QUADRATIC FUNCTIONS 1)
Simplify the following. a) (3 x + 1)( x x 2) d) (3 x 1)(2 x + 2) g) (5 x + 6)(3 x 10) j) (3 x)(4 x)(4 x) x) m) ( x x + 1)( x x + 3)( x x 5)
2)
b) ( x x 9)( x x 7) e) (4 x + 3)(2 x 3) h) (5 2 x)( x)( x x + 1) 2 k) ( x x + 2 x + 1)( x x 1) 3 n) (2 x + 1) .
c) ( x x 100)( x x 10) f) ( x x 3)(3 x 6) i) (3 x 4)(5 2 x) x) l) ( x x 1)( x x 2)( x x 3)
a) Copy Copy and and com compl plet etee thi thiss tabl tablee of of val value uess for for the the func functi tion on y y = x2 4 x. x. x 2
y = x
2
0
1
1
2
3
4
5
6 12
12
4 x
Plot the graph of y of y = x2 4 x for values of x of x between between 2 and and 6. b) Use your graph graph to find the values of x of x of the points where the curve crosses the x the xaxis.
3)
a) Copy Copy and and com compl plet etee thi thiss tabl tablee of of val value uess for for the the func functi tion on y y = x2 5 x + 2.
2
y = x
x
1
8
5 x + x + 2
0 2
1
2
3
4
5
6 8
Plot the graph of y of y = x2 5 x + 2 for for valu values es of x of x between between 1 and and 6. 6. b) The equation of the line of symmetry symmetry of the curve is x is x = a for some number a. Use your your graph graph to to find a.
4)
Copy Copy and and com compl plet etee thi thiss tab table le of valu values es for for the the func functi tion on y y = 6 2 x2. x
3 2
y = 6 2 x
2
1
0
1
2
12
3 -12
Plot the graph of y of y = 6 2 x2 for values of x of x between between 3 and and 3. 3.
5)
i) Copy Copy and and com compl plet etee the the fol follo lowi wing ng tabl tablee of of val value uess for for the the fun funct ctio ion n of of y = 10 + x + x – – 2 x2. x 2 2 x y = 10 + x + x
–3 –11
–2
–1
0
1
2
ii) Plot the graph of y = y = 10 + x + x – – 2 x2 for values of x of x between between 3 and and 4. 4. iii) Using the same axes, plot the graph of the straight line y = 2 x + 1. iv) Write down the x -coordinates -coordinates of the points where your 2 graphs meet. www.mathsguru.co.uk
3
4
6)
Factorise the following quadratic expressions. a) x2 + 7 x + 10 e) x2 + 8 x + 12 i) y2 + 15 y + 36 m) x2 2 x 35 q) x2 8 x + 15
7)
10)
*11)
d) x2 + 10 x + 21 h) y2 + 10 y + 25 l) z 2 + z 6 p) y2 5 y + 6 t) b2 4b 21.
b) 2 x2 + 7 x + 3 f) 2 x2 + 7 x + 5 j) 3 x2 17 x 28 n) 6 y2 + 7 y 3 r) 12 x2 + 23 x + 10
c) 3 x2 + 7 x + 2 g) 3 x2 5 x 2 k) 6 x2 + 7 x + 2 o) 10 x2 + 9 x + 2 s) 4 y2 23 y + 15
d) 2 x2 + 11 x + 12 h) 2 x2 x 15 l) 3 x2 11 x + 6 p) 6 x2 19 x + 3 t) 6 x2 27 x + 30.
Solve the following equations by factorising and sketch the appropriate graphs. a) x2 + 7 x + 12 = 0 d) x2 + x 6 = 0 g) x2 5 x + 6 = 0 j) 2 x2 3 x 2 = 0 m) 6 x2 13 x + 6 = 0 p) y2 15 y + 56 = 0 s) 6a2 a 1 = 0
9)
c) x2 + 8 x + 15 g) y2 + 11 y + 24 k) a2 a 12 o) x2 6 x + 8 s) a2 + 14a + 45
Factorise the following quadratic expressions. a) 2 x2 + 5 x + 3 e) 3 x2 + 8 x + 4 i) 2 x2 + x 21 m) 3 y2 11 y + 10 q) 8 x2 10 x 3
8)
b) x2 + 7 x + 12 f) y2 + 12 y + 35 j) a2 3a 10 n) x2 5 x 24 r) a2 a 6
b) x2 + 7 x + 10 = 0 e) x2 8 x + 12 = 0 h) x2 4 x 5 = 0 k) 3 x2 + 10 x 8 = 0 n) 4 x2 29 x + 7 = 0 q) 12 y2 16 y + 5 = 0 t) 4a2 3a 10 = 0.
c) x2 + 2 x 15 = 0 f) x2 + 10 x + 21 = 0 i) x2 + 5 x 14 = 0 l) 2 x2 + 7 x 15 = 0 o) 10 x2 x 3 = 0 r) y2 + 2 y 63 = 0
Complete the square for each of the following. a) x2 + 8 x, b) x2 12 x, c) x2 + 4 x, e) x2 + 12 x, f) x2 4 x + 4, g) x2 6 x + 9, i) x2 + 4 x + 1, j) x2 6 x 9, k) x2 + 2 x 15, m) x2 + 6 x + 15, n) x2 8 x 3, o) x2 10 x + 1, q) x2 + 8 x + 11, r) x2 6 x 19, s) x2 6 x,
d) x2 10 x, h) x2 + 6 x + 9, l) x2 + 16 x + 5, p) x2 + 2 x +1, t) x2 + 10 x + 30.
Complete the square for each of the following. a) 2 x2 + 8 x, b) 2 x2 8 x + 1, c) 2 x2 12 x, e) 2 x2 8 x 3, f) 5 x2 10 x + 6, g) 3 x2 + 12 x + 1, i) 4 x2 8 x + 1, j) 3 x2 + 6 x 2, k) 3 x2 12 x 5, m) 4 x2 + 16 x + 3, n) 2 x2 + 20 x + 18, o) 2 x2 + 4 x 3,
d) 3 x2 + 6 x + 1, h) 2 x2 4 x 2, l) 2 x2 + 8 x + 20, p) 3 x2 + 12 x 7.
Complete the square for the following. (Each one will require fractions; -do not use decimals!) a) x2 + x, b) x2 + 3 x, c) x2 + 5 x + 1, d) x2 + x 2, e) x2 x, f) x2 3 x + 1, g) x2 5 x + 3, h) x2 x 1, i) 2 x2 + 6 x, j) 2 x2 2 x + 1, k) 3 x2 9 x 5, l) x2 + 12 x.
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12)
Solve the following equations by firstly completing the square and secondly by factorising. i) x 2
iii) x 2
6x
v) 2 x 2
40
10 x
+
16
7x
4
+
0,
=
= =
0,
ii) x 2
+
8x
iv) x 2
+
12 x
20 =
0,
=
13,
0.
13)
Show that the equation x2 + 6 x + 10 = 0 has no solutions. {Hint: try and solve it by completing the square and see what happens!}
14)
Use the quadratic formula to solve the following equations. Give your answers to 2 decimal places where appropriate. a) 2 x2 + 11 x + 5 = 0 d) 3 x2 10 x + 3 = 0 g) 2 x2 + 5 x 1 = 0 j) 3 x2 7 x 20 = 0
15)
b) 3 x2 + 11 x + 6 = 0 e) 5 x2 7 x + 2 = 0 h) 3 x2 + x 3 = 0 k) 2 x2 7 x 15 = 0
c) 6 x2 + 7 x + 2 = 0 f) 6 x2 11 x + 3 = 0 i) 3 x2 + 8 x 6 = 0 l) 2 x 6 x2 = 0.
Use the formula to show that the equation x2 + 3 x + 3 = 0 has NO solutions. {Hint: simply use the formula and remember that you cannot square-root a negative number!}
16)
Rearrange the following as quadratic equations and solve. a) x d)
17)
+
1
=
2 x
1 + 1
3, 1
1
2 x x
+ +
3 4
=
x
x
+
5 1
,
c)
12 x
+
2
1 x
=
2,
= 3.
