Download CHAPTER 14 SOLVING QUADRATIC EQUATIONS

CHAPTER 14 SOLVING QUADRATIC EQUATIONS
EXERCISE 56 Page 119
1. Solve by factorization: x2 – 16 = 0
Since x 2 − 16 =
0 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 x 2 − 16 =
0 then x = –4 and x = 4
2. Solve by factorization: x2 + 4x – 32 = 0
Since x 2 + 4 x − 32 =
0 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 x 2 + 4 x − 32 =
0 then x = 4 and x = –8
3. Solve by factorization: (x + 2)2 = 16
Since ( x + 2 ) =
16 then x 2 + 4 x + 4 =
16
2
x 2 + 4 x + 4 − 16 =
0
and
x 2 + 4 x − 12 =
0
i.e.
Thus,
(x + 6)(x – 2) = 0
from which,
x+6=0
i.e.
and
x–2=0
i.e. x = 2
Hence, if
( x + 2)
2
x = –6
=
16 then x = –6 and x = 2
4. Solve by factorization:
4x2 − 9 =
0
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© 2014, John Bird
Since 4 x 2 − 9 =
0 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
3x 2 + 4 x =
0
5. Solve by factorization:
Since 3 x 2 + 4 x =
0 then
x(3x + 4) = 0
from which,
x=0
or
3x + 4 = 0
i.e.
3x = –4
and x = –
6. Solve by factorization:
8 x 2 − 32 =
0
Since 8 x 2 − 32 =
0 then
(4x – 8)(2x + 4) = 0
4
3
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:
x 2 − 8 x + 16 =
0
Since x 2 − 8 x + 16 =
0 then
from which,
(x – 4)(x – 4) = 0
x – 4 = 0 (twice)
i.e.
8. Solve by factorization:
x 2 + 10 x + 25 =
0
Since x 2 + 10 x + 25 =
0
then
from which,
Since x 2 − 2 x + 1 =
0
(x + 5)(x + 5) = 0
x + 5 = 0 (twice)
9. Solve by factorization:
x = 4 (twice)
i.e.
x = –5 (twice)
x 2 − 2 x + 1 =0
then
(x – 1)(x – 1) = 0
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© 2014, John Bird
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
from which,
x–2=0
i.e.
x=2
and
x+1=0
i.e.
x = –1
(x – 2)(x + 1) = 0
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
212
© 2014, John Bird
15. Solve by factorization: x2 + 8x + 16 = 0
Since x2 + 8x + 16 = 0
from which,
then
(x + 4)(x + 4) = 0
x + 4 = 0 (twice)
i.e.
x = –4 (twice)
16. Solve by factorization: x2 – 4x + 4 = 0
Since x 2 − 4 x + 4 =
0 then (x – 2)(x – 2) = 0
from which,
x–2=0
17. Solve by factorization:
i.e.
x=2
(twice)
x2 + 6x + 9 =
0
Since x 2 + 6 x + 9 =
0 then (x + 3)(x + 3) = 0
from which,
x+3=0
i.e.
x = –3
(twice)
Hence, if x 2 − 4 x + 4 =
0 then x = –3
18. Solve by factorization:
Since x2 – 9 = 0
then
x2 − 9 =
0
(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:
Since 3x2 + 8x + 4 = 0
3x 2 + 8 x + 4 =
0
then
(3x + 2)(x + 2) = 0
from which,
3x + 2 = 0
i.e. 3x = –2
and
x+2=0
i.e.
20. Solve by factorization:
i.e. x = −
2
3
x = –2
4 x 2 + 12 x + 9 =
0
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© 2014, John Bird
Since 4 x 2 + 12 x + 9 =
0
from which,
then
(2x + 3)(2x + 3) = 0
2x + 3 = 0 (twice)
21. Solve by factorization:
4z2 −
i.e. 2x = –3
i.e. x = –1.5 (twice)
1
=
0
16
1 
1

 2 z −  2 z +  = 0
4 
4

Since 4 z 2 −
1
0 then
=
16
from which,
2z –
1
=0
4
i.e. 2x =
and
2z +
1
=0
4
i.e. 2x = –
1
4
i.e. x =
1
4
1
8
i.e. x = −
1
8
and
3
1
or 1 or 1.5
2
2
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 2 x 2 − x − 3 =
0 then (2x – 3)(x + 1) = 0
from which,
2x – 3 = 0
and
x+1=0
i.e.
2x = 3
x=
i.e. x = –1
Hence, if 2 x 2 − x − 3 =
0 then x = 1.5 and x = –1
24. Solve by factorization: 6x2 – 5x + 1 = 0
Since 6 x 2 − 5 x + 1 =
0
then (3x – 1)(2x – 1) = 0
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© 2014, John Bird
from which,
3x – 1 = 0
and
2x – 1 = 0
Hence, if 6 x 2 − 5 x + 1 =
0 then x =
i.e.
3x = 1
and
i.e. 2x = 1
and
1
3
1
x=
2
x=
1
1
and x =
2
3
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= −
and
2x – 1 = 0
i.e.
2x = 1
and
x=
7x = –1
and
x= −
4
5
1
2
26. Solve by factorization: 21x2 – 25x = 4
Since 21x 2 − 25 x =
4 then 21x 2 − 25 x − 4 =
0
and
(7x + 1)(3x – 4) = 0
from which,
7x + 1 = 0
i.e.
and
3x – 4 = 0
i.e. 3x = 4
and
x=
and
x=
1
7
4
1
= 1
3
3
27. Solve by factorization: 8x2 + 13x – 6 = 0
Since 8x2 + 13x – 6 = 0 then
(8x – 3)(x + 2) = 0
from which,
8x – 3 = 0
i.e.
8x = 3
and
x+2=0
i.e.
x = –2
3
8
28. Solve by factorization: 5x2 + 13x – 6 = 0
Since 5x2 + 13x – 6 = 0
then
(5x – 2)(x + 3) = 0
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© 2014, John Bird
from which,
5x – 2 = 0
i.e.
5x = 2
and
x+3=0
i.e.
x = –3
and
x=
2
5
and
x=
1
4
or 1
3
3
29. Solve by factorization: 6x2 – 5x – 4 = 0
Since 6x2 – 5x – 4 = 0
then
(3x – 4)(2x + 1) = 0
from which,
3x – 4 = 0
and
2x + 1 = 0
i.e.
3x = 4
i.e. 2x = –1
x= −
and
1
2
30. Solve by factorization: 8x2 + 2x – 15 = 0
Since 8 x 2 + 2 x − 15 =
0 then
(4x – 5)(2x + 3) = 0
from which,
4x – 5 = 0
and
2x + 3 = 0
i.e.
4x = 5
i.e. 2x = –3
and
x=
and
5
4
x= −
3
2
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.
x 2 − 3x − x + 3 =
0
i.e.
x2 − 4x + 3 =
0
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.
x 2 + 5 x − 2 x − 10 =
0
i.e.
x 2 + 3 x − 10 =
0
216
© 2014, John Bird
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.
x2 + x + 4x + 4 =
0
i.e.
x2 + 5x + 4 =
0
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 
1

 x −  x +  = 0
2 
2

i.e.
x2 +
1
5
5
x− x− =
0
2
2
4
x2 − 2x −
and
5
=
0
4
4 x2 − 8x − 5 =
0
or
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
x 2 − 6 x + 6 x − 36 =
0
i.e.
x 2 − 36 =
0
i.e.
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.
x 2 + 0.7 x − 2.4 x − 1.68 =
0
i.e.
x 2 − 1.7 x − 1.68 =
0
217
© 2014, John Bird
EXERCISE 57 Page 121
1. Solve, correct to 3 decimal places, by completing the square: x2 + 4x + 1 = 0
x 2 + 4 x + 1 =0
x2 + 4 x =
−1
i.e.
x 2 + 4 x + ( 2 ) =−1 + ( 2 )
2
and
( x + 2)
Hence,
from which,
2
2
adding to both sides (half the coefficient of x) 2
=3
3 = ±1.7321
(x + 2) =
Thus,
x = –2 + 1.7321 = –0.268
and
x = –2 – 1.7321 = –3.732
Hence, if x 2 + 4 x + 1 =
0 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
2 x2 + 5x − 4 =
0
x2 +
or
5
x−2=
0
2
x2 +
i.e.
5
2
x=
2
2
and
5
5
5
x + x+  =
2+ 
2
4
4
2
2
adding to both sides (half the coefficient of x) 2
2
Hence,
from which,
5
25 32 25 57

