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1
NCS Mathematics
DVD Series
Solving
Equations and
Inequalities
Outcomes for this DVD
In this DVD you will:
• Solve quadratic equations LESSON1.
• Solve quadratic inequalities LESSON 2.
• Solve general inequalities LESSON 3.
• Solve systems of equations LESSON 4.
3
NCS Mathematics
DVD Series
Lesson 1
Solving
Quadratic
Equations
What is a Quadratic Equation?
Definition of quadratic equation:
2
A quadratic equation is an equation that can be written
as 0 where , and and 0.ax bx c a b c a
2 0 is also known as the standard form
of the quadratic equation.
ax bx c
5
Write the quadratic equation in standard form:
2 0; 0ax bx c a
Factorize if possible: ( )( ) 0a x xor
Use Formula:
2 4
2 2
b b ac bx
a a
Roots
The roots are solution's of this equation
x - a x 0 0Either or
Strategy for solving Quadratic Equations
2x y ax bx ci.e. - intercepts for the graph of +
Roots – The Graphical Interpretation
2 2 3y x x 2 1y x
Example 1: Solving a Quadratic Equation
1. ( 2)( 1) 0x x
This is already given in factorised form:
Thus; ( 2)( 1) 0
2 0 or 1 0
x x
x x
2 or 1x x
Example 2: Solving a Quadratic Equation
23 27 0 (Standard form)x
2 9 0 (Divide by 3)x
( 3)( 3) 0 (Factorise)x x
3 or 3 (Solutions- Check!)x
22. Solve: 3 27x
Example 3: Solving a Quadratic Equation
5 or 2x x
22 6 20 0x x 2 3 10 0x x
( 5)( 2) 0x x
3. Solve: 2 ( 3) 20x x
Tutorial 1: Solving Quadratic Equations
2
2
2
Solve the following quadratic equations:
1. 12
2. 2 15
3. 3 14 5
x x
k k
x x
PAUSE DVD
• Do Tutorial 1
• Then View Solutions
Tutorial 1 Example 1: Suggested Solution
21. Solve: 12 x x 2 12 0x x
( 4)( 3) 0x x
4 or 3x x
Tutorial 1 Example 2: Suggested Solution
22. Solve: 2 15k k
2 1 or
5 3k k
215 2 0k k
(5 2)(3 1) 0k k
Tutorial 1 Example 3: Suggested Solution
23. Solve: 3 14 5x x
(5 1)( 3) 0x x
25 14 3 0x x
1 or 3
5x x
Solving a Quadratic Equation by
Completing the Square
2
When a quadratic equation does not factorise,
we can solve the equation by comple
Especially when:
4 is positive but
t
n
in
ot
g the
a pe
squ
rfect square.
are.
b ac
Note: Completing the square means that we add and subtract
a number to an existing quadratic expression so that part
of the expression becomes a perfect square.
Example 1: Creating a Perfect Square
2(1) What must be added to 6 ,
to make this expression a perfect square?
x x
2
2
2 2
22 2
We rewrite 6 as:
66 6 9
2
6 3 9 ( 3) 9
x x
x x x x
x x x
Assume that leading
coefficient is 1
2
Number to be added and subtracted
can be calculated as follows:
1 Coefficient of Middle Term
2
Example 2: Convert the Quadratic Expression
into a Perfect Square
2
2
2
(2) To make 4 a perfect square
1 we add and subtract 4 2 4 to it.
2
x x
2
2 2 2 2
4 can be rewritten as:
4 4 2 4 ( 2) 4
x x
x x x x x
17
Solving Quadratic Equations by
Completing the Square 2Instruction: Solve: 2 4 7 0x x
2Take constant to RHS 2 4 7x x
2 2 7Divide both sides by coeff. of 2
2x x x
2
21 7Add coeff. of to both sides 2 1 1
2 2x x x
2
2
The 3 terms on LHS ( 1) 1 3,5
form a perfect square 1 4,5
x
x
Take square roots on both sides: 1 4,5
Write final solution: 1 4,5
x
x
Solving ax2 + bx+ c = 0 by completing the square
2 22
2 24 4
b b b cx x
a a a a
2 2
bx
a a
2 2
2
4
2 4
b b acx
a a
2 4
2 2
b b acx
a a
2 2 b cax bx c x x
a a
2Solve 0
by completing the square.
ax bx c
2
bx
a
This is the formula to solve
any quadratic equation.
Tutorial 2: Solve quadratic equations by
completing the square
2
2
Solve for by completing the square.
(Write answer(s) correct to 2 decimal
digits where necessary.)