Complete the square for the following. Hence sketch the curves and write down the CO-ORDINATES of min/max points as appropriate. a) y = x2 + 8 x + 20, d) y = x2 + 6 x, g) y = 5 + 6 x x2, j) y = x2 2 x 3, m) y = x2 4 x + 7
18)
b)
b) y = x2 8 x + 3, e) y = 2 x2 + 8 x + 7, h) y = x2 + 2 x + 1, k) y = 4 x2 8 x + 11, n) y = 8 x2
c) y = x2 6 x + 11, f) y = 4 2 x x2, i) y = x2 9 l) y = 10 x x2 o) y = x2 + 2 x 4.
i) Complete the square for x2 + 6 x + 10. ii) Use i) to show that there are NO solutions to the equation x2 + 6 x + 10 = 0. iii) Use i) to sketch the graph of y = x2 + 6 x + 10 and explain how this also shows that the equation x2 + 6 x + 10 = 0 has NO solutions.
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19)
i) Complete the square for x2 + 4 x + 6 and hence write down the MINIMUM value of x2 + 4 x + 6. ii) Use i) to explain why the equation x2 + 4 x + 6 = 1 has NO solutions.
20)
21)
Solve the following inequalities.
{Sketch the graphs first to be safe!}
a) x2 + 3 x 10 0 d) x2 5 x + 4 > 0
b) x2 + 5 x + 4 0 e) x2 6 x 8
g) x2 < 64
h) x2 > 64
j) ( x + 3)( x 7) > 0
k) 2 x2 5 x + 2 < 0
Factorise the expression 2 x 2
+
5x
c) x2 2 x > 0 f) x2 + 5 x < 6 1 i) x 2 < 9 l) 2 x( x 1) < 3 – x.
3 and hence sketch the graph of y = 2
2
+
5 x
3.
2
Solve the inequality 2 x + 5 x > 3.
2
2
2
22)
Sketch the graph of y = x + 4 x 12 and hence solve the inequality 2 x + 5 x < x + x + 12.
23)
Draw sketch graphs to solve each of the following inequalities: 2 a) x + 6 x + 8 > 3,
24)
b) 3
+
4 x
2
+
3.
Solve the following pairs of simultaneous equations. {Use either algebraic or graphical methods.} a) x + y = 3, y = x + 1 d) 3 x + y = 9, y = x + 1 g) x + 2 y = 7, y = x + 2 j) 3 x + 2 y = 15, y = 2 x 3 m) 5 x + 3 y = 41, y = 2 x 1 p) 5 x + 3 y = 18, x = 2 y + 1 s) 7 x 2 y = 1, y = 2 x + 1 u) 2 x 3 y = 13, y = 2 x 1 w) 5 x 3 y = 6, y = 2 x 3
25)
2
b) x + y = 3, y = x 1 e) 3 x + 2 y = 11, y = x + 3 h) 2 x + y = 7, y = x 2 k) 3 x + 4 y = 20, y = 2 x 6 n) 3 x + 4 y = 27, x = y + 2 q) 5 x 3 y = 1, y = x + 1 t) 9 x 2 y = 4, y = 3 x + 1 v) 3 x 2 y = 5, y = 3 x 7 x) 3 x 2 y = 7, y = x 1.
c) 2 x + y = 8, y = x 1 f) 2 x + 3 y = 11, y = x 3 i) 2 x + y = 9, y = 2 x + 1 l) 5 x + 2 y = 51, y = 2 x + 3 o) 5 x + 3 y = 37, x = y + 1 r) 7 x 3 y = 6, y = x + 2
Solve the following pairs of simultaneous equations. a)
d)
g)
x2 + y2 = 25 x + y = 7 2 2 16 x + y = 65 x + y = 1
y = x2 x + 3 y = 2 x + 1
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b)
e)
h)
y = x2 3 x 8 y = x 3 2 2 2 x + 3 xy + y = 8 y x = 2
x2 + y2 = 26 y = 3 x + 2
c)
x2 y2 = 8 y = 2 x
f)
y = 2 x2 3 x 1 y 3 x = 7.
i)
y = x2 5 y = 2 x 2.
26)
Establish the nature of the roots of the following quadratic equations by investigating suitable discriminants. a) x2 + 5 x + 3 = 0 b) x2 5 x + 7 = 0 c) 3 x2 x 1 = 0 d) 4 x2 + 12 x + 9 = 0 e) 2 x2 = 3 x 4 f) x2 + ax + a2 = 0. 2
{Hint: remember to rearrange each equation into the form ax + bx + c = 0.}
27)
The roots of the equation 3 x2 + k x + 12 = 0 are equal. Find the possible values of k.
28)
Find the range, or ranges, of values K can take for the equation Kx2 4 x + (5 K ) = 0 to have 2 distinct real roots.
29)
If ax2 8 x + 2 = 0 has a repeated root, find the value of a.
30)
Find the range(s) of values b can take for 9 x2 + bx + 4 = 0 to have 2 real distinct roots.
31)
Find the range(s) of values k can take for x2 + (k + 1) x + 1 = 0 to have 2 distinct roots.
32)
Find the range(s) of values k can take for 2 x2 + (3 k ) x + k + 3 = 0 to have 2 real distinct roots.
33)
Show that the function y = 3 x2 5 x + 4 is always positive for any value of x. 2
{Hint: show that the equation 3 x 2 of y = 3 x 5 x + 4.}
5 x + 4 = 0 has no solutions. Now think about the graph
Miscellaneous questions. The following questions contain many past exam questions.
34)
The diagrams shows that part of the graph of y
=
3
for which
0, and the graph of y = 2 x + 1.
The graphs intersect at the point A( x, y).
= 2 x + 1
y
2
a) Show that x satisfies the equation 2 x + x www.mathsguru.co.uk
=
3 x
3 = 0 and hence find the co-ordinates of A.
b) Show that the line
35)
=
1
x
does not intersect with the graph of the curve y
=
3
.
i) Sketch, on the same axes, the graphs of the curve y = x2 + 3 x and the line y = x + k where k is a constant. ii) Given that the curve given by y = x2 + 3 x and the line y = x + k intersect at two distinct points, show that k > 1. {Hint: solve simultaneously to get x2 + 2 x k = 0 and now use the discriminant etc.}
36)
iii) Interpret, in terms of the above graphs, the case when k
=
1.
i) Solve the simultaneous equations y = x2 3 x + 2,
y = 3 x 7.
ii) Interpret your solution to part i) geometrically.
Factorise ( x
38)
The quadratic polynomial x2 10 x + 17 is denoted by f ( x). Express f ( x) in the form ( x a)2 + b, stating the values of a and b. {Hint: the question is asking you to complete the square!}
+
3)
2
37)
16 .
Hence find the least possible value that f ( x) can take, and the corresponding value of x.
39)
Solve the simultaneous equations x + y = 1, x2 xy + y2 = 7.
40)
Find the set of values of a for which the equation ax 2
41)
Using the substitution y = x2, or otherwise, solve the equation x4 + 5 x2 36 = 0.
42)
4 2 Solve the equation 2 x 3 x + 1 = 0 .
43)
Solve the equation 2 t 3
44)
Solve the simultaneous equations 2 x + y = 3,
2
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6
+
a
0 has two distinct real roots.
{c.f. example 9.2}
1
=
1
3 t 3 + 1 = 0 by means of the substitution u = t 3 .
2 2 x xy = 10.