= +
=
x+  = 2+
4
16 16 16 16

x+
5
=
4
 57 
  = ±1.88746
 16 
Thus,
x = 1.88746 – 1.25 = 0.637
and
x = –1.88746 – 1.25 = –3.137
Hence, if 2 x 2 + 5 x − 4 =
0 then x = 0.637 or –3.137, correct to 3 decimal places
218
© 2014, John Bird
3. Solve, correct to 3 decimal places, by completing the square: 3x2 – x – 5 = 0
3x2 – x – 5 = 0
1
5
0
x2 − x − =
3
3
or
1
5
x2 − x =
3
3
i.e.
2
1
 1 5  1
x2 − x +  −  = +  − 
3
 6 3  6
and
2
adding to both sides (half the coefficient of x) 2
2
1
5 1 60 + 1 61

=
x−  = + =
6
3 36
36
36

Hence,
from which,
1

x−  =
6

61
= ± 1.301708
36
Thus,
x=
1
+ 1.301708 = 1.468
6
and
x=
1
– 1.301708 = –1.135
6
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
5x2 − 8x + 2 =
0
8
2
x2 − x + =
0
5
5
or
8
2
x2 − x =
−
5
5
i.e.
2
and
2
8
4 4 2
x2 − x +   =   −
5
5 5 5
adding to both sides (half the coefficient of x) 2
2
Hence,
from which,
Thus,
4
16 2 16 − 10 6

=
−
=
x−  =
5
25 5
25
25

x−
4
=
5
 6 
  = ±0.4899
 25 
x = 0.4899 + 0.8 = 1.290
219
© 2014, John Bird
and
x = –0.4899 + 0.8 = 0.310
Hence, if 5 x 2 − 8 x + 2 =
0 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
4 x 2 − 11x + 3 =
0
x2 −
or
11
3
x+ =
0
4
4
x2 −
i.e.
11
3
x=
−
4
4
2
and
x2 −
2
11
 11   11  3
x+  =   −
4
8 8 4
adding to both sides (half the coefficient of x) 2
2
11 
121 3 121 48 73

− =
− =
x−  =
8
64 4 64 64 64

Hence,
x−
from which,
11
=
8
 73 
  = ±1.0680
 64 
11
= 2.443
8
Thus,
x = 1.0680 +
and
x = –1.0680 +
11
= 0.307
8
Hence, if 4 x 2 − 11x + 3 =
0 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
x2 +
i.e.
5
x=
1
2
2
and
x2 +
5
5
5
x+  =
1+  
2
4
4
2
adding to both sides (half the coefficient of x) 2
2
Hence,
from which,
5
25 16 25 41

= +
=
 x +  = 1+
4
16 16 16 16

x+
5
=
4
 41 
  = ± 1.60078
 16 
220
© 2014, John Bird
Thus,
x = 1.60078 –
and
x = –1.60078 –
5
= 0.351
4
5
= –2.851
4
Hence, if 2x2 + 5x = 2 then x = 0.351 or –2.851, correct to 3 decimal places
221
© 2014, John Bird
EXERCISE 58 Page 122
1. Solve, correct to 3 decimal places, using the quadratic formula: 2x2 + 5x – 4 = 0
If 2 x 2 + 5 x − 4 =
0 then
2
−5 ± ( 5 ) − 4 ( 2 )( −4 )  −5 ± 57


