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Introduction

By Joe Gill

Introduction

There are several ways of solving Quadratic equations. These are

FACTORISING

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.

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

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   ) 3.

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.

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