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03
Chapter 3 – Algebra III
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
ACTIVE MATHS 4
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Algebra III
Surd Equations
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+7
2
= 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 + 2 + 𝑥 + 2
𝑥 + 7 = 27 − 10 𝑥 + 2 + 𝑥
Again, isolate the surd term on
one side of the equation.
10( 𝑥 + 2) = 20
𝑥+2=2
𝑥+2=4
ACTIVE MATHS 4
𝑥=2
Squaring both sides
Check (in the original equation)
𝐼𝑓 𝑥 = 2
𝐿𝐻𝑆 = 2 + 7 + 2 + 2 = 3 + 2 = 5
𝑅𝐻𝑆 = 5
𝑥 = 2 Which is true.
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Algebra III
Linear Inequalities
Solve the inequality
and show the solution set on the numberline:
−1 ≤
2𝑥 + 4
<2
3
Multiply every term by 3
−3 ≤ 2𝑥 + 4 < 6
−3 − 4 ≤ 2𝑥 < 6 − 4
−7 ≤ 2𝑥 < 2
−7
≤𝑥<1
2
−4 −3.5 −3
ACTIVE MATHS 4
−2
−1
0
1
2
3
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Algebra III
Quadratic and Rational Inequalities
QUADRATIC INEQUALITY
Solve 𝒙2 + 7𝒙 + 12 ≤ 0, x ∈ R.
Let 𝑥2 + 7𝑥 + 12 = 0 and solve.
𝑥2 + 7𝑥 + 12 = 0
(𝑥 + 3)(𝑥 + 4) = 0
𝑥 = –3 OR 𝑥 = – 4
Sketch the quadratic function.
RATIONAL INEQUALITY
𝟐𝒙 + 𝟒
Solve
< 𝟑, 𝒙 ∈ 𝑹, 𝒙 ≠ −𝟏
𝒙+𝟏
Multiply both sides by (𝑥 + 1)2
𝑥 + 1 2 (2𝑥 + 4)
< 3(𝑥 + 1)2
𝑥+1
(𝑥 + 1)(2𝑥 + 4) < 3(𝑥 + 1)2
2𝑥 2 + 6𝑥 + 4 < 3𝑥 2 + 6𝑥 + 3
0 < 𝑥2 − 1
𝑥2 − 1 > 0
Sketch the quadratic function 𝑥 2 − 1.
As the function is ≤ 0 the solution
lies on or below the x-axis.
The solution is −4 ≤ 𝑥 ≤ −3
ACTIVE MATHS 4
As the function is > 0
the solution lies
above the x-axis).
The solution is
𝑥 < –1 OR 𝑥 > 1
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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)
Leave one modulus term on one side
And the other terms on the other side.
Method 2 (Graph)
We plot f(x) = 3|x + 1| and g(x) = |x + 5|.
8 y
7
3|x + 1| = |x + 5|
Squaring both sides removes
the modulus notation.
9(x + 1)2 = (x + 5)2
This simplifies to
x2 + x – 2 = 0
Solving gives
x = –2 OR x = 1
ACTIVE MATHS 4
𝑓 𝑥 = 3|𝑥 + 1|
6
(1,6)
5
4
(−2,3)
3
2
𝑔 𝑥 = |𝑥 + 5|
1
x
-12
-10
-8
-6
-4
-2
2
-1
The two functions intersect at (–2,3) and (1,6).
∴ x = –2 OR x = 1
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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
x2 + 2x + 1 > 4(x2 + 6x + 9)
We plot f(x) = |x + 1| and g(x) = 2|x + 3|.
x2 + 2x + 1 > 4x2 + 24x + 36
3x2
8 y
+ 22x + 35 < 0
6
Let
+ 22x + 35 = 0.
(3x + 7)(x + 5) = 0
∴x= −
5
𝑓 𝑥 = |𝑥 + 1|
3x2
7
3
7
𝑔 𝑥 = 2|𝑥 + 3|
4
3
2
1
OR x = –5
x
-12
-10
-8
-6
-4
-2
2
-1
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 < −
The solution is −5 < 𝑥 < −
ACTIVE MATHS 4
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3
7
3
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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 + 25 + 4b2 ≥ 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.
ACTIVE MATHS 4
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Algebra III
Discriminants
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
When using the formula
b2 – 4ac is called the discriminant.
2𝑎
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
Real Roots
b2 – 4ac = 0
b2 – 4ac < 0
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
ACTIVE MATHS 4
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