x=
=
2 ( 2)
4
=
−5 + 57
−5 − 57
or
4
4
= 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
2
−2.86 ± ( 2.86 ) − 4 ( 5.76 )( −1.35 )  −2.86 ± 39.2836


x=
=
2 ( 5.76 )
11.52
=
−2.86 + 39.2836
−2.86 − 39.2836
or
11.52
11.52
= 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 2 x 2 − 7 x + 4 =
0 then
2
− − 7 ± ( −7 ) − 4 ( 2 )( 4 )  7 ± 17


x=
=
2 ( 2)
4
=
7 + 17
7 − 17
or
4
4
= 2.781 or 0.719, correct to 3
decimal places
4. Solve, correct to 3 decimal places, using the quadratic formula: 4x + 5 =
3
4x + 5 =
x
i.e. 4 x 2 + 5 x =
3 and
3
x
4 x2 + 5x − 3 =
0
222
© 2014, John Bird
2
−5 ± ( 5 ) − 4 ( 4 )( −3)  −5 ± 73


x=
=
2 ( 4)
8
Hence,
=
−5 + 73
−5 − 73
or
8
8
= 0.443 or –1.693, correct to 3 decimal places
5. Solve, correct to 3 decimal places, using the quadratic formula: (2x + 1) =
5
x −3
( 2 x + 1) =
i.e.
i.e.
5
( 2 x + 1)( x − 3) =
and
5
x −3
2 x2 − 6 x + x − 3 =
5
2
− − 5 ± ( −5 ) − 4 ( 2 )( −8 )  5 ± 89


=
2 x2 − 5x − 8 =
0 from which, x =
2 ( 2)
4
=
5 + 89
5 − 89
or
4
4
= 3.608 or –1.108, correct to 3 decimal places.
6. Solve, correct to 3 decimal places, using the quadratic formula: 3 x 2 − 5 x + 1 =
0
If 3 x 2 − 5 x + 1 =
0
2
− − 5 ± ( −5 ) − 4 ( 3)(1)  5 ± 13


then x =
=
2 ( 3)
6
=
5 + 13
5 − 13
or
6
6
= 1.434 or 0.232, correct to 3 decimal places
7. Solve, correct to 3 decimal places, using the quadratic formula: 4 x 2 + 6 x − 8 =
0
If 4 x 2 + 6 x − 8 =
0
2
−6 ± ( 6 ) − 4 ( 4 )( −8 )  −6 ± 164


then x =
=
2 ( 4)
8
=
−6 + 164
−6 − 164
or
8
8
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© 2014, John Bird
= 0.851 or –2.351, correct to 3 decimal places
8. Solve, correct to 3 decimal places, using the quadratic formula: 5.6 x 2 − 11.2 x − 1 =0
If 5.6 x 2 − 11.2 x − 1 =0
2
− − 11.2 ± ( −11.2 ) − 4 ( 5.6 )( −1)  11.2 ± 147.84


then x =
=
2 ( 5.6 )
11.2
=
11.2 + 147.84
11.2 − 147.84
or
11.2
11.2
= 2.086 or –0.086, correct to 3 decimal places
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 3 x 2 + 6 x + 2 x 2 − 8 x =
8
5x2 − 2 x − 8 = 0
i.e.
2
− − 2 ± ( −2 ) − 4 ( 5 )( −8 )  2 ± 164


Hence, x =
=
2 ( 5)
10
=
2 + 164
2 − 164
or
10
10
= 1.481 or –1.081, correct to 3 decimal places
10. Solve, correct to 3 decimal places, using the quadratic formula: 4 x 2 − x(2 x + 5) =
14
If 4 x 2 − x(2 x + 5) =
14
then 4 x 2 − 2 x 2 − 5 x =
14
i.e.
2 x 2 − 5 x − 14 =
0
2
− − 5 ± ( −5 ) − 4 ( 2 )( −14 )  5 ± 137


