1. Chapter 3 – Algebra III 03 Learning Outcomes
In this chapter you have learned to:
Solve equations containing surds
Find solutions to inequalities of the following forms: linear, quadratic and rational
Understand absolute value (modulus) and use its notation, |x|
Find solutions to modulus inequalities
Use discriminants to determine the nature of roots of quadratic equations
Prove algebraic inequalities

2. 03 Algebra III Surd Equations 𝑺𝒐𝒍𝒗𝒆 𝒙+𝟕 + 𝒙+𝟐 =𝟓
𝑺𝒐𝒍𝒗𝒆 𝒙+𝟕 + 𝒙+𝟐 =𝟓
As we have more than one surd term, we leave one surd term on one side of the equation  and move every other  term onto the other side, to simplify the arithmetic. 
𝑥+7 =5− 𝑥+2
𝑥 = 5− 𝑥+2 2
Squaring both sides
𝑥 + 7=(5− 𝑥+2 )(5− 𝑥+2 )
𝑥+7=25−5 𝑥+2 −5 𝑥+2 +( 𝑥+2 ) 2
𝑥+7=25−10 𝑥 𝑥+2
𝑥+7=27−10 𝑥 𝑥
Again, isolate the surd term on
one side of the equation.
Check (in the original equation)
𝐼𝑓 𝑥=2
10( 𝑥+2 ) = 20
𝐿𝐻𝑆= =3+2=5
𝑥+2 = 2
Squaring both sides
𝑅𝐻𝑆=5
𝑥+2=4
𝑥=2 Which is true.
𝑥=2

3. 03 Algebra III Linear Inequalities 1 2 3 −1 −2 −3 −4 −3.5
Solve the inequality
and show the solution set on the numberline:
Multiply every term by 3
−1≤ 2𝑥+4 3 <2
−3≤2𝑥+4<6
−3−4≤2𝑥<6−4
−7≤2𝑥<2
−7 2 ≤𝑥<1 1 2 3 −1 −2 −3 −4
−3.5

4. Sketch the quadratic function.03
Algebra III
Quadratic and Rational Inequalities
QUADRATIC INEQUALITY
RATIONAL INEQUALITY
Solve
𝟐𝒙+𝟒 𝒙+𝟏 <𝟑, 𝒙∈𝑹, 𝒙≠−𝟏
Solve 𝒙2 + 7𝒙 + 12 ≤ 0, x ∈ R.
Let 𝑥2 + 7𝑥 + 12 = 0 and solve.
Multiply both sides by (𝑥+1) 2
𝑥2 + 7𝑥 + 12 = 0
𝑥+1 2 (2𝑥+4) 𝑥+1 <3 (𝑥+1) 2
(𝑥 + 3)(𝑥 + 4) = 0
𝑥 = –3 OR
𝑥 = – 4
(𝑥+1)(2𝑥+4)<3 (𝑥+1) 2
Sketch the quadratic function.
2 𝑥 2 +6𝑥+4<3 𝑥 2 +6𝑥+3
0< 𝑥 2 −1
𝑥 2 −1>0
Sketch the quadratic function 𝑥 2 −1.
As the function is > 0
the solution lies above the x-axis).
As the function is ≤ 0 the solution
lies on or below the x-axis.
The solution is
𝑥 < –1 OR 𝑥 > 1
The solution is −4≤𝑥≤−3

5. Solve for x if 3|x + 1| – |x + 5| = 0.03
Algebra III
Absolute Value (Modulus)
The Modulus of a number is the non-negative value of the number. Example |−7| = 7
Geometrically, the absolute value is how far away the number is from zero on the numberline.
Solve for x if 3|x + 1| – |x + 5| = 0.
Method 1 (Algebra)
Method 2 (Graph)
Leave one modulus term on one side
And the other terms on the other side.
We plot f(x) = 3|x + 1| and g(x) = |x + 5|.
-12
-10 -8 -6 -4 -2 2 -1 1 3 4 5 6 7 8 x y
3|x + 1| = |x + 5|
𝑓 𝑥 =3|𝑥+1|
(1,6)
Squaring both sides removes
the modulus notation.
(−2,3)
9(x + 1)2 = (x + 5)2
𝑔 𝑥 =|𝑥+5|
This simplifies to
x2 + x – 2 = 0
Solving gives
The two functions intersect at (–2,3) and (1,6).
x = –2 OR x = 1
∴ x = –2 OR x = 1

6. Solve the inequality |x + 1| > 2|x + 3|, x ∈ R.03
Algebra III
Modulus Inequalities
Solve the inequality |x + 1| > 2|x + 3|, x ∈ R.
Method 1 (Algebra)
Method 2 (Graph)
|x + 1|2 > 4|x + 3|2
We plot f(x) = |x + 1| and g(x) = 2|x + 3|.
x2 + 2x + 1 > 4(x2 + 6x + 9)
x2 + 2x + 1 > 4x2 + 24x + 36
-12
-10 -8 -6 -4 -2 2 -1 1 3 4 5 6 7 8 x y
𝑔 𝑥 =2|𝑥+3|
3x2 + 22x + 35 < 0
𝑓 𝑥 =|𝑥+1|
Let 3x2 + 22x + 35 = 0.
(3x + 7)(x + 5) = 0
∴ x = − OR x = –5
As f(x) > g(x)
As the function is
< 0 the solution
lies below the x-axis.
As f(x) > g(x) the solution is the part of the
Graph where f(x) lies above g(x).
The solution is −5<𝑥<− 7 3
The solution is −5<𝑥<− 7 3

7. 03 Algebra III Inequalities: Proofs
Prove that if a and b are real numbers, then a2 + b2 ≥ 2ab.
a2 + b2 ≥ 2ab
⇒ a2 + b2 – 2ab ≥ 0
⇒ (a – b)2 ≥ 0. True, as (a – b) ∈ R and (real)2 ≥ 0
∴ a2 + b2 ≥ 2ab
If a, b ∈ R, prove that a2 + 4b2 – 10a + 25 ≥ 0.
a2 – 10a b2 ≥ 0
(a – 5)(a – 5) + (2b)2 ≥ 0
⇒ (a – 5)2 + (2b)2 ≥ 0
True, as a, b ∈ R, and (real)2 + (real)2 ≥ 0.

8. 03 Algebra III Discriminants b2 – 4ac > 0 b2 – 4ac = 0
When using the formula
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
b2 – 4ac is called the discriminant.
The value of the discriminant, b2 – 4ac , can be used to determine whether the graph of the function (the parabola) cuts, touches or does not cut the x-axis.
b2 – 4ac > 0
b2 – 4ac = 0
b2 – 4ac < 0
Real Roots
Equal Roots
No Real Roots
Find the value of k if x2 + 6x + k has two equal roots.
Equal roots: b2 – 4ac = 0
a = 1, b = 6, c = k
(6)2 – 4(1)(k) = 0
k = 9