Matriculation QS015 2014

S.Y.Chuah

June 24, 2014

Chapter 2 : Equations, Inequalities & Absolute Values

2.1 Equations

(a) Solve equations involving surds, indices and logarithms

2.2 Inequalities

(a) Relate the properties of inequalities

(b) Solve linear inequalities

(c) Solve quadratic inequalities by graphical and analytical approach.

(d) Solve rational inequalities involving linear expressions.

2.3 Absolute Values

(a) Use the properties of absolute values.

i. |a| 0ii. | a| = |a|iii. |a + b| = |b + a|iv. |a b| = |b a|v. |ab| = |a||b|; andvi.

ab

= |a||b|(b) Solve absolute equations of the forms.

i. |ax + b| = cii. |ax + b| = cx + d

iii. |ax + b| = |cx + d|; andiv. |ax2 + bx + c| = d

(c) Solve absolute inequalities of the forms.

i. |ax + b| > cii. |ax + b| > |cx + d|

iii.

ax + bcx + d > e; and

iv. |ax2 + bx + c| > d

1

2.2 Inequalities

2.2.1 Relate the properties of inequalities

If a > b,

(i) a + c > b + c for any c

(ii) a c > b c for any c(iii) ac > bc if c > 0

(iv) ac < bc if c < 0

(v)1

a

4

2x > 2

2.2.2 Solve Linear Inequalities

After solving the inequalities, the answers can be written in the form of solution set orinterval.

(a, b] = {x : a < x b} (a, b) = {x : a < x < b} [a, b] = {x : a x b}

EXAMPLE 1 Give the solution sets for the following inequalities.

(A) 2x 4 > 8 (B) 3x + 5 x 7 (C) 4x 5 < 2x + 9

EXAMPLE 2 Give the solution intervals for the following inequalities.

(A) 7 < 2 + 3x < 8 (B) 2 x + 42

9 (C) 6 < 4 x 12

**Special Cases

2x 3 7 7x 3x + 7This inequality can be split into two parts.

2x 3 7 7x and 7 7x 3x + 7

2.2.3 Solve Quadratic Inequalities

Quadratic InequalitiesA quadratic inequality is an inequality of the form ax2 + bx+ c < 0 where a, b and c arereal number with a 6= 0. A quadratic inequality can also be referred to as ax2+bx+c > 0,ax2 + bx + c 0 or ax2 + bx + c 0.

Quadratic inequalities can be solved by using graphical or analytical methods.

(A) Graphical ApproachTo solve quadratic inequalities by graphical method, we have to master the skill in sketch-ing a quadratic function.Let us review back what you have learned before in Form 5 in sketching a quadraticfunction.

For example, f(x) = x2 + 2x 8

In solving quadratic inequalities, we do not have to draw the quadratic graph in details.

Example is given as below:-

x2 + 2x 8 < 0(x + 4)(x 2) < 0

Hence, we get the roots x = 4 and x = 2. Since we want x2 + 2x 8 < 0, therefore

The solution set is .

EXAMPLE 3 Solve these inequalities by graphical approach.

(A) x2 3x 4 0 (B) m2 2m 1

(B) Analytical Approach

(i) Analytical approach by using theorem.

(ii) Analytical approach by using number line.

(iii) Analytical approach by using table of sign.

Theorem 1ab > 0 if and only if a > 0 and b < 0 or a < 0 and b < 0.ab < 0 if and only if a > 0 and b < 0 or a < 0 and b > 0.

Now we will look at all these methods by using an example.

EXAMPLE 4 Solve the inequality x2 + 2x < 8.(i)By Theorem

x2 + 2x < 8

x2 + 2x 8 < 0(StandardForm)(x 2)(x + 4) < 0

In this example, we are looking for values of x that will make the quadratic on the leftside less than 0 (negative).From Theorem 2, we know that (x 2)(x + 4) < 0 iff

(x 2) > 0 and (x + 4) < 0 or (x 2) < 0 and (x + 4) > 0

(ii) By Table of sign

Before drawing the table of sign, we need to identify the zeroes of the quadratic equationor in other words, the roots of the quadratic equation.

x2 + 2x 8 = 0(x 2)(x + 4) = 0

Next, plot the zeroes on a number line and determine the intervals as below:

Lastly, draw the table of sign to determine the answer.

(iii) By using Number line

2.2.4 Solving Rational Inequalities

The steps for solving quadratic inequalities can, with slight modification, be used to solverational inequalities such as

x 3x 5 > 0 or

x2 + 5x 65 x 3

If after a suitable operations on an equality, the right side is 0 and the left side is of the

formP

Q, where P and Q are nonzero polynomials, then the inequality is said to be a

rational inequalit in standard form.For example,

2

x 1 0,

1. |x| = p is equivalent to x = p or x = p2. |x| < p is equivalent to p < x < p [x > p and x < p]3. |x| > p is equivalent to x > p or x < p.

Theorem 3: Properties of Equations & inequalities involving |ax+ b| For p > 0,

1. |ax + b| = p is equivalent to ax + b = p or ax + b = p2. |ax + b| < p is equivalent to p < ax + b < p [ax + b > p and ax + b < p]3. |ax + b| > p is equivalent to ax + b > p or ax + b < p.

Solve Absolute Values Equations & Inequalities of Different Form

i. Solve |ax + b| = c , |ax + b| > c(A) |3x + 5| = 4(B) |2x 1| 3(C) |7 3x| 2

ii. Solve |ax + b| = cx + d , |ax + b| > cx + d.(A) |x + 4| = 3x 8(B) |x 2| 2x + 1

(C) |x + 2| < 12

(6 x)

iii. Solve |ax + b| = |cx + d| , |ax + b| > |cx + d|. If x is any real number,

|x2| |x|2 x2 x2 = |x| meansis equivalent to

|x| = |a| x2 = a2|x| > |a| x2 > a2

ifa 6= 0, |x| < |a| x2 < a2

(A) |x 2| = |2x 1|(B) |x 2| |2x 3|(C) |4x + 1| |4x 1|

iv. Solve

ax + bcx + d > e.

(A)

2xx + 3 1

v. Solve |ax2 + bx + c| = d , |ax2 + bx + c| > d(A) |x2 x 1| = 1(B) |x2 2x 10| 2

Examples & Solutions of Absolute Values Inequalities

(A) |x 1| 3

(B) |2x 1| 1

(C) |3 x| 2x + 9

(D) |3x + 7| 2x + 9