Transcript Solving Quadratics
SOLVING QUADRATICS
SOLVING QUADRATICS
•General Form:
y
ax
2
bx
c
•Where
a
,
b
and
c
are constants
To solve a quadratic equation, the equation must be expressed in the form:
ax
2
bx
c
0 That is, all variables and constants must be on one side of the equals sign with zero on the other.
Methods for Solving Quadratic equations
ax
2
bx
c
0 •Method 1 : Factorisation •Method 2 : Quadratic Formula
x
b
b
2 4
ac
2
a
•Method 3 : Completing the Square •Method 4 : Graphic Calculator
•Method 1 : Factorisation Example 1 :
x
2 12
x
20 0 Step 1: Equation in the right form Step 2: Factorise
x
2 12
x
20
x
10
x
2 0 0 Step 3: Separate the two parts of the product either (
x
10 ) 0 or (
x
2 ) 0 Step 4: Solve each equation
x
10 or
x
2
Some points to note Example 1 :
x
2
x
12
x
10
x
x
2 12
x
20 0 20 2 0 0 Need to remember your factorisation skills The equation is expressed as the product of two factors being equal to zero, therefore, one (or both) of the factors must be zero.
either
x
10 0 or
x
2 0
x
10 or
x
2 To check your answers are correct you can substitute them one at a time into the original equation.
Checking solutions Example 1 :
x
2 12
x
20 0
x
10 10 2 12 ( 10 ) 20 100 120 20 0
x
2 2 2 12 ( 2 ) 20 4 24 20 0
•Method 1 : Factorisation Example 2 :
x
2 3
x
28 Step 1: Equation in the right form Step 2: Factorise Step 3: Separate the two parts of the product
x
2 3
x
28
x
7
x
4 0 0 either
x
7 0 or
x
4 0 Step 4: Solve each equation
x
7 or
x
4
•Method 2 : Quadratic Formula Example 1 :
x
2 8
x
5 0
x
Step 1: Determine the values of
a, b
and
c
b
b
2 4
ac
2
a
a = 1 b =
– 8 and
Step 2: Substitute the values of
a, b
and
c
Step 3: Simplify This step is optional. Students need to have covered surds
c =
– 5
x
8
x x x
( 8 ) 2 4 ( 1 )( 5 ) 8 2 ( 1 ) 64 20 8 2 8 2 2 2 84 21 See next slide
Note that 2 is a factor of the numerator Once 2 is factored out, it can be cancelled with the 2 in the denominator
x
4 21
x
x
8 2 2 4 2 2 21 21 Exact answer
x
4 4 21 21 8 .
58 0 .
58 Decimal approximation
•Method 2 : Quadratic Formula Example 2 : 3
x
2 8
x
2 0
x
Step 1: Determine the values of
a, b
and
c
b
b
2 4
ac
2
a
a = 3 b
= 8 and
c
= 2
Step 2: Substitute the values of
a, b
and
c
Step 3: Simplify This step is optional. Students need to have covered surds
x x
x x
8 8 2 4 ( 3 )( 2 ) 8 2 ( 3 ) 64 24 8 6 8 2 6 6 40 10 See next slide
Note that 2 is a factor of the numerator Once 2 is factored out, it can be cancelled with the 6 in the denominator
x
4 3 10
x x
8 2 2 4 6 6 10 10 Exact answer
x
4 4 3 3 10 10 0 .
28 Decimal approximation 2 .
39
•Method 2 : Quadratic Formula Example 3 : 5
x
2 7
x
6 0
x
Step 1: Determine the values of
a, b
and
c
b
b
2 4
ac
2
a
a = 5 b =
– 7 and
Step 2: Substitute the values of
a, b
and
c
Step 3: Simplify Problem - you cannot find the square root of a negative number
c
= 6
x
7 ( 7 ) 2 4 ( 5 )( 6 )
x
7 2 ( 5 ) 49 120 10
x
7 71 10
2 Imaginary Solutions
That was a lot of work to find that there was no solution!
It would be useful to be able to “test” the equation before we start. For this we use the DISCRIMINANT.
The Discriminant
b
2 4
ac
The discriminant is a quick way to check how many real solutions exist for a given quadratic equation. As shown above the symbol for the
b
4
ac
Summary of Results using Discriminant
b
2 4
ac
> 0 The equation has
two
real solutions = 0 The equation has
one
real solutions < 0 The equation has
no
real solutions
Relating the Discriminant to graphs
b
2 4
ac
> 0
y y) y x x x
The graph cuts the
x
-axis in two places. These are the 2 real solutions to the quadratic equation.
Relating the Discriminant to graphs
b
2 4
ac
= 0
y y y) x x x
These graphs have their turning point on the
x
-axis and hence there are 2 equal solutions.
Relating the Discriminant to graphs
b
2 4
ac
< 0
y) y y x x x
There are 2 imaginary solutions in this case because the graphs do not intersect with the
x
-axis.
•Method 3:
Completing the Square Technique
Example 1 :
x
2 6
x
3 0 (
x
This value is half
b
Add 6 to both sides (
x
Take the square root of both sides 3 ) (
x
2 6
x
3 ) 2 3 0 6 Subtract 3 from both sides 3 ) 2 9 3
x
0 3
Subtract the square of the number in the bracket
6 6
•Method 3:
Completing the Square Technique
x
3 6 This result gives us the exact answers.
x
3 6 or
x
3 6 Use your calculator to find decimal approximations accurate to two decimal places.
x
0 .
55 or
x
5 .
45
•Method 3:
Completing the Square Technique
Example 2 :
x
2 5
x
8 0
This value is half
b
(
x
Add to both sides 4 Take the square root of both sides Subtract 2.5 from both sides 5 2 ) 2 (
x
25 4 8
x
2 .
5 5 2 ) 2 (
x
57 4 5 2 ) 2 0 57 4
x
0 5
Subtract the square of the number in the bracket
57 2 2 57
Example 3: Solve
x
2 7
x x
2 10 7
x
0 10 0
Factor out the coefficient of x²
(
x
3 .
5 ) 12 .
25 10
This value is half
b
[(
x
3 .
5 ) 2 (
x
3 .
5 ) 2 2 .
25 ] 2 .
25 0 0 (
x
(
x
3 .
5 ) 2 3 .
5 ) 2 0
Subtract the square of the number in the bracket
2 .
25 2 .
25 See next slide
(
x
3 .
5 ) 2 2 .
25
x
3 .
5 2 .
25 i.e.
x
3 .
5 2 .
25
x
3 .
5 2 .
25 or
x
3 .
5 2 .
25
•Method 4 Graphic Calculator Graph the function and find the value of the
x
-intercepts Use a solver function for a polynomial of degree 2
We wish to thank our supporters: