• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15

Quadratic Equations.

Extracts from this document...

Introduction

GCSE Maths Coursework Quadratic Equations

By Joe Gill

Introduction

There are several ways of solving Quadratic equations. These are                                              

FACTORISING

THE QUADRATIC FORMULA

COMPLETING THE SQUARE

TRIAL & IMPROVEMENT

Here is an example of how to solve a quadratic equation using factorising:-

Factorising a quadratic means putting the equation into two brackets. The standard format for quadratics is ax² + bx +c = 0.

How to Answer a Quadratic Equation by factorising

Solve x2-x+12=0 by factorising

1. Firstly rearrange it x²-x=12

2. a=1 so the initial brackets are :- (x    ) (x   )

3. We now want to look at all the pairs of numbers that multiply to give c (=12), but which also add or subtract to give the value of b:

1 x 12 gives: 13 or 11

2 x 6   give: 8 or 4

3 x 4    give 7 or 1 (this is the value of b)

4. So 3 & 4 will give b + or – 1, so put them in the brackets: - (x  3) (x   4)= 0

5. Now fill in the +/- signs so that the 3 &4 add/subtract to give -1 (=b),so we must have +3 and -4 so we’ll have (x+3) (x-4)

6. Now check by expanding the brackets out and see of they give x²-x = 12.

7. Now we have to work out the roots. A simple way to work out the roots is to just switch the +/- signs on the two numbers in the bracket i.e. x=-3 or +4

Examples of some quadratic equations Solved by Factorising

Here are a few examples of factorising & solving equations where a=1:-

1.)    X² +7x+10= 0

...read more.

Middle

 put these values into the Quadratic formula and write down each stage:

x= -7 +/- √7² -4x3x-1

                 2x3

= -7 +/- √49 +12

              6

= -7+/- √61   =  -7 +/- 7.81

         6                     6

= 0.1350 or =2.468

X= -0.14 or -2.47 (2dp)

4. Finally check by putting the values back into the original equation

How to solve Quadratics by completing the square

1. Rearrange The quadratic into the standard format ax²+bx+c=0

2. If a is not 1 then divide the whole equation by a to make sure it is

3.Write out the initial bracket  

4. Multiply out the brackets and compare to the original

Example

Express x²-6x-7=0 as a completed Square, and hence solve it

(x-3)²        

Square out the brackets x²-6x+9 and compare it to the original :   x²-6x-7

To make it like the original equation it needs – 16 on the end, hence we get (x-3)²-16=0 as the alternative version of x²-6x-7

Now we have to solve it.

Take the 16 over to get (x-3)²=16

Then square root both sides to give us (x-3)=+/- 4

Take the 3 over to get x= +/-4+3

So x=7 or -1

Factorising  Quadratics when a does not equal one

Example

Solve 3x² + 7x=6 by factorising

1. Firstly put the formula into the basic format 3X²+7x-6=0

2. Now because a=3 The two x-terms in the brackets will have to multiply to give 3x² so the initial brackets will have to be: (3x   )(x   )image00.png

3.

...read more.

