Transcript Quadratic Theory - Bearsden Academy

Quadratic Theory
Introduction
This algebraic expression: x2 + 2x + 1
is a polynomial of degree
2
Expressions like this in which the highest power of x is x2 are also called
quadratics.
Quadratic Equations look like this:
3x  4 x  8  0
2
What values of x make the quadratic expression 0?
What are the roots of this quadratic equation?
Solving Quadratic Equations
Quadratic equations may be solved by:
 Factorising
 Completing the Square
 Using the quadratic formula
Solve x2 – 2x – 8 = 0 using each of the above methods
 Factorising
x2  2 x  8  0
( x  4)( x  2)  0
x  4 or x  2
When factorising quadratics
check for:
 common factors
 difference of 2 squares
 double brackets
 Completing the Square
x  2x  8  a( x  b)  c
2
2
 1.( x 1) 1  8
2
 ( x  1)2  9
( x  1)2  9  0
( x 1)  9
2
x  1  3 or x  1  3
x  4 or x  2
 Using the quadratic formula
Reminder:
ax2  bx  c  0
x2  2 x  8  0
the roots are
b  b 2  4ac
x
2a
(2)  (2) 2  4.1.(8)
x
2.1
2  36
x
2
x  4 or x  2
b2  4ac
The Discriminant
Equation
864210
54321– 1
Solve
x2 – 6x + 8 = 0
x 
x2 – 6x + 9 = 0
6  36 108642 32
54321– 1
2
 3 1
x 
–1
Roots
x 
6  36 108642 36
54321– 1
2
 3 0
y
y
Sketch
x2 – 6x + 10 = 0
y
10
10
10
8
8
8
6
6
6
4
4
4
2
2
2
1
2
3
4
x = 2 or x = 4
5 x
–1
6  36  40
2
4
 3
2
1
2
x=3
3
4
5 x
–1
1
2
x=?
3
4
5 x
As we can see from the above examples, the nature of the roots of a
quadratic can be found using the discriminant
b2  4ac
 If
b2  4ac  0
 If
b2  4ac  0
 If
b2  4ac  0
there are two distinct roots
there are two identical roots
(i.e. one root)
there are no real roots
Examples:
Tangents to Curves
To determine whether a straight line cuts, touches or does not meet a
curve the equation of the line is substituted into the equation of the curve.
When a quadratic equation results, the discriminant can be used to find
the number of points of intersection.
Two Points of Intersection
y  x2  5x  4 meets y  x  1 where
y
x2  5x  4  x  1
4
3

2
1
–4
–3
–2
–1
1
x
x2  4 x  3  0
cp. w. ax2  bx  c  0
–1
–2

–3
a 1 b  4 c  3
b2  4ac  16  12
b2  4ac  0


2 distinct real roots
2 distinct points of intersection
One Point of Intersection
y  x2  5x  4 meets y  x
where
y
x2  5x  4  x
4
3

2
1
–4
–3
–2
–1
1
x
x2  4 x  4  0
cp. w. ax2  bx  c  0
–1
–2

–3
a 1 b  4 c  4
b2  4ac  16  16
b2  4ac  0


1 real root
1 point of intersection

Line is a tangent to the curve
No Points of Intersection
y  x2  5x  4 meets y  x  1
where
y
x2  5x  4  x  1
4
3

2
1
–4
–3
–2
–1
1
x
x2  4 x  5  0
cp. w. ax2  bx  c  0
–1
–2

–3
a 1 b  4 c  5
b2  4ac  16  20
b2  4ac  0


no real roots
no points of intersection

The line and the curve do not intersect
Examples involving Tangents to Curves:
Quadratic Inequalities
A quadratic inequality is an expression such as:
3x2  5x  64  0
The problem is to find the values of x for which such an expression is true
A quadratic inequality can be solved using a sketch of the quadratic function
We can then easily see where the graph is positive or negative.
515
10
21– 4321
Find the values of x for which:
a) 12  5 x  2 x 2 > 0
b) 12  5 x  2 x 2 < 0
y
15
First sketch the curve.
y intercept at: (0,12)
10
5
x intercepts at: (1½ ,0) and (-4,0)
–4
–3
–2
–1
a) From the graph 12 – 5x – 2x2 is positive (i.e. above the x axis)
when -4 < x < 1½
b) From the graph 12 – 5x – 2x2 is negative (i.e. below the x axis)
when x < -4 or 1½ < x
1
2
x
Examples involving solution of
Inequalities: