Quadratic Theory(Introduction) (i) f(x) = ax2 + bx + c, a ≠ 0, is a quadratic function
(ii) 4x2 + 11x - 3 is a quadratic expression, with a = 4, b = 11 & c = -3
(iii) 4x2 + 11x - 3 = 0 is a quadratic equation, which in this case can be solved by factors:
(4x - 1)(x + 3) = 0
4x - 1 = 0 or x + 3 = 0
x = ¼ or x = -3 Roots of the function

Quadratic Theory(Solving Equations) :

Quadratic Theory(Solving Equations) Quadratic equations may be solved by:
☻ Factorising
☻ The Quadratic Formula
☻ Completing the Square

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Example 1 (factorising) Solve:
6x2 + 2x - 8 = 0
2(3x2 + x - 4) = 0
2(3x + 4)(x - 1) = 0
3x + 4 = 0 or x - 1 = 0
x = - or x = 1 When factorising always remember the priority of:
Common factors
Difference of two squares
Double brackets

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Example 2 (The quadratic Formula) Solve: 5x2 + 3x - 7 = 0, correct to 2 dp a b c x = 0.92 or x = -1.52

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Example 3 (Completing the square) Solve: x2 + 4x - 3 = 0 expression in the form: a(x +p)²+ q x² + 4x - 3 = 0
(x + 2)² - 4 - 3 = 0
(x + 2)² - 7 = 0
(x + 2)² = 7
(x + 2) =√7 or -√7
x = √7 - 2 or x = -√7 - 2
x = 0.65 or x = -4.65

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Example 4 x = 4.30 or x = 0.70 Multiply both sides by x Take 5x over to the other side Quadratic Formula

Quadratic Theory(The Discriminant) :

Quadratic Theory(The Discriminant) Discriminant

Slide 9:

Roots:
x = 2 or x = 6 Roots:
x = 4 Roots:
x = ??

Slide 10:

From the previous examples we noted that the discriminant can help us describe the nature of the roots.
i.e. b2 - 4ac discriminates between the different types of roots.
There are three different situations as follows:

Slide 11:

b2 - 4ac > 0
(the discriminant is positive) Two real and distinct roots
(cuts x-axis at 2 points)

Slide 12:

b2 - 4ac = 0
(the discriminant is equal to zero) The roots are equal and real
(cuts the x-axis at 1 point)

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b2 - 4ac < 0
(the discriminant is negative) There are no real roots
(does not cut the x-axis)

Slide 14:

Example 5 Decide the nature of the roots of the equation:
3x2 - 6x - 5 = 0
Consider the discriminant:
b2 - 4ac
= -62 - (4 3 -5)
= 96
b2 - 4ac > 0
Therefore the roots are real and distinct a b c

Slide 15:

Example 6 Decide the nature of the roots of the equation:
4x2 + 12x + 9 = 0
Consider the discriminant:
b2 - 4ac
= 122 - (4 4 9)
= 0
b2 - 4ac = 0
Therefore the roots are equal a b c

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Example 7 Decide the nature of the roots of the equation:
9x2 - 12x + 7 = 0
Consider the discriminant:
b2 - 4ac
= -122 - (4 9 7)
= -108
b2 - 4ac < 0
Therefore there are no real roots a b c

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Example 8 Find the value of a if the equation below has equal roots, and hence find the value of that root:
ax2 + 14x + 1 = 0
If roots are equal: b2 - 4ac = 0
142 - (4 a 1) = 0
196 - 4a = 0
4a = 196
a = 49
Therefore the equation is : 49x2 + 14x + 1 = 0 a b c

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Therefore the equation is : 49x2 + 14x + 1 = 0

Quadratic Theory(Tangents to Curves) :

Quadratic Theory(Tangents to Curves) A straight line can either:
cut a quadratic curve
touch at only one point (tangent)
or miss the curve completely

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To determine if a line is a tangent (i.e. cuts the curve at one point), the equation of the line is substituted into the equation of the curve. Using the resultant quadratic equation, the discriminant can be used to find the number of points of intersection.
b2- 4ac > 0 the line intersects the curve at two points
b2- 4ac = 0 the line intersects the curve at one point (the line is a tangent)
b2- 4ac < 0 the line does not intersect

Slide 21:

Example 9 Show that the line y = 4x - 10 is a tangent to the curve y = x2 - 6x + 15 and find its point of contact.
x2 - 6x + 15 = 4x - 10 (y co-ordinates are equal)
x2 - 10x + 25 = 0 (quadratic equation)
b2 - 4ac
= -102 - (4 1 25)
= 0
As b2 - 4ac = 0, the line y = 4x - 10 is a tangent

Slide 22:

To establish the point of contact solve:
x2 - 10x + 25 = 0 Substitute x = 5 into y = 4x - 10
i.e. y = 4 5 - 10
y = 10
Point of contact is (5,10)

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Example 10 Show that the line y = 8x + c meets the parabola y = x2 + 6x - 5 where x2 - 2x + (-5 - c) = 0
Find c for the line to be a tangent
Find the point of contact

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Show that the line y = 8x + c meets the parabola y = x2 + 6x - 5 where x2 - 2x + (-5 - c) = 0 When the line and the parabola intersect the y co-ordinates will be equal.
Therefore:
x2 + 6x - 5 = 8x + c
x2 + 6x - 8x - 5 - c = 0
x2 - 2x + (-5 -c) = 0

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b) Find c for the line to be a tangent x2 - 2x + (-5 -c) = 0
If the line is a tangent then
b2 - 4ac = 0
-22 - (4 1 (-5 -c) =0
4 - (4 (-5 - c)) = 0
4 - (-20 - 4c) = 0
24 + 4c = 0
c = -6

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c) Find the point of contact x2 - 2x + (-5 -c) = 0 substitute c = -6
x2 - 2x + 1 = 0 Tangent line: y = 8x + c
Therefore: y = 8x - 6
Substitute x = 1 into above
y = 8 - 6
y = 2
Point of contact (1,2)

Quadratic Theory(Inequalities) :

Quadratic Theory(Inequalities) Opposite is the graph of y = x2 - 8x + 12
To solve x2 - 8x + 12 = 0 the roots of the equation were found (x = 2 or 6) i.e. the y co-ordinate of the points on the graph when x = 2 or x = 6 are zero.
We now consider where the y co-ordinates are not equal to zero.

Quadratic Theory(Inequalities) :

Quadratic Theory(Inequalities) To solve x2 - 8x + 12 > 0 the part of the graph above y = 0 (x axis) is considered. In this example it is when x is less than 2 but greater than 6
i.e. x < 2 or x > 6

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To solve x2 - 8x + 12 ≤ 0 the value of the roots and the part of the graph below y = 0 (x axis) are considered. In this example it is when x is equal to 2 or 6 and between the values of 2 and 6
i.e. 2 ≤ x ≤ 6

Slide 32:

Example 11 Make a sketch of the graph y = (x + 3)(x - 5) and solve the inequality
(x + 3)(x - 5) ≥ 0
Roots of the graph:
x + 3 = 0 or x - 5 = 0
x = -3 or x = 5

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i.e. when the graph crosses the x-axis and above.
When x is equal to and less than -3
or when x is equal to and greater than 5
x ≤ -3 or x ≥ 6 For (x + 3)(x - 5) ≥ 0

Slide 34:

Example 12 Make a sketch of the graph y = x(10 - x) and solve the inequality
x(10 - x) > 0
Roots of the graph:
x = 0 or 10 - x = 0
x = 0 or x = 10

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i.e. when the graph is above the x axis
When x is greater than 0 but less than 10
0 < x < 10 For x(10 - x) > 0

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