Hence, x =
=
2 ( 2)
4
=
5 + 137
5 − 137
or
4
4
= 4.176 or –1.676, correct to 3 decimal places
11. Solve, correct to 3 decimal places, using the quadratic formula:
224
5
2
+
=
6
x −3 x −2
© 2014, John Bird
If
5
2
then
+
=
6
x −3 x −2
( x − 3)( x − 2)
5
2
+ ( x − 3)( x − 2)
=6( x − 3)( x − 2)
x −3
x−2
i.e.
5(x – 2) + 2(x – 3) = 6( x 2 − 2 x − 3 x + 6 )
i.e.
5x – 10 + 2x – 6 = 6 x 2 − 30 x + 36
i.e.
0 = 6 x 2 − 30 x + 36 – 5x + 10 – 2x + 6
i.e.
6 x 2 − 37 x + 52 = 0
2
− − 37 ± ( −37 ) − 4 ( 6 )( 52 )  37 ± 121


Hence, x =
=
2 (6)
12
=
37 + 11
37 − 11
or
12
12
= 4 or 2.167, correct to 3 decimal places
12. Solve, correct to 3 decimal places, using the quadratic formula:
If
3
+ 2 x =7 + 4 x
x−7
then
( x − 7)
3
+ 2 x =7 + 4 x
x−7
3
+ 2 x( x − 7) =7( x − 7) + 4 x( x − 7)
x−7
i.e.
3 + 2 x 2 − 14 x =7 x − 49 + 4 x 2 − 28 x
i.e.
0 = 7 x − 49 + 4 x 2 − 28 x − 3 − 2 x 2 + 14 x
i.e.
2 x 2 − 7 x − 52 = 0
2
− − 7 ± ( −7 ) − 4 ( 2 )( −52 )  7 ± 465


Hence, x =
=
2 ( 2)
4
=
7 + 465
7 − 465
or
4
4
= 7.141 or –3.641, correct to 3 decimal places.
13. Solve, correct to 3 decimal places, using the quadratic formula:
If
x +1
=x–3
x −1
x +1
= x – 3 then x + 1 = (x – 1)(x – 3)
x −1
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© 2014, John Bird
i.e.
x + 1 = x 2 − 3x − x + 3
i.e.
0 = x 2 − 3x − x + 3 – x – 1
i.e.
x2 − 5x + 2 = 0
2
− − 5 ± ( −5 ) − 4 (1)( 2 )  5 ± 17


Hence, x =
=
2 (1)
2
=
5 + 17
5 − 17
or
2
2
= 4.562 or 0.438, correct to 3 decimal places
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© 2014, John Bird
EXERCISE 59 Page 124
1. The angle a rotating shaft turns through in t seconds is given by: θ = ωt +
1 2
αt . Determine the
2
time taken to complete 4 radians if ω is 3.0 rad/s and α is 0.60 rad/s2
1
2
θ= ωt + α t 2 and if θ = 4, ω = 3 and α = 0.60
then
or
4= 3t +
1
( 0.60 ) t 2
2
i.e. 4= 3t + 0.03t 2
0.03t 2 + 3t − 4 =
0
Using the quadratic formula:
2
−3 ± ( 3) − 4 ( 0.30 )( −4 )  −3 ± 13.8


t=
=
2 ( 0.30 )
0.60
=
−3 + 13.8
−3 − 13.8
or
0.60
0.60
= 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 – 8I 2 and when P = 2.5 W,
i.e.
i.e.
2.5 = 10I – 8I 2
8 I 2 − 10 I + 2.5 =
0 and using the quadratic formula
2
− − 10 ± ( −10 ) − 4 ( 8 )( 2.5 )  10 ± 20