1. 4 2 0
2. 3 2 2 0
x
x x
x x
PAUSE DVD
• Do Tutorial 2
• Then View Solutions
Tutorial 2 Problem 1: Suggested Solution
2
2
2 2 2
2
1. 4 2 0
4 2
4 ( 2) ( 2) 2
( 2) 2
2 2
2 1, 41
3, 41 or 0,59
x x
x x
x x
x
x
x
x
Tutorial 2 Problem 2: Suggested Solution
2 2
2
2 2
2
2 2
2. 3 2 2 0 3 2 2
2 2
3 3
2 1 1 2
3 3 3 3
1 1 2 1 1 6
3 9 3 3 9
1 7
3 3
1 7 1 2,65 1,22 or 0,55
3 3
x x x x
x x
x x
x x
x
x x
Solving a Quadratic Equation by
using the Formula
2 4
2 2
b b ac bx
a a
2If 0, then by
completing the square we have:
ax bx c
This is the general formula
for solving quadratic equations.
Using the Formula to Solve Quadratic Equations
2
Solve for by by using the general quadratic formula:
2 7 1 0
x
x x
Know that:
2
7
1
a
b
c
2 4
2
b b acx
a
2
7 41
4
bx
a
x
7 41 7 41 or
4 4
0,15 or 3,35
x x
x
2
2
4
7 4 2 1
49 8 41
b ac
Tutorial 3: Solving quadratic equations
2
Solve the following quadratic equations
using any suitable method:
1) 8 12
2) 2 5 0
(Answers correct to 2 decimal places, where applicable)
x x
x x
PAUSE DVD
• Do Tutorial 3
• Then View Solutions
Tutorial 3 Problem 1: Suggested Solution
2
1. ( 8) 12
8 12 0
x x
x x
2 8 12 0
( 2)( 6) 0
2 or 6
x x
x x
x
2 28 4 1 12 64 48 4
2
is a perfect square
8 12 can be factorised.x x
Tutorial 3 Problem 2: Suggested Solution
2 2 5 0
2 24
2(1)
2 4,899
2
1,45 or 3,45
x x
x
x
x
is not a perfect square
Use formula
2
2
2) 2 5 0
2 4 1 5 24
x x
27
NCS Mathematics
DVD Series
Lesson 2
Solving
Quadratic
Inequalities
a,b a b Suppose and
Sub-intervals of the Real Number Line
{ : }x a x b Open Interval
Closed Interval { : }x a x b
( , )a b
[ , ]a b
( , )a ( , )b
( , )
a b
a b
x<a and x>b
x<a or x>b
,a b a b Suppose and
x>a and x<b
x>a or x<b
Empty set!
The use of OR / AND
Notice the exclusion
Procedure for Graphical Solution of Non-linear
Quadratic Inequalities
2 2
Write all non-zero terms on left hand side of inequality.
e.g. 0 or 0ax bx c ax bx c
2 Sketch the parabola y ax bx c
2 0
For which values does the
parabola lie below or on the axis?
ax bx c
x
x
2 0
For which values does the
parabola lie above the axis?
ax bx c
x
x
31
2 2 3y x x
1 and 3x x
2 2 3 0x x
This happens when...
