3 decimal places

12. Solve, correct to 3 decimal places, using the quadratic formula: 3 2 7 47

x xx

+ = +−

If 3 2 7 47

x xx

+ = +−

then 3( 7) 2 ( 7) 7( 7) 4 ( 7)7

x x x x x xx

− + − = − + −−

i.e. 2 23 2 14 7 49 4 28x x x x x+ − = − + − i.e. 0 = 2 27 49 4 28 3 2 14x x x x x− + − − − + i.e. 22 7 52x x− − = 0

Hence, x = ( ) ( )( )

( )

27 7 4 2 52 7 4652 2 4

− − ± − − − ± =

= 7 4654

+ or 7 4654

−

= 7.141 or –3.641, correct to 3 decimal places.

13. Solve, correct to 3 decimal places, using the quadratic formula: 11

xx+−

= x – 3

If 11

xx+−

= x – 3 then x + 1 = (x – 1)(x – 3)

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i.e. x + 1 = 2 3 3x x x− − +

i.e. 0 = 2 3 3x x x− − + – x – 1

i.e. 2 5 2x x− + = 0

Hence, x = ( ) ( )( )

( )

25 5 4 1 2 5 172 1 2

− − ± − − ± =

= 5 172

+ or 5 172

−

= 4.562 or 0.438, correct to 3 decimal places

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1. The angle a rotating shaft turns through in t seconds is given by: θ = ωt + 12αt2. Determine the

time taken to complete 4 radians if ω is 3.0 rad/s and α is 0.60 rad/s2

212

t tθ ω α= + and if θ = 4, ω = 3 and α = 0.60

then ( ) 214 3 0.602

t t= + i.e. 24 3 0.03t t= +

or 20.03 3 4 0t t+ − =

Using the quadratic formula: t = ( ) ( )( )

( )

23 3 4 0.30 4 3 13.82 0.30 0.60

− ± − − − ± =

= 3 13.80.60

− + or 3 13.80.60

− −

= 1.191 s (or –11.191 which is neglected)

Hence, the time taken to complete 4 radians is 1.191 s

2. The power P developed in an electrical circuit is given by P = 10I – 8I2, where I is the current in amperes. Determine the current necessary to produce a power of 2.5 watts in the circuit. P = 10I – 28I and when P = 2.5 W, 2.5 = 10I – 28I i.e. 28 10 2.5 0I I− + = and using the quadratic formula

i.e. I = ( ) ( )( )

( )

210 10 4 8 2.5 10 202 8 16

− − ± − − ± =

= 10 2016+ or 10 20

16−

= 0.905 A or 0.345 A

Hence, the current necessary to produce a power of 2.5 watts is 0.905 A or 0.345 A

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3. The area of a triangle is 47.6 cm2 and its perpendicular height is 4.3 cm more than its base length. Determine the length of the base correct to 3 significant figures.

Area of a triangle = 1 base height2× × and if base = b then perpendicular height h = b + 4.3

Hence, 47.6 = 1 ( 4.3)2

b b× × +

i.e. 2(47.6) = 2 4.3b b+ Hence, 95.2 = 2 4.3b b+ and 2 4.3b b+ – 95.2 = 0

i.e. b = ( ) ( )( )

( )

24.3 4.3 4 1 95.2 4.3 399.292 1 2

− ± − − − ± =

= 4.3 399.292

− + or 4.3 399.292

− −


The latter result has no meaning, hence length of base = 7.84 cm

4. The sag l metres in a cable stretched between two supports, distance x m apart is given by:

l = 12x

+ x. Determine the distance between supports when the sag is 20 m.

12l xx

= + and when sag l = 20 m, 1220 xx

= +

i.e. 20x = 12 + 2x or 2x – 20x + 12 = 0

Using the quadratic formula: t = ( ) ( )( )

( )

220 20 4 1 12 20 3522 1 2

− − ± − − ± =

= 20 3522

+ or 20 3522

−

= 19.38 m or 0.619 m

Hence, the distance between supports when the sag is 20 m is 19.38 m or 0.619 m

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5. The acid dissociation constant Ka of ethanoic acid is 1.8 × 10–5 mol dm–3 for a particular solution.