I=
=
2 (8)
16
=
10 + 20
10 − 20
or
16
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
227
© 2014, John Bird
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 =
Hence,
47.6 =
1
× base × height
2
and if base = b then perpendicular height h = b + 4.3
1
× b × (b + 4.3)
2
2(47.6) = b 2 + 4.3b
i.e.
95.2 = b 2 + 4.3b
Hence,
and
b 2 + 4.3b – 95.2 = 0
2
−4.3 ± ( 4.3) − 4 (1)( −95.2 )  −4.3 ± 399.29


b=
=
2 (1)
2
i.e.
=
−4.3 + 399.29
−4.3 − 399.29
or
2
2
= 7.84 or –8.22, correct to 3 decimal places
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=
l
=
i.e.
12
+ x. Determine the distance between supports when the sag is 20 m.
x
12
+x
x
and when sag l = 20 m, 20
=
20x = 12 + x 2
Using the quadratic formula:
or
12
+x
x
x 2 – 20x + 12 = 0
2
− − 20 ± ( −20 ) − 4 (1)(12 )  20 ± 352


t=
=
2 (1)
2
=
20 + 352
20 − 352
or
2
2
= 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
228
© 2014, John Bird
5. The acid dissociation constant K a of ethanoic acid is 1.8 × 10–5 mol dm–3 for a particular solution.
Using the Ostwald dilution law: K a =
x2
, determine x, the degree of ionization, given that
v(1 − x)
v = 10 dm3
Ka =
then
i.e.
x2
and when K=
1.8 ×10−5 mol dm −3 and v = 10 dm3
a
v (1 − x )
x2
1.8 ×10−5 =
10 (1 − x )
i.e. (10 )(1 − x )(1.8 ×10−5 ) =
x2
1.8 ×10−4 − 1.8 ×10−4 x =
x2
or
x 2 + 1.8 ×10−4 x − 1.8 ×10−4 =
0
Using the quadratic formula:
2
−1.8 ×10−4 ± (1.8 ×10−4 ) − 4 (1)( −1.8 ×10−4 )  −1.8 ×10−4 ± 720.0324 ×10−6


x=
=
2 (1)
2
=
−1.8 ×10−4 + 720.0324 ×10−6
−1.8 ×10−4 − 720.0324 ×10−6
or
2
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 x 2 = 30x + 22x + 4 x 2 = 52x + 4 x 2
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© 2014, John Bird
Since the area of the path is 60.0 m 2 then 60.0 = 52x + 4 x 2
i.e.
4 x 2 + 52x – 60.0 = 0
Using the quadratic formula:
2
−52 ± ( 52 ) − 4 ( 4 )( −60.0 )  −52 ± 3664


x=
=
2 ( 4)
8
=
−52 + 3664
−52 − 3664
or
8
8
= 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 2π r 2 + 2π rh , where r is its radius
and h its height.
If the surface area is 20.0 m 2 and h = 2.80 m,
then
20.0 = 2π r 2 + 2π r (2.80)
i.e.
2π r 2 + 5.60π r − 20.0 =
0 or
i.e.
r 2 + 2.80r −
10
π
r 2 + 2.80r −
20
0 by dividing by 2π
=
2π
0
=
Using the quadratic formula:

2
 10  
−2.80 ± ( 2.80 ) − 4 (1)  −  
 π   −2.80 ± 20.5724

r=
=
2 (1)
2
=
−2.80 + 20.5724
−2.80 − 20.5724
or
2
2
= 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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© 2014, John Bird
8. The bending moment M at a point in a beam is given by: M =
3 x(20 − x)
where x metres is the
2
distance from the point of support. Determine the value of x when the bending moment is 50 Nm.
M=
3 x ( 20 − x )
2
and when M = 50 Nm, then
i.e.
100 = 3x(20 – x)
or
3 x 2 – 60x + 100 = 0
Using the quadratic formula:
i.e.
50 =
3 x ( 20 − x )
2
100 = 60x – 3 x 2
2
− − 60 ± ( −60 ) − 4 ( 3)(100 )  60 ± 2400


x=
=
2 ( 3)
6
=
60 + 2400
60 − 2400
or
6
6
= 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 x 2 = 4 x 2 + 70x
231
© 2014, John Bird
Now, shaded area = 950 – (11 × 24) = 686 m 2
i.e.
4 x 2 + 70x = 686
Using the quadratic formula:
or
4 x 2 + 70x – 686 = 0
2
−70 ± ( 70 ) − 4 ( 4 )( −686 )  −70 ± 15876