Conclusion

          X3-2x2-5x+6 =0

5. Finally work out the roots

       a=1 b=-2 c=-5 d=6

Examples on Building a Cubic Equation

1. X=2,-1&3

   (X-2)(x+1)(X-3)=0

  (X2+x -2x -2)(X-3)=0

X³+x²-2x2-2x -3x2-3x+6x+5=0

 Equation = x³-4x² +3x+5 =0

  a=1 b=-4 c=3 d=+5

2. X= 4,-2&-3

       (X-4)(x+2)(x+3)=0

(X2+2x-4x-8)=0

       (X2-2x-8)(x+3)=0

X3-2x2-8x +3x2 -5x -28=0

Equation= x3+x2-13x-24=0

a=1 b=1 c=-13 d=-24

3. 2x=3 x=-1 x=2

     x=3/2

   (2x-3)(x+1)(X-2)=0

    (2x2-x-3)(x-2)=0

   2x3-x2-3x-4x2+2x+6

Equation= 2x3 -5x2 –x + 6=0

a=2 b=5 c=1 d=6

4. x= -5-7 4

  (x+5)(x+7)(x-4)=0

    (x2+5x + 7x + 35)(x-4)=0

        x3+5x2+7x2+35 -4x2-20x -28x- 140=0

Equation= x3 +8x2-48x- 105=0

a=1 b=8 c=-48 d=-105

5. 5x = 10 x= 2,-5, 10

 (5x-10)(x-10)(x+5)=0

           (5x2-60x+100)(x+5)=0

           5x3-60x2+100x +25x2-300x +500=0

Equation= 5x3-35x2-200x+500=0

Na=5 b=-35 c=-200 d= 500

A Table of Results for Cubic Equations

Equation

Roots

a

b

c

d

Sum

Product

X3-4x2+3x+5

2,-1,3

1

4

3

5

4

6

X3+x2-13x-24

4,-2,-3

1

1

-13

-24

-1

-24

2x3-5x2-x+6

3/2,-1,2

2

5

1

6

2.5

-3

X3+8x2-48x-105

-5,-7,4

1

8

-48

-105

-8

140

5x3-35x2- 200+500

2,-10,5

5

-35

-200

500

-3

-100

Rules Shown from The Table

  • The product divided or subtracted by d gives a
  • The sum = b divided by a
  • When the roots are substituted back into the equation they equal themselves or d.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Numerical solutions of equations

    x + 2 1/2 (x+21/2)1/2 + ln (x+21/2) (x+21/2) -1/2 (x+21/2)1 2 1 + ln(x+21/2) (x+21/2)1/2 2(x+21/2) 1/2 = 2+ ln((x+21/2) 2(x+21/2) 1/2 The iterative formula for the Newton-Raphson method is: xn+1 = xn = f'(xn) f'(xn) x n+1 = xn - 2(x+21/2)ln(x+21/2) 2 + ln(x+21/2) I choose my starting value of x (x1)

  2. Numerical Solutions of Equations

    This can be shown by replacing the lower and upper error bounds into y = x5-3x+1. When x = (0.334735 +0.00005) = 0.334735, y = -2.52 x 10-6 (y is negative). When x = (0.334735 - 0.00005) = 0.334725, y = 2.69 x 10-5 (y is positive).

  1. Arctic Research (Maths Coursework)

    Plane Velocities; 320, 340, 360 After, I will keep the plane velocity constant at 340 km/h and vary the strength of the wind. Wind Velocities; 25, 30, 35, Total flight time summary Plane Velocity (km/h) Wind Velocity (km/h) 320 340 360 25 2.373 30 2.539 2.384 2.249 35 2.39 The

  2. Numerical Solutions of Equations.

    = 0.49269, y is +0.000005. When x = (0.492685 - 0.000005) = 0.49268, y is -0.000037. This change of sign means the root lies between this two values for x. Problems with decimal search Although decimal search was successful in finding this particular root there are problems with the method in certain situations that result in no root being detected.

  1. Three ways of reading The Bloody Chamber.

    claim that the 'real' meaning is to be found at the second level. Keys are not simply keys nor are wolves simply wolves. The most widely accepted mythic second level structure is Freudianism. And, of course, what Freud claimed to be doing was precisely the uncovering the real meaning behind certain phenomena.

  2. Investigate the relationships between the lengths of the 3 sides of the right angled ...

    = 4 x 100� + 6 x 100 + 2 = 10000 + 200 = 400000602 Area of the triangle: 6 , 30 , 84 , 180 , 330 24 , 54 , 96 , 150 30 42 54 12 12 F (n)

  1. 2D and 3D Sequences Project

    So we will now sub. in the values to the equation If we take sequence 2 for our value of n with our present values we will get: _(8) - 2(4) + c = 8 this can then be simplified to, _(2)

  2. Methods of Advanced Mathematics (C3) Coursework.

    If the intervals were smaller then it would have worked. Newton Raphson Method In this method an estimate of the route is taken a line equal to x is taken up until it hits the curve and then a tangent is drawn down to the x-axis.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work