45)
a)
O
The diagram shows the graphs of y = x 1 and y = kx2, where k is a positive constant. The graphs intersect at two distinct points A and B. Write down the quadratic equation satisfied by the xcoordinates of A and B, and hence show 1 k < . 4 2 2 Solve the inequality x 1 > x . 9 b) Describe briefly the relationship between the graphs of y = x 1 and y = kx2 in each of the 1 1 following cases i) k = , ii) k > . 4 4 c) Show, by means of a graphical argument or otherwise, that when k is a negative constant, the equation x 1 = kx2 has two real roots, one of which lies between 0 and 1. 2
{Hint: draw a sketch graph showing y = x 1 and y = kx . Remember that k is negative and that the solutions of the equation x 1 = kx2 are the xcoordinates of where the 2 graphs meet!}
46)
The diagram shows the graphs of y = 6 2 x and y = k x2, where k is a positive constant. The graphs intersect at two distinct points A and B.
k = k x
2
B = 6 2 x www.mathsguru.co.uk
a) Write down the quadratic equation satisfied by the x-coordinates of A and B, and hence show that k > 5. b) Solve the inequality 9 x2
6 2 x.
c) Describe, briefly, the relationship between the graphs of y = 6 2 x and y = k x2 in each of the following cases. i) k = 5, ii) k < 5.
47)
A quadratic function is defined by f ( x)= x2 + kx + 9, where k is a constant. It is given that the equation f ( x) = 0 has two distinct real roots. Find the set of values that k can take. For the case where k = 4 3 , i) express f ( x) in the form ( x + a)2 + b, stating the values of a and b, and hence write down the least value taken by f ( x), {Hint: the question is asking you to complete the square!} ii) solve the equation f ( x) = 0, expressing your answers in terms of surds, simplified as far as possible.
48)
Solve the equation x 2 as possible.
(6 3 ) x + 24 = 0, giving your answers in terms of surds, simplified as far
i) solve the inequality x 2
Hence,
(6 3 ) x + 24 < 0, {Hint: sketch a graph!}
ii) find all four solutions of the equation x 4 answers to 2 decimal places. 2
{Hint: put y = x to give y 2
(6
(6 3 ) x 2 + 24 = 0, giving your
3 ) y + 24 = 0 and now use your earlier
answers to find y etc.}
49)
A biologist claims that the average height, h metres, of trees of a certain species after t months’ growth is given by
h =
1 5
2
t3
+
1 8
1
t 3 .
For this model, i) find the average height of trees of this species after 64 months, ii) find the number of months that the trees have been growing when the average height is 10 1
metres. {Hint: put h = 10, rearrange to get 8u2 + 5u 400 = 0 where u = t 3 etc.}
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ANSWERS. 1)
6)
7)
8)
a) 3 x2 5 x 2 b) x2 16 x + 63 c) x2 110 x + 1000 e) 8 x2 6 x 9 f) 3 x2 15 x + 18 g) 15 x2 32 x 60 i) 6 x2 + 23 x 20 j) x2 7 x + 12 k) x3 + x2 x 1 m) x3 x2 17 x 15 n) 8 x3 + 12 x2 + 6 x + 1. a) ( x + 2)( x + 5) b) ( x + 3)( x + 4) c) ( x + 3)( x + 5) e) ( x + 6)( x + 2) f) ( y + 7)( y + 5) g) ( y + 8)( y + 3) i) ( y + 12)( y + 3) j) (a 5)(a + 2) k) (a 4)(a + 3) m) ( x 7)( x + 5) n) ( x 8)( x + 3) o) ( x 4)( x 2) q) ( x 3)( x 5) r) (a 3)(a + 2) s) (a + 9)(a + 5) a) (2 x + 3)( x + 1) b) (2 x + 1)( x + 3) c) (3 x + 1)( x + 2) e) (3 x + 2)( x + 2) f) (2 x + 5)( x + 1) g) (3 x + 1)( x 2) i) (2 x + 7)( x 3) j) (3 x + 4)( x 7) k) (2 x + 1)(3 x + 2) m) ( y 2)(3 y 5) n) (2 y + 3)(3 y 1) o) (5 x + 2)(2 x + 1) q) (4 x + 1)(2 x 3) r) (3 x + 2)(4 x + 5) s) (4 y 3)( y 5) a) 3, 4 b) 2, 5 c) 3, 5 d) 2, 3 e) 2, 6 g) 2, 3 h) 5, 1 i) 7, 2 j) ½ , 2 k) 23 , 4 2 3
m) s) 9)
10)
11)
12)
14)
17)
1 3
,½
n) ¼, 7
o)
3 5
, ½
p) 7, 8
q)
5 6
d) ( x + 7)( x + 3) h) ( y + 5)2 l) ( z + 3)( z 2) p) ( y 3)( y 2) t) (b 7)(b + 3). d) (2 x + 3)( x + 4) h) (2 x + 5)( x 3) l) (3 x 2)( x 3) p) (6 x 1)( x 3) t) (2 x 5)(3 x 6). f) 3, 7 l) 1½, 5
,½
r) 7, 9
t) 1¼, 2.
