## Presentation on theme: "QUADRATIC FUNCTIONS AND INEQUALITIES"— Presentation transcript:

Integrated Programme/mainstream Secondary Three Mathematics Name : _____________( ) Class:______ Date : ___________ to _____________ Term 1: Unit 4 Notes QUADRATIC FUNCTIONS AND INEQUALITIES Factoring Completing Square General Quadratic Formula Quadratic Graphs Quadratic Inequalities Discriminant and Nature of roots At the end of the unit, students should be able to solve quadratic equations (1) by factorization (recall Sec 2 work) (2) by completing the square, (3) by formula . understand relationships between the roots and coefficients of the quadratic equation form quadratic equations in the product form given two roots and , apply substitution to solve some higher order algebraic equations, understand and use discriminant to determine the nature of roots of a quadratic equation, use discriminant to determine when is always positive (or always negative) solve intersection problems between line and curve and discuss the nature of roots. find the maximum or minimum value by using completing the square, sketching of graphs of quadratic functions given in the form (1) y = a(x-h)2 + c ,a > 0 or a < 0 (2) y = a(x – b)(x – c) , a > 0 or a < 0 . solve quadratic inequalities using algebraic and graphical methods, representing the solution set on the number line.

Contains a x2 term Solving Quadratic Equations by Factorising
Sec 2 Revision Solving Quadratic Equations by Factorising A quadratic equation is an equation like: y = x2 y = x2 + 2 y = x2 + x – 4 y = x2 + 2x – 3 Contains a x2 term There are several methods of solving these but one methods that you must know is called FACTORISING

3 x 2 = x 2 = x 0 = 0 A x B = 0 What can you say about A or B (x + 3)(x – 2) = 0 means (x + 3) x (x – 2) What can you say about (x + 3) or (x – 2) x + 3 = 0 x = -3 x - 2 = 0 x = 2 or

(x + 3)(x + 2) x(x + 2) + 3(x + 2)  x X (x + 2) + 3 X (x + 2)
You try (x + 5)(x + 2) (x – 2)(x + 3) (x + 2)(x – 4) (x – 3)(x – 2) x(x + 2) + 3(x + 2)  x X (x + 2) + 3 X (x + 2)  x X x + x X X x + 3 X 2  x2 + 2x + 3x + 6  x2 + 5x + 6

(x )(x ) What goes with the x? Solve by factorising: 0 = x2 + 7x + 12
Write down all the factor pairs of 12. (x )(x ) What goes with the x? From this list, choose the pair that adds up to 7 3 + 4 = 7 0 = (x + 3)(x + 4) x = – 3 and – 4 Put these numbers into brackets

Solve by factorising: 0 = x2 + x - 6
Write down all the factor pairs of – 6 From this list, choose the pair that adds up to 1 (3) + (-2) = 1 3 – 2 = 1 0 = (x + 3)(x - 2) x = – 3 and 2 Put these numbers into brackets

Copy and fill in the missing values when you factorise x2 + 8x + 12 = 0
Find all the factor pairs of x 12 = 12 2 x _ = 12 3 x 4 = 12 From these choose the pair that add up to 8 _ + 6 = 8 Put these values into the brackets (x + _)(x + _) = 0 x = -2 and - 6 Solve by factorising 1. x2 + 3x + 2 = 0 2. x2 + x – 12 = 0 3. x2 – 12x – 20 = 0 4. x2 – x – 12 = 0

1 x2 + 5x + 6 = 0 2 x2 - x – 6 = 0 3 x2 + 8x + 12 = 0 4 x2 + x – 12 = 0 5 x2 - 8x + 15 = 0 6 x2 + 3x – 21 = 0 7 x2 - 3x – 18 = 0 8 x2 - 10x – 24 = 0 9 x2 + 8x + 16 = 0 10 x2 - 4x – 60 = 0

