• 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. In this coursework I will be looking at equations that cannot be solved algebraically

    from the x-axis, this x-value is called x0. At the point where this line hits the curve, a tangent to the curve is drawn until it hits the x-axis. The point of interception is the first iterative value (x1). Another line is drawn up from x1 until it hits the curve, and the method is repeated.

  2. Numerical solutions of equations

    I can see from the graph that two roots need to be found using this method. The graph of this is below (or see Figure 3). The iterative formula for the "Newton-Raphson" method is: xn+1=xn- f(xn) f'(xn) For my equation, f'(xn)

  1. Functions Coursework - A2 Maths

    x f(x) 1.879380 -0.0000398162 1.879381 -0.0000322200 1.879382 -0.0000246238 1.879383 -0.0000170275 1.879384 -0.0000094313 1.879385 -0.0000018350 1.879386 0.0000057612 1.879387 0.0000133575 1.879388 0.0000209538 1.879389 0.0000285501 1.879390 0.0000361464 The root therefore lies in the interval [1.879385,1.879386]. Therefore the root of the equation f(x)=0, in the interval [1,2] is: x=1.87939 to five decimal places.

  2. Methods of Advanced Mathematics (C3) Coursework.

    1.54074074 0.077437779 20.74492848 1.537007887 1.53700789 0.000695055 20.37311978 1.536973771 1.53697377 5.76884E-08 20.36973796 1.536973768 1.53697377 0 20.36973768 1.536973768 The three routes I have found using this method are all highlighted in green. They can also be seen on the below graph In this method I also looked at the error bounds of my solutions.

  1. Solving Equations Numerically

    1.6 6.736 1.16 -0.2167 1.176 -0.01772 1.7 8.773 1.17 -0.09279 1.177 -0.00515 1.8 10.992 1.18 0.032632 1.178 0.007428 1.9 13.399 1.19 0.159559 1.179 0.020022 2 16 1.2 0.288 1.18 0.032632 x f(x) x f(x) x f(x) 1.177 -0.00515 1.1774 -0.00012177 1.1774 -0.000121767 1.1771 -0.00389 1.17741 4.0137E-06 1.177401 -0.000109189 1.1772 -0.00264

  2. Numerical solution of equations

    They are 2.81691, 0.33761 and -3.15452 which are the same as the result from the graph. The only advantage of this method is that the upper bound and lower are found automatically. However, the procedure is long and bored. Therefore, mistakes are easily made if you didn't check it.

  1. Solutions of equations

    I will know find the values for the function between 0.6 and 0.7. Table 3 x F(x) 0.6 -0.392 0.61 -0.377 0.62 -0.349 0.63 -0.306 0.64 -0.247 0.65 -0.171 0.66 -0.075 0.67 0.0412 0.68 0.1797 0.69 0.3418 0.7 0.529 Table 4 x F(x)

  2. Numerical Solutions of Equations

    -1.41 0.03391633 -1.411 0.01682557 -1.412 -0.000321374 -1.413 -0.017524621 -1.414 -0.034784292 -1.415 -0.052100505 -1.416 -0.069473381 -1.417 -0.086903041 -1.418 -0.104389604 -1.419 -0.121933191 -1.42 -0.139533923 x Y=x^5-2.7x+1.8 -1.4119 0.001395852 -1.41191 0.001224155 -1.41192 0.001052452 -1.41193 0.000880743 -1.41194 0.000709029 -1.41195 0.000537309 -1.41196 0.000365584 -1.41197 0.000193853 -1.41198 2.21161E-05 -1.41199 -0.000149626 -1.412 -0.000321374 I have now found the root to an accuracy of five decimal places.

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