Graphical View of an Inequality
Tutorial 4: Quadratic inequalities
PAUSE DVD
• Do Tutorial 4
• Then View Solutions
2
2
2
Solve the following quadratic inequalities:
1. 3 48 (leave answer in set-builder notation)
2. 3 10 (leave answer in interval notation)
3. 1 2 (draw graphical solution)
x
x x
x x
Tutorial 4 Problem 1: Suggested Solution
2
2
2
1. 3 48
3 48 0
16 0
x
x
x
( 4)( 4) 0
4 or 4
Solution Set
{ : 4 or 4; }
; 4 4;
x x
x x
x x x x
2 16y x
Tutorial 4 Problem 2: Suggested Solution
22. 3 10 0
( 2)( 5) 0
x x
x x
2 5
Solution 2;5
x
x
Tutorial 4 Problem 3: Suggested Solution
2
2
3. Write in standard form and factorize:
1 2
2 1 0
(2 1)( 1) 0
x x
x x
x x
From sketch:
(2 1)( 1) 0
11 and
2
11;
2
x x
x x
x
Sketch parabola defined by
2 1 1y x x
Tutorial 4 Problem 3: Graphical Solution
2
2
We interpret the inequality 1 2 graphically
by finding the value/s of for which the graph of
1 is below the graph of 2 . We see from the
1graph that these values lie between 1 and
2
x x
x
y x y x
1y x
22y x
37
NCS Mathematics
DVD Series
Lesson 3
Solving
General
Inequalities
Strategy to Solve General Inequalities
Determine the sign of each linear factor between
zero points, and hence the sign of the overall expression
3 22To solve inequalities like 0; 0;
3 4
we use the following method:
x xx
x x
Write all terms on LHS of inequality and 0 on RHS
Write expression on LHS as a factorized fraction
Find the value for which each factor in the
expression is equal to zero
39
Signs of Linear Factors
Consider the graphs of linear functions
defined by 3 and 4y x y x
3 0 when 3
Then 3 0 when 3
3 0 when 3
x x
x x
x x
4 0 when 4
and 4 0 when 4
4 0 when 4
x x
x x
x x
Example 1: Solve a Quadratic Inequality
Algebraically
2
1. Determine the values of such that
2 15
x
x x
2
Solution:
2 15 0 5 3 0x x x x
5 :x
3 :x
5 3 :x x
3 5
00
0 0
3;5x
2. Determine the values of such that
3 2 0
4
x
x x
x
3 :x
2 :x
3 2:
4
x x
x
2 40
0
0
0
4 :x
3
0
+ ++
2;3 4;x
Example 2: Solve a Quadratic Inequality
Algebraically
Tutorial 5: Solve General Inequalities
2
Solve:
1. 2 3 0
32. 0
2 4
x x
x
x x
PAUSE DVD
• Do Tutorial 5
• Then View Solutions
Tutorial 5 Problem 1: Suggested Solution
2
1. Determine the values of such that
2 3 0
x
x x
2
Solution:
2 3 0 2 3 0x x x x
:x
2 3 :x
2 3 :x x
0 1.5
00
0 0
30;
2x
Tutorial 5 Problem 2: Suggested Solution
2. Determine the values of such that
3 0
2 4
x
x
x x
3 :x
2 :x
3:
2 4
x
x x
2 40
0
UD
0
4 :x
3
0 UD
+ ++
; 2 3;4x
45
NCS Mathematics
DVD Series
Lesson 4
Solving
Systems of
Equations
Simultaneous Algebraic Solution of two
Linear Equations in Two Unknowns
Now this value for in (1):
So 3( 2) 2 1
x
y y
substitute
Solve for and simultaneously:
3 2 1 and 2 1 2
x y
x y x y
in (2):
So we have: 2x y
Make x the subject of the formula
3 6 2 1 5y y y
So 5 2 3
; 3; 5
x
x y
Graphical Solution of two Simultaneous
Linear Equations in Two Unknowns
From previous slide we know that:
; : 3 2 1 ; : 2 3; 5x y x y x y x y
This implies that the two straight line intersect in the
point 3; 5 or have the point 3; 5 in common.
Solving Simultaneously a Linear and a
Quadratic equation in Two Unknowns
We can now use similar methods in solving 2 equations
simultaneously, where one is linear and the other is quadratic
2Solve for and if 3 and 21 27x y x y x y
From : 1 33y x
Substitute int3 o 2 :
2 23 27 30 0x x x x
6 5 0 6 or 5x x x x
Back substitute values into :(3)x
6 3 9 or 5 3 2y y
Solution Set 6; 9 ; 5;2
49
Solving Simultaneously a Linear and a
Quadratic Equation Graphically
2
From previous slide:
; : 3 ; : 27
6; 9 ; 5;2
x y x y x y x y
Tutorial 6 : Solving Systems of Equations
2
2
Solve the following systems
of simultaneous equations:
1. 3
2 4 (algebraicly)
2. 4
2 (graphically)
y x
y x
y x
y x
PAUSE DVD
• Do Tutorial 6
• Then View Solutions
Tutorial 6 Problem 1: Suggested Solution
21. 3 (A)
2 4 (B)
y x
y x
( 1)( 1) 0
1 (only)
x x
x
2
2
Substitute for in equation (A)
We have: 2 4 3
2 1 0
y
x x
x x
Hence, 2(1) 4 2
Solution Set {(1; 2)}
y
Tutorial 6 Problem 2: Suggested Solution
2 4y x
2y x
Solution are points and .
Solution Set
2;0 ; 1;3
A B
53
End of the DVD on Solving Equations
and Inequalities
REMEMBER!
•Consult text-books for additional examples.
•Attempt as many as possible other similar examples
on your own.
•Compare your methods with those that were
discussed in the DVD.
•Repeat this procedure until you are confident.
•Do not forget:
Practice makes perfect!
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