1. 1.1
Solve for x:
1. 1.1.1
x(x - 7) = 0
(2)
1. 1.1.2
x2 - 6x + 2 = 0 (Correct to TWO decimal places)
(3)
1. 1.1.3
x - 1 + 1 = x
(5)
1. 1.1.4
3x + 3 - 3x + 2 = 486
(4)
2. 1.2
Given f(x) = x2 + 3x - 4
1. 1.2.1
Solve for x if f(x) = 0
(2)
1. 1.2.2
Solve for x if f(x) < 0
(2)
1. 1.2.3
Determine the values of x for which f`(x) ≥ 0
(2)
2. 1.3
Solve for x and y: x = 2y and x2 - 5xy = -24
(4)
(24)

1. 1.1
1. 1.1.1
x(x - 7) = 0
x = 0 or x = 7
2. 1.1.2
x2 - 6x + 2 = 0
x =
 6 ± √(-6)2 - 4(1)(-2) 2(1)

=
 6 ± √28 2

x = 0.35 or x = 5.65
3. 1.1.3
x - 1 + 1 = x
x - 1 = x -1
x - 1 = x2 - 2x + 1
x2 - 3x + 2 = 0
(x - 2)(x - 1) = 0
x = 2 or x = 1

4. 1.1.4
3x + 3 - 3x + 2 = 486
3x33 - 3x32 = 486
3x(33 - 32) = 486
3x = 27
3x = 33
x = 3
2. 1.2
1. 1.2.1
f(x) = x2 + 3x - 4
0 = (x + 4)(x - 1)
x = -4 or x = 1
2. 1.2.2
x2 + 3x - 4 < 0
(x + 4)(x - 1) < 0 -4 < x < 1
3. 1.2.3
2x + 3 ≥ 0
x ≥ -32
f'(x) ≥ 0 when f is increasing

The turning point occurs at x =
 -4 + 1 2

x ≥ -3/2
3. 1.3
x = 2y and x2 - 5xy = -24

(2y)2 - 5(2y)(y) = -24
4y2 - 10y2 = -24
-6y2 = -24
y2 = 4
y = -2 or y = 2
x = -4 or x = 4
Msaki, 26 Mar 2018 10:46
4.  