S3T3 Solving Quadratic Equations by Factorisation/Formula

Recap

To solve for x, we ‘put’ x on the LHS (left-hand-side) and the number value to the RHS (Right-hand-side),

and we find the value of x.

Example 

            Solve for 7x + 3 = 17

                           7x = 17 – 3

                                = 14

                             x = 14/7 = 2

Solving Quadratic Equation

(1)  What is the meaning of  Solve x² + 3x + 2 = 0

=> to find the values (two or less) of x where it will make the equation = 0.

(2) Any number multiply by 0 is equal to 0.

=> For any (x + a)(x – b) = 0, as long as any one of (x + a) or (x – b) = 0, equation is solved because 

When x = -a

             (-a + a)( x- b) 

            = 0 x (x – b) = 0

When x = b

             (x+ a)( b - b) = 0 

              = (x + a) x 0 = 0

             = 0

(3) Solving the quadratic equation

            (x + P)(x – Q) = 0

        When (x + P) = 0

x + P = 0

x = -P

When (x – Q) = 0

    x – Q = 0

    x = Q

Values of x are 

         x = -P or x = Q

Quadratic equations are usually solved by putting the equation into the form

         (Jx + P)(Kx + Q) = 0 where J, K, P and Q are numerical values.

A quadratic equation can have 0, 1 or 2 values.

Recap : Factorisation  (x ² + bx + c)

Factorise x ² + 5x + 6

=> To find P and Q where   x ² + bx + c = x ² + 5x + 6 = (x + P)(x + Q)

     => b = 5 = P + Q, c = 6 = PQ 

Step 1 : Draw the cross X

<< Since is it x²  , fill in x² >>

                     x                [What is P?]

                            \ /

                            / \

                      x               [What is Q?]

Step 2 :  Find factors of c (P X Q) for b (P + Q)

When there is more than 1 set of factors,

Find the factor of c = 6 (P x Q)

6 = 1 x 6, 

          = 2 x 3

Using the X to cross-multiply, 

                     x          2                   x        1

                           \ /                             \ /

                           / \                             / \

                      x         3                    x        6

 x x 3 = 3 x, 2 x x = 2 x                x x 1 = 3 x, x x 6= 2 x

2 x + 3 x = 5 x                               x + 6 x = 7 x

P = 2 , Q = 3

Step 3 : Factorise

  =>              (  x        +2 )       

                            \ /

                            / \

                      ( x         +3)

 Thus,

                 x ² + 5x + 6 = (x + 2) (x + 3) 

Solving Quadratic Equation by Factorisation                                    

Example:

Solve y² + 3y + 2 = 0

            (y + 2)(y + 1) = 0                       By Factorisation     y  \  /        2

            (y + 2) = 0 or (y+ 1) = 0                                           y   /  \        1

            y = -2 or y = -1

Solving by Formula

For Equation:         ax2 + bx + c = 0

         formula x = -b +   √b2 – 4ac

                                          2a

Example:

Solve 2x²+ 6x + 1 = 0              (ax2 + bx + c = 0)

a = 2 , b = 6, c = 1

                      x = -6 +   √6²– 4 x 2 x 1

                                         2x2      

                        = -6 + √28

                                    4

                        = -6 + 2/4√7

                        = -6 + 7/2

S1G1 : Angles

TYPES OF ANGLES

           

PROPERTIES OF ANGLES

         

        ㄥ at a point = 360^o       ㄥon a straight line=180^o          Sum ㄥ of triangle = 180^o

                                                                                                 a + b+ c = 180

                                                                                       b + d = 180(ㄥ on str line)

      

Vertically opposite angles           Alternate Angle         Corresponding Angles

ㄥA = ㄥC,                                           ㄥa = ㄥd                         ㄥA = ㄥE, ㄥC=ㄥG, 

ㄥD = ㄥB                                            ㄥc = ㄥd                        ㄥB = ㄥF , ㄥD = ㄥH 

Complimentary angles
For a right angle, two or more angles add up to 90^o 

NA/O Level

Triangles

                             

            Perimeter = adding the three side 

                    Area = ½ x Base  x Height


                        

                       Area of triangle ABC = ½ ab sinC 

Circles
                                 
                      Angle at centre is twice angle at circumference    

N621-G23 Cartesian coordinates in two dimensions y = ax + b

This is called a cartesian graph. 

(1) A Cartesian coordinate graph: two axes ("axes" plural of "axis"):

                           

• Horizontal axis is called the x-axis. 
• Vertical one is the y-axis. 

(2) Coordinates are points on a graph 

       Format: (x, y)

(3) The centre point of the graph is called the origin (0, 0). 

      It is the zero point of both the x-axis and the y-axis. 

Plotting a coordinate: 

Step1: locate on x-axis

Step2: locate on y-axis

Step3: "Move"/plot for both to "meet"

Example: 

     Plot coordinate (5,10) on the graph                          

Practice on same graph:

     Plot coordinates (-5, 5), (10,0), (0,5)

Linear functions y = ax + b

A function that produces a straight line when on a graph with a constant rate of change.