Using the Ostwald dilution law: Ka = 2

(1 )x

v x−, determine x, the degree of ionization, given that

v = 10 dm3

( )2

1a

xKv x

=−

and when 5 31.8 10 moldmaK − −= × and 310dmv =

then ( )

251.8 10

10 1x

x−× =

− i.e. ( )( )( )5 210 1 1.8 10x x−− × =

i.e. 4 4 21.8 10 1.8 10 x x− −× − × = or 2 4 41.8 10 1.8 10 0x x− −+ × − × = Using the quadratic formula:

x = ( ) ( )( )

( )

24 4 44 61.8 10 1.8 10 4 1 1.8 10 1.8 10 720.0324 10

2 1 2

− − −− −

− × ± × − − × − × ± × =

= 4 61.8 10 720.0324 10

2

− −− × + × or 4 61.8 10 720.0324 10

2

− −− × − ×

= 0.0133 (or –0.0135 which is neglected)

Hence, the degree of ionisation, x, is 0.0133

6. A rectangular building is 15 m long by 11 m wide. A concrete path of constant width is laid all the way around the building. If the area of the path is 60.0 m2, calculate its width correct to the nearest millimetre. The concrete path is shown shaded in the sketch below.

Shaded area = 2(15x) + 2(11x) + 4 2x = 30x + 22x + 4 2x = 52x + 4 2x

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Since the area of the path is 60.0 2m then 60.0 = 52x + 4 2x i.e. 4 2x + 52x – 60.0 = 0

Using the quadratic formula: x = ( ) ( )( )

( )

252 52 4 4 60.0 52 36642 4 8

− ± − − − ± =

= 52 36648

− + or 52 36648

− −

= 1.066 m (or –14.066 m, which has no meaning)

Hence, the width of the path is 1.066 m, correct to the nearest millimetre.

7. The total surface area of a closed cylindrical container is 20.0 m3. Calculate the radius of the cylinder if its height is 2.80 m. From Chapter 29, the total surface area of a closed cylinder is 22 2r rhπ π+ , where r is its radius and h its height. If the surface area is 20.0 2m and h = 2.80 m, then 20.0 = 22 2 (2.80)r rπ π+

i.e. 22 5.60 20.0 0r rπ π+ − = or 2 202.80 02

r rπ

+ − = by dividing by 2π

i.e. 2 102.80 0r rπ

+ − =

Using the quadratic formula: r = ( ) ( )

( )

2 102.80 2.80 4 12.80 20.5724

2 1 2π

− ± − − − ± =

= 2.80 20.57242

− + or 2.80 20.57242

− −

= 0.8678 m (or –3.668 m, which has no meaning)

Hence, the radius of the cylinder is 86.78 cm

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8. The bending moment M at a point in a beam is given by: M = 3 (20 )2

x x− where x metres is the

distance from the point of support. Determine the value of x when the bending moment is 50 Nm.

M = ( )3 202

x x− and when M = 50 Nm, then 50 = ( )3 20

2x x−

i.e. 100 = 3x(20 – x) i.e. 100 = 60x – 3 2x or 3 2x – 60x + 100 = 0


( )

260 60 4 3 100 60 24002 3 6

− − ± − − ± =

= 60 24006

+ or 60 24006

−

= 18.165 m or 1.835 m

Hence, when M is 50 Nm, the values of x are 18.165 m or 1.835 m

9. A tennis court measures 24 m by 11 m. In the layout of a number of courts an area of ground must be allowed for at the ends and at the sides of each court. If a border of constant width is allowed around each court and the total area of the court and its border is 950 m2, find the width of the borders.

The tennis court with its shaded border is shown sketched below.

Shaded area = 2(11x) + 2(24x) + 4 2x = 4 2x + 70x

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Now, shaded area = 950 – (11 × 24) = 686 2m i.e. 4 2x + 70x = 686 or 4 2x + 70x – 686 = 0


( )

270 70 4 4 686 70 158762 4 8

− ± − − − ± =

= 70 158768

− + or 70 158768

− −

= 7 m (or 24.5 m, which has no meaning)

Hence, the width of the border is 7 m

10. Two resistors, when connected in series, have a total resistance of 40 ohms. When connected in parallel their total resistance is 8.4 ohms. If one of the resistors has a resistance Rx ohms: (a) show that Rx

2 – 40Rx + 336 = 0 and (b) calculate the resistance of each. (a) Let resistor values be 1R and xR In series: 1R + xR = 40 (1)

In parallel: 1

18.4x

x

R RR R

=+

(2)