x=
=
2 ( 4)
8
−70 + 15876
−70 − 15876
or
8
8
=
=7m
(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 R x2 – 40R x + 336 = 0 and
(b) calculate the resistance of each.
(a) Let resistor values be R1 and Rx
In series:
In parallel:
From (1),
R1 + Rx = 40
R1 Rx
= 8.4
R1 + Rx
(1)
(2)
R1 = 40 – Rx
Substituting in (2) gives:
( 40 − Rx ) Rx
( 40 − Rx ) + Rx
= 8.4
40 Rx − Rx 2
= 8.4
40
i.e.
from which,
40 Rx −=
Rx 2 (40)(8.4)
= 336
Rx 2 − 40 Rx + 336 =
0
i.e.
(b) Solving Rx 2 − 40 Rx + 336 =
0 using the quadratic formula gives:
− − 40 ± ( −40 ) − 4 (1)( 336 )  40 ± 256


=
2 (1)
2
2
Rx
232
© 2014, John Bird
=
40 + 256
40 − 256
or
2
2
= 28 Ω or 12 Ω
From equation (1), when Rx = 28 Ω, R1 = 12 Ω
and when Rx = 12 Ω, R1 = 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 t 2 . Determine the times, correct to 3 significant figures, when the height is
12 m.
If h = 25t – 4 t 2 when height h = 12 m, then
i.e.
12 = 25t – 4 t 2
4 t 2 – 25t + 12 = 0
− − 25 ± ( −25 ) − 4 ( 4 )(12 )  25 ± 433


Hence,
=
t =
2 ( 4)
8
2
=
25 + 433
25 − 433
or
8
8
= 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: =
X ωL −
1
ωC
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 =
X ωL −
1
ωC
and X = 220, L = 800 ×10−3 and C = 25 ×10−6
1
25 ×10−6 ω
Then
220 = 800 ×10−3 ω −
i.e.
220(25 ×10−6 ω ) = (25 ×10−6 ω )(800 ×10−3 ω ) – 1
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© 2014, John Bird
i.e.
5.5 ×10−3 ω =
2 ×10−5 ω 2 – 1
i.e.
2 ×10−5 ω 2 − 5.5 ×10−3 ω – 1 = 0
− − 5.5 ×10−3 ± ( −5.5 ×10−3 ) − 4 ( 2 ×10−5 )( −1)  5.5 ×10−3 ± 1.1025 ×10−4


=
−
5
2 ( 2 ×10 )
4 ×10−5
2
Hence,
ω
=
5.5 ×10−3 + 1.1025 ×10−4
5.5 ×10−3 − 1.1025 ×10−4
or
4 ×10−5
4 ×10−5
= 400
or –125 (which has no meaning)
Hence, the angular velocity ω is 400 rad/s
234
© 2014, John Bird
EXERCISE 60 Page 125
1. Solve the simultaneous equations: y = x2 + x + 1
y=4–x
Equating the y values gives:
x2 + x + 1 = 4 – x
x2 + 2x − 3 =
0
i.e.
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:
15 x 2 + 21x − 11 = 2x – 1
i.e.
15 x 2 + 19 x − 10 =
0
Factorizing gives:
(5x – 2)( 3x + 5) = 0
5x – 2 = 0
and
3x + 5 = 0 i.e. x = −
When x =
2
,y=
5
i.e. x =
2
5
from which,
5
3
5
1
2
2   −1 = −
and when x = − , y =
3
5
5
Thus, the solutions to the simultaneous equations are x =
13
 5
2  −  − 1 =−
3
 3
1
2
2
1
, y = − and x = −1 , y = −4
5
5
3
3
3. Solve the simultaneous equations: 2x2 + y = 4 + 5x
x+y=4
235
© 2014, John Bird
From the first equation,
y =4 + 5 x − 2 x 2
and from the second equation, y = 4 – x
Equating the y values gives:
4 + 5 x − 2 x 2 =4 − x
i.e.
6x − 2x2 =
0
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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© 2014, John Bird