1 x2 + 5x + 6 = 0 (x + 3)(x + 2) 2 x2 - x – 6 = 0 (x – 3)(x + 2) 3
4 x2 + x – 12 = 0 (x – 3)(x + 4) 5 x2 - 8x + 15 = 0 (x – 3)(x – 5) 6 x2 + 3x – 21 = 0 (x + 7)(x – 4) 7 x2 - 3x – 18 = 0 (x – 6)(x + 3) 8 x2 - 10x – 24 = 0 (x - 12)(x + 2) 9 x2 + 8x + 16 = 0 (x + 4)(x + 4) 10 x2 - 4x – 60 = 0 (x - 10)(x + 4)

1 x2 + 5x + 6 = 0 (x + 3)(x + 2) -3 and -2 2 x2 - x – 6 = 0
4 x2 + x – 12 = 0 (x – 3)(x + 4) 3 and -4 5 x2 - 8x + 15 = 0 (x – 3)(x – 5) 3 and 5 6 x2 + 3x – 21 = 0 (x + 7)(x – 4) -7 and 4 7 x2 - 3x – 18 = 0 (x – 6)(x + 3) 6 and -3 8 x2 - 10x – 24 = 0 (x - 12)(x + 2) 12 and -2 9 x2 + 8x + 16 = 0 (x + 4)(x + 4) -4 and -4 10 x2 - 4x – 60 = 0 (x - 10)(x + 4) 10 and -4

This is often called the difference between two squares
-1 x 4 = -4 -2 x 2 = -4 4 x -1 = -4 = 0 x2 – 4 x2 + 0x – 4 (x – 2)(x + 2) Notice that x2 – 4 could be written as x2 – 22 (x – 2)(x + 2) This is often called the difference between two squares x2 – 25 (x + 5)(x – 5)

1 x2 - 9 2 x 3 x2 - 36 4 x2 - 49 5 x2 - 81 6 x2 - 64 7 x2 - 18 8 x2 - 24

1 x2 - 9 (x + 3)(x – 3) 2 x (x + 10)(x – 10) 3 x2 - 36 (x + 6)(x – 6) 4 x2 - 49 (x + 7)(x – 7) 5 x2 - 81 (x + 9)(x – 9) 6 x2 - 64 (x + 8)(x – 8) 7 x2 - 18 (x + √18)(x – √18) 8 x2 - 24 (x + √24)(x – √24)

Completing Square For quadratic equations that are not expressed as an equation between two squares, we can always express them as If this equation can be factored, then it can generally be solved easily.

If the equation can be put in the form then we can use the square root method described previously to solve it. “Can we change the equation from the form to the form ?”

The procedure for changing is as follows
The procedure for changing is as follows. First, divide by , this gives Then subtract from both sides. This gives

Recall that If we let we can solve for to get

Substituting in we get Using the symmetric property of equations to reverse this equation we get

Now we will return to where we left our original equation. If we add to both sides of we get or

We can now solve this by taking the square root of both sides to get

Example: [By completing square method]

The Quadratic formula allows you to find the roots of a quadratic equation (if they exist) even if the quadratic equation does not factorise. The formula states that for a quadratic equation of the form : ax2 + bx + c = 0 The roots of the quadratic equation are given by :

Example : Use the quadratic formula to solve the equation : x 2 + 5x + 6= 0 Solution: x 2 + 5x + 6= 0 a = 1 b = 5 c = 6 x = - 2 or x = - 3 These are the roots of the equation.

Example : Use the quadratic formula to solve the equation : 8x 2 + 2x - 3= 0 Solution : 8x 2 + 2x - 3= 0 a = 8 b = 2 c = -3 x = ½ or x = - ¾ These are the roots of the equation.

Example : Use the quadratic formula to solve for x to 2 d.p : 2x 2 +3x - 7= 0 Solution: 2x 2 + 3x – 7 = 0 a = 2 b = 3 c = - 7 x = or x = These are the roots of the equation.