a) ( x + 4)2 16 e) ( x + 6)2 36 i) ( x + 2)2 3 m) ( x + 3)2 + 6 q) ( x + 4)2 5 a) 2( x + 2)2 8 e) 2( x 2)2 11 i) 4( x 1)2 3 m) 4( x + 2)2 13
b) ( x 6)2 36 f) ( x 2)2 j) ( x 3)2 18 n) ( x 4)2 19 r) ( x 3)2 28 b) 2( x 2)2 7 f) 5( x 1)2 + 1 j) 3( x + 1)2 5 n) 2( x + 5)2 32
c) ( x + 2)2 4 g) ( x 3)2 k) ( x + 1)2 16 o) ( x 5)2 24 s) ( x 3)2 9 c) 2( x 3)2 18 g) 3( x + 2)2 11 k) 3( x 2)2 17 o) 2( x + 1)2 5
d) ( x 5)2 100 h) ( x + 3)2 l) ( x + 8)2 59 p) ( x + 1)2 t) ( x + 5)2 + 5. d) 3( x + 1)2 2 h) 2( x 1)2 4 l) 2( x + 2)2 + 12 p) 3( x + 2)2 19.
) 14 2 e) ( x 12 ) 14 2 i) 2 ( x + 32 ) 92
b) ( x
) 94 2 f) ( x 32 ) 54 2 j) 2 ( x 12 ) + 12
c) ( x
) 214 2 g) ( x 52 ) 134 2 k) 3 ( x 32 ) 474
d) ( x
+
1 2 2
)
9 4
h) ( x
1 2 2
5 4
i) x = 10 or 4 1 v) x = or 4. 2 a) ½, 5
ii) x = 2 or 10
iii) x = 2 or 8
iv) x = 1 or 13
a) ( x
f) 15) 16)
, 1½
d) 6 x2 + 4 x 2 h) 2 x2 + 3 x + 5 l) x3 6 x2 + 11 x 6
1 3
+
1 2 2
, 1½
b)
+
2 3
3 2 2
, 3
g) 019, 269
+
c) ½,
5 2 2
2 3
,
h) 085, 118
l) ( x
d)
1 3
+
) 1 2 ) 4
,3
i) 061, 328,
1 16
.
e)
2 5
,1
j) 1 23 , 4
k) 1½, 5 l) 23 , ½. This question gives you a simple technique for checking when a quadratic equation HAS solutions. a) 018, 282 b) No solutions c) 031, 319, d) 13 or 13 . a) y = ( x + 4)2 + 4, Min (4, 4) c) y = ( x 3)2 + 2, Min (3, 2) e) y = 2( x + 2)2 1, Min (2, 1) g) y = 14 ( x 3)2, Max (3, 14) i) y = x2 9, Min (0, 9) k) y = 4( x 1)2 + 7, Min (1, 7) m) y = ( x 2)2 + 3, Min (2, 3) o) y = ( x + 1)2 5, Min (1, 5).