A linear (straight line) equation has a standard form:  

                            y = ax + b 

Example: 

     y = 2x +c 

     y = -x + 5 

     y = 4

     x = 3 are all linear equations.

Equation of the Line and its Coordinates

On the graph, the equation of the line y = mx+c = the LINE drawn.
-    y and x are the variables, and a and b are constant values (a number)

-    a is the slope (rate of change) 

-    b crosses the y-axis called the y-intercept with coordinate (0,b)

-  Coordinates are points on the line that satisfies/can be substituted into the equation.

    => the coordinates (x,y) can be substituted into the equation.

                  Equation of the line y = 2x + 8

Examples

To test if the following coordinates are on the line y = 2x + 8. 

a. (-4,0),        b. (-3,2)        c. (5, 5) 

Equation of the line is y = 2x + 8

(a) Substitute (-4,0) into y = 2x + 8 

                        =>  0 = 2(-4) +8

      (-0, -4) is on the line.

(b) Substitute (-3,2) into y = 2x + 8 

                        =>  2 = 2(-3) +8

      (-3, 2) is on the line.

(c) Substitute (5,5) into y = 2x + 8

                  => 5 ≠ 2(5) + 8

(5, 5) is a coordinate on the graph but not on the line.

Graph Below : y = 2x + 8, and the coordinates (-3,2), (-4,0) and (5,5).

Name the other coordinate on the line (satisfy the equation).

=> (1,10)

If the line passes through a coordinate (x,y), => can be used for the (x,y) of the equation : y = mx + c

Practice: What are the coordinates of (x₁, y₁), (x₂, y₂), (x₃, y₃)?

What is c (the y-intercept)?

c = y-intercept 

=> where the equation(line) intercept/‘cuts’ the y-axis. 

     Coordinate : (0, y), (0, c). 

         

         Line cuts y-axis at 5          

         c = 5

Finding the value of c from the equation

Equation of a straight line:

              y = mx + c 

c= the y-intercept; coordinate (0,c)

Example:

(1)       y = 3x + 5  => c = 5, (0,5)

(2)       y = -x – 3   => c = -3 (0,-3)

(3)       2y = 5x + 6      (must change equation of format: y = mx + c)

             y = 5x + 6

                        2

               = 5x + 3,  => c = 3 (0, 3)

                   2

(4)       -3y = -2x + 9    (must change equation format to: y = mx + c)

            2x - 9 = 3y

             3y = 2x - 9

                          3

                  = 2/3 x - 3,  => c = -3 (0, -3)

What is m?

                      y = mx + c

Gradient(m) is the ‘steepness’ or slope of a graph. 

Formula: 

                    Gradient = y₂ – y₁

                                       x₂ – x₁

We need 2 sets of coordinates (x, y) on the line to find the gradient(m). 

Example: 

Find the gradients of the following graphs.

(x₁, y₁) = (0, -2), (x₂, y₂) = (5,8)                  (x₁, y₁) = (-5, 7), (x₂, y₂) = (1,-5)              

   Gradient = -2 – 8                                                   Gradient = 7 – (-5) 

                        0 - 5                                                                        -5 - 1 

                   = -10/-5                                                                     =12/-6

                   = 2                                                                             = -2

 

Observation:

1. The gradient/"slope" is positive (2) for  (0, -2), (5,8) as it is  "ascending" from left to right.

  Slope Upward => positive gradient(+m)

2. The gradient/"slope" is -2 for (-5,7), (1-5) as it is "descending" from left to right.

          Slope downward => -m

3.  When calculating the value of m, the (x,y) must be of the same "set order" =>

y₁ - y₂ or y₂ - y₁  NOT  y₁ - y₂

                x₁ - x₂     x₂ - x₁            x₂ - x₁

Positive and negative gradient equation

To find the Value of m from the equation

Equation of a straight line:

            y = mx + c 

Example:

(1)       y = 3x + 5 , m = 3

(2)       y = -x – 3, => y = (-1)x - 3, m = -1

(3)       2y = 5x + 6    (must change equation of format: y = mx + c)

             y = 5x + 6

                        2

               = 5x + 3, m = 5/2

                  2                   

(4)       -3y = 2x + 9  (must change equation of format: y = mx + c)

             -2x - 9 = 3y

                   3y = -2x - 9

                               3

                        = -2/3 x - 3,  m = -2/3

Equation Of Vertical And Horizontal Lines

(1) x = numeric value 

      -> a vertical line graph

Example: 

                        x=5

                     

         Equation of the line : x = 5

(2) y = numerical value 

         => a horizontal line graph. 

Example:

                         y=5

                          

   Equation of the line : y = 5

To find intercept of 2 lines

(1) Use simultaneous equation or 

(2) the point of interception of the 2 lines on the graph 

2 linear equations:  y = 2x – 2

                               y = -2x + 4

From the graph, the intersection coordinate is (3/2, 1)

=> x = 3/2 and y = 1

Practice

1.  Draw a simple graph. Plot and label the point A(-2, 1) and B (3, 5).

2.  Find the gradient of the line AB   [17/I/92,2/T]

3.  Find the gradient of the line joining the points A (2, 6) and B(7,3)

4.   Find the gradient of the line joining the points A(2, 1) and B(4, 6)