Discriminant

x2 - 8x + 16 = 0 b=-8; c=16 b2-4ac=(-8)2-4(1)(16) =64-64 b2-4ac=0
Example : x2 - 8x + 16 = 0 a=1; b=-8; c=16 b2-4ac=(-8)2-4(1)(16) =64-64 b2-4ac=0 real, rational, equal

2x2 + 5x – 3 = 0 b2-4ac=52-4(2)(-3) =25+24 b2-4ac=49
Example : 2x2 + 5x – 3 = 0 a= ? b= ? c= ? b2-4ac=52-4(2)(-3) =25+24 b2-4ac=49 Real, rational, unequal

x2 + 5x + 3 = 0 =25-12 b2-4ac=13 real, irrational, unequal Example :

Example : x2 – x + 2 = 0 b2-4ac=12-4(1)(2) =1-8 b2-4ac=-7 imaginary

b2 - 4ac = 0 : real, rational, equal.
b2 - 4ac > 0 : perfect square , real, rational, unequal.

b2 - 4ac > 0 : not a perfect square – real, irrational, unequal.
b2 - 4ac < 0 : imaginary, complex, no solution.

Quadratic Graphs The graph of is a parabola. The graph looks like
if a > if a < 0

Key features of the graph:
The maximum or minimum point on the graph is called the vertex. The x-coordinate of the vertex is:

The y-intercept; the y-coordinate of the point where the graph intersects the y-axis. The y-intercept is: When x = 0, y = c The x-intercepts; the x-coordinates of the points, if any, where the graph intersects the x-axis. To find the x-intercepts, solve the quadratic equation

Example: Sketch the graph of vertex: min. point y-intercept: x-intercepts:

Example: Sketch the graph of Vertex: y-intercept: x-intercept(s):

Example: Sketch the graph of Vertex: y-intercept: x-intercept(s):

What do they look like? Here are some examples:

When solving inequalities we are trying to find all possible values of the variable which will make the inequality true. Consider the inequality We are trying to find all the values of x for which the quadratic is greater than zero or positive.

We can find the values where the quadratic equals zero by solving the equation,

You may recall the graph of a quadratic function is a parabola and the values we just found are the zeros or x-intercepts. The graph of is (-2,0) (3,0)

From the graph we can see that in the intervals around the zeros, the graph is either above the x-axis (positive) or below the x-axis (negative). So we can see from the graph the interval or intervals where the inequality is positive. But how can we find this out without graphing the quadratic? We can simply test the intervals around the zeros in the quadratic inequality and determine which make the inequality true.

For the quadratic inequality, we found x = 3 and x = –2 by solving the equation . Put these values on a number line and we can see three intervals that we will test in the inequality. We will test one value from each interval. -2 3

Interval Test Point Evaluate in the inequality True/False

Thus the intervals make up the solution set for the quadratic inequality, In summary, one way to solve quadratic inequalities is to find the x-intercept/s and test a value from each of the intervals surrounding the zeros to determine which intervals make the inequality true.

Example : Solve Step 2: Sketch the quadratic graph Step 1: Solve

Quadratic with linear Solve: x2 – 8x + 16 > 2x +7 Estimate ?
y = 2x + 10 y = x2 – 8x +16

Example: Solve: x2 – 8x + 16 > 2x +7 Algebraically:
1. Rearrange first 2. Solve like the others x2 – 8x + 16 > 2x +7 x2 – 10x + 16 > 7 (-2x) (-7) x2 – 10x + 9 > 0 Like the ones we did (x-9)(x-1) > 0 x>9 or x<1

Try this one Solve: x2 + x + 4 > 4x +14 Algebraically:
First: try a sketch Algebraically: 1. Rearrange first 2. Solve like the others x2 + x + 4 > 4x +14 x2 – 3x + 4 > 14 (-4x) (-14) x2 – 3x - 10 > 0 (x+2)(x-5) > 0 x<-2 or x>5

Summary In general, when solving quadratic inequalities
Find the zeros by solving the equation you get when you replace the inequality symbol with an equals. Find the intervals around the zeros using a number line and test a value from each interval in the number line. The solution is the interval or intervals which make the inequality true.

Practice Problems No solution