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b) y = ( x 4)2 13, Min (4, 13) d) y = ( x + 3)2 9, Min (3, 9) f) y = 5 ( x + 1)2, Max (1, 5) h) y = ( x + 1)2, Min (1, 0) j) y = ( x 1)2 4, Min (1, 4) l) y = 25 ( x 5)2, Max (5, 25) n) y = 8 x2, Max (0, 8)
18) 19) 20)
21) 23) 24)
25)
i) ( x + 3)2 + 1. i) ( x + 2)2 + 2 ii) Because the least value of x2 + 4 x + 6 is 2. a) 5 x 2 b) 4 x 1 c) x < 0 or x > 2 d) x < 1 or x > 4 e) x 2 or x 4 f) 6 < x < 1 g) 8 < x < 8 h) x < 8 or x > 8 i) 1/3 < x < 1/3 j) x < 3 or x > 7 k) ½ < x < 2 l) 1 < x <3/2. 6 < x < 2. x < 3 or x > ½.. 22) a) x < 5 or x > 1 b) 3 x ½. a) x = 1, y = 2. b) x = 2, y = 1. c) x = 3, y = 2. d) x = 2, y = 3. e) x = 1, y = 4. f) x = 4, y = 1. g) x = 1, y = 3. h) x = 3, y = 1. i) x = 2, y = 5. j) x = 3, y = 3. k) x = 4, y = 2. l) x = 5, y = 13. m) x = 4, y = 7. n) x = 5, y = 3. o) x = 5, y = 4. p) x = 3, y = 1. q) x = 2, y = 3. r) x = 3, y = 5. s) x = 1, y = 3. t) x = 2, y = 7. u) x = 4, y = 7. v) x = 3, y = 2. w) x = 3, y = 3. x) x = 5, y = 4. a) x = 3, y = 4 or x = 4, y = 3. b) x = 5, y = 2 or x = 1, y = 4. 49 c) x = 3, y = 1. d) x = 2, y = 1 or x = 32 17 , y = 17 . e)
x = 2, y = 0 or x =
1 3
, y =
7 3
.
f)
x = 1, y = 4 or x = 4, y = 19.
37) 38) 40) 42)
g) x = 2, y = 5 or x = 1, y = 3. h) x = 1, y = 5 or x = 2.2, y = 4.6. i) x = 3, y = 4 or x = 1, y = 4. a) Discriminant = 13. 2 distinct real roots. b) Discriminant = 3. No real roots. c) Discriminant = 13. 2 distinct real roots. d) Discriminant = 0. Equal roots. e) {Rearrange to get 2 x2 3 x + 4.} Discriminant = 23. No real roots. f) Discriminant = 3a2. No real roots. k = 12 or k = 12. 28) K < 1 or K > 4. 29) a = 8. b < 12 or b > 12. 31) k < 3 or k > 1. 32) k < 1 or k > 15. 2 The equation 3 x 5 x + 4 = 0 has discriminant 23 and thus has no solutions. This means that the graph of y = 3 x2 5 x + 4 lies completely above the x-axis and hence y = 3 x2 5 x + 4 is always positive. a) (1, 3). iii) If k = 1, then the line y = x + k touches the curve y = x2 + 3 x exactly once, i.e. y = x + k is a tangent to y = x2 + 3 x. i) x = 3, y = 2, ii) The line y = 3 x 7 is the tangent to the curve y = x2 3 x + 2 at the point (3, 2) since the 2 graphs touch at the single point (3, 2). ( x 1)( x + 7). ( x 5)2 8; a = 5, b = 8. Least f ( x) = 8. 39) When x = 2, y = 1, when x = 1, y = 2. 41) x = 2 or x = 2. 3 < a < 3. 1 1 x = 2 or 2 or 1 or 1. 43) t = 18 or t = 1.
44)
When x = 54 , y =
11 2
45)
a) kx2 x + 1 = 0 ;
3 2
26)
27) 30) 33)
34) 35) 36)
y = kx2.
, when x = 2, y = 1. < x < 3. b) i) The 2 graphs touch exactly once, i.e. y = x 1 is a tangent to
ii) The 2 graphs do not intersect.
46)
b) 1 x 3. c) i) The line y = 6 2 x is a tangent to the curve y = k x2. ii) The 2 graphs do not intersect.
47)
k < 6 or k > 6. i) ( x 2 3 )2 3; least value of f ( x) = 3, ii) x =
48) 49)
x = 2 3 or 4 3 . i) 2 3 < x < 4 3 . ii) x = 1.86 or 1.86 or 2.63 or 2.63 (2 decimal places). i) 3.7 m. ii) 309.67 months (2 decimal places).
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3 or 3 3 .