From (1), 1R = 40 – xR

Substituting in (2) gives: ( )( )

408.4

40x x

x x

R RR R−

=− +

i.e. 240 8.4

40x xR R−

=

from which, 240 (40)(8.4) 336x xR R− = = i.e. 2 40 336 0x xR R− + = (b) Solving 2 40 336 0x xR R− + = using the quadratic formula gives:

( ) ( )( )

( )

240 40 4 1 336 40 2562 1 2

xR − − ± − − ± = =

© 2014, John Bird

233

= 40 2562

+ or 40 2562

−

= 28 Ω or 12 Ω

From equation (1), when xR = 28 Ω, 1R = 12 Ω

and when xR = 12 Ω, 1R = 28 Ω

Hence, the two values of resistance are 12 ohms and 28 ohms

11. When a ball is thrown vertically upwards its height h varies with time t according to the

equation h = 25t – 4 2t . Determine the times, correct to 3 significant figures, when the height is

12 m. If h = 25t – 4 2t when height h = 12 m, then 12 = 25t – 4 2t i.e. 4 2t – 25t + 12 = 0

Hence, ( ) ( )( )

( )

225 25 4 4 12 25 4332 4 8

t − − ± − − ± = =

= 25 4338

+ or 25 4338

−

= 5.73 or 0.52

i.e. the times when the height of the ball is 12 m are 5.73 s and 0.52 s

12. In an RLC electrical circuit, reactance X is given by: 1X LC

ωω

= −

X = 220 Ω, inductance L = 800 mH and capacitance C = 25 μF. The angular velocity ω is

measured in radians per second. Calculate the value of ω.

If 1X LC

ωω

= − and X = 220, L = 800 310−× and C = 25 610−×

Then 220 = 800 36

11025 10

ωω

−−

× −×

i.e. 220(25 610−× ω ) = (25 610−× ω )(800 310−× ω ) – 1

© 2014, John Bird

234

i.e. 3 5 25.5 10 2 10ω ω− −× = × – 1

i.e. 5 2 32 10 5.5 10ω ω− −× − × – 1 = 0

Hence, ( ) ( )( )

( )

23 3 53 4

5 5

5.5 10 5.5 10 4 2 10 1 5.5 10 1.1025 102 2 10 4 10

ω− − −

− −

− −

− − × ± − × − × − × ± × = =× ×

= 3 4

5

5.5 10 1.1025 104 10

− −

−

× + ××

or 3 4

5

5.5 10 1.1025 104 10

− −

−

× − ××

= 400 or –125 (which has no meaning)

Hence, the angular velocity ω is 400 rad/s

© 2014, John Bird

235


1. Solve the simultaneous equations: y = x2 + x + 1

y = 4 – x Equating the y values gives: 2 1x x+ + = 4 – x i.e. 2 2 3 0x x+ − = Factorizing gives: (x + 3)( x – 1) = 0 from which, x + 3 = 0 i.e. x = –3 and x – 1 = 0 i.e. x = 1 When x = –3, y = 7 and when x = 1, y = 3 (from either of the two equations) Thus, the solutions to the simultaneous equations are x = –3, y = 7 and x = 1, y = 3 2. Solve the simultaneous equations: y = 15x2 + 21x – 11

y = 2x – 1 Equating the y values gives: 215 21 11x x+ − = 2x – 1 i.e. 215 19 10 0x x+ − = Factorizing gives: (5x – 2)( 3x + 5) = 0

from which, 5x – 2 = 0 i.e. x = 25

and 3x + 5 = 0 i.e. x = 53

−

When x = 25

, y = 22 15

−

= 15

− and when x = 53

− , y = 5 132 13 3

− − = −

Thus, the solutions to the simultaneous equations are x = 25

, y = 15

− and x = 213

− , y = 143

−

3. Solve the simultaneous equations: 2x2 + y = 4 + 5x

x + y = 4

© 2014, John Bird

236

From the first equation, 24 5 2y x x= + − and from the second equation, y = 4 – x Equating the y values gives: 24 5 2 4x x x+ − = − i.e. 26 2 0x x− = i.e. 2x(3 – x) = 0 from which, x = 0 or x = 3 When x = 0, y = 4 and when x = 3, y = 1 (from either of the two equations) Thus, the solutions to the simultaneous equations are x = 0, y = 4 and x = 3, y = 1

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