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f ( x )  ax  bx  c
2
a  0 , is a quadratic function
Reminders
2
y

ax
 bx  c
(1) The graph of a quadratic function
is a parabola.
If a > 0 the parabola is
a minimum .
If a < 0 the parabola is
a maximum .
shaped and the turning point is
shaped and the turning point is
(2) To sketch and annotate a parabola y  ax  bx  c
we need to identify where possible :
2
• whether the shape is
(a > 0) or
(a < 0)
• the co-ordinates of the y-axis intercept (0 , c)
• the roots (zeros) of the function by solving
• the equation of the axis of symmetry
• the co-ordinates of the turning point
Note to find the axis of symmetry
when there are two roots x1 and x2
carry out the following :x1
x1  x 2
x3  2
ax 2  bx  c  0
c
( x3 , d )
x3
x2
(3) When the equation
in the form
y  ax 2  bx  c
y  a( x  p)2  q
the axis of symmetry
is x = - p and the turning point is (-p , q)
(-p , q)
x = -p
is written
(4) Quadratic equations may be solved by
• using the graph
• factorising
• completing the square
• using the quadratic formula
(5)
(see point (5) below)
If ax 2  bx  c  0 then
 b  b 2  4ac
x
where a  0
2a
The Discriminant
(Nature of the roots of a quadratic Equation)
Example Solve the following equations using the
quadratic formula
Graphical
Implications
(1) 4 x 2  2 x  1  0
a4 

b  2
c  1 
b2  4ac  (-2)2 - 4 x 4 x (-1)
= 4 - (-16)
= 20
Roots are given by :
x1 
2  20
8
x
x
Or
x
2 20
8
x2 
2  20
8
Two distinct
real roots
( 2) 4 x 2  4 x  1  0
a  4

b  4
c  1 
b2  4ac  42 - 4 x 4 x 1
Graphical
Implications
= 16 - 16
=0
Roots are given by :
x
4 0
8
x   48   12 (twice)
One real root
(repeated/equal roots)
x
OR
x
( 3) 4 x 2  2 x  1  0
a4 

b  2
c  1 
Graphical
Implications
b2  4ac  (-2)2 - 4 x 4 x 1
= 4 - 16
= -12
Roots are given by :
x
2  12
8
Since 12) is not a real number
there are NO REAL ROOTS
x
OR
x
Conclusion :-
If b2  4ac is
(i) positive
2 real roots
(ii) zero
1 real root
(iii) negative
no real roots
b 2  4ac discriminates between real and non-real roots
and so is called the DISCRIMINANT
SUMMARY
For the quadratic equation ax 2  bx  c  0
(i) If b2  4ac  0 the roots are real and distinct (unequal)
TWO ROOTS
(ii) If b2  4ac  0 the roots are real and equal
ONE ROOT
(iii) If b2  4ac  0 the roots are non-real
NO ROOTS
Example 1 Determine the nature of the roots of the following
equations:-
(a ) x 2  6 x  8  0
a1 

b  6
c  8 
2
b2  4ac  (-6) - 4 x 1 x 8
= 36 - 32
=4
Since b2  4ac  0 roots are real and distinct
(b) x 2  6 x  9  0
a1 

b  6
c  9 
2
b2  4ac  (-6) - 4 x 1 x 9
= 36 - 36
=0
Since b2  4ac  0 roots are real and equal
(c ) x 2  6 x  10  0
a1 

b  6
c  10 
2
b2  4ac  (-6) - 4 x 1 x 10
= 36 - 40
= -4
Since b2  4ac  0 roots are non-real
Example 2 Find p given that 2 x 2  4 x  p  0 has
real roots
Solution
Note This could be distinct roots
or equal roots.
2
a  2  For real roots
b

4ac

0
2

4 - 4 x 2 x p 
b  4
16 - 8p 
c  p 
16  p
8p  16
p2
So equation has
equal roots when
p2
Example 3
1
q
Find the value of q so that 2
x  x 1
has equal roots.
Solution
Cross-multiplying gives
q( x 2  x  1)  1
qx 2  qx  q  1
qx 2  qx  q  1  0
For equal roots
aq 

b  q 
c  q  1
So
b2  4ac  0
(-q) 2  4q(q - 1)  0
q 2  4q 2  4q  0
4q  3q 2  0
q(4 - 3q)  0
So equation has equal roots when
q0 q
4
3
Example 4
Given that k is a real number , show that the roots of the equation
kx 2  5 x  5  k
are always real numbers.
Solution
ak 

b5 
c  5  k 
kx 2  5 x  5  k
kx 2  5 x  (5  k )  0
b2  4ac  25 - 4k(5 - k)
 25 - 20k  4k 2
 4k 2  20k  25
 ( 2k - 5) 2
Since ( 2k - 5)2 is a square it has minimum value 0 ,
therefore
b2  4ac  0 and so the roots of the equation
are always real.
Example 5
For what values of p does the equation
x2 - 2px + (2 - p) =0
have non-real roots?
Solution
x2 - 2px + (2 - p) =0
For non-real roots
a1


b  2 p 
c  2  p
b2  4ac  0
(-2p) 2 - 4 x 1 x (2 - p) < 0
4p2 - 8 + 4p < 0
4p2 + 4p - 8 < 0
4(p2 + p - 2) < 0
This is a quadratic INEQUATION .
To solve this carry out the following steps :-
Step 1 Factorise the quadratic
So
can be written as
Step 2
4(p2 + p - 2) < 0
4(p + 2)(p - 1) < 0
Next consider solving
4(p + 2)(p - 1) = 0
So p = - 2 or p = 1
Step 3 Then consider other values of p either side of these two
values of p and also between these values :- setting
working out in a table as follows:-
Table of Values
To solve 4(p + 2)(p - 1) < 0
0
-3
p
(p + 2)
(p - 1)
4(p + 2)(p - 1)
+
-2
0
2
1
-
+
-
+
0
+
+
0
-
0
+
So 4(p + 2)(p - 1) < 0 for values of p in the range
-2 < p < 1
So the equation has non-real roots when -2 < p < 1