(x+9)(x+3)
x= -9 or – 3
5.) x²+14x +40 = 0
(x+4)(x+10)
x=-4 or-10
6.) x²+ 15x + 36 = 0
(x+12)(x+3)
x= -12 or – 3
7.) x²-3x-18=0
(x+3)(x-6)
x= -3 or + 6
8.) x² -7x -30=0
(x+3)(x-10)
x=-3 or +10
9.) x²+ 4x – 32=0
(x+4) (x-8)
x= -4 or +8
A Table of Results for Quadratic Equations
Here is a table to summarise the Quadratic equations on the previous page.
Equation A B C Roots
Rules Shown from The Table
- The two roots multiplied always give C
- The two roots added always give B
- Both roots are factors of C
- A always = 1
Other Ways of Solving Quadratic Equations
The Quadratic Formula
The Solutions to any quadratic equation ax² +bx+c = 0 are given by this formula
X= -b+/- √b²– 4ac
2a
Example of how to use the quadratic formula
“Find the roots of 3x² + 7x=1 to 2 decimal places
1. First get it into the for ax² + bx+c=0
2.Then carefully identify a,b &c
3. 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. We now need to look at all the pairs of numbers that multiply with each other to give c=6 (ignoring the minus sign for now) i.e 1x6 3x2
4.Now we need to find the combination that does this
(3x 1)(x 6) gives 18x and 1x which add/subtract to give 17x or 19x
(3x 6)(x 1) gives 3x & 6x which add/subtract to give 9x or 3x
(3x 3)(x 2) gives 6x & 3x which add/subtract to give 9x or 3x
(3x 2)(x 3) gives 9x & 2x which add subtract too give 11x or 7x
5. So (3x 2)(x 3) is the combination which gives b=7(give or take +/-)
6. Now fill in the +/- signs so that the combination will add/subtract to give +7(=b) so the final brackets are (3x-2)(x+3)
7.Check
8. The last step is to get the roots from the equation : (3x-2)(x+3)=0
Which you do by separately putting each bracket =0
i.e (3x-2)=0=>x=2/3 or (x+3)=0=> x=-3
Examples of some Quadratic Equations Where A does not equal 1 solved by factorising
1.) 2x² +6x=20
rearrange into normal form 2x²+6x-20=0
set up brackets (2x )(x )
factorises to give 4&5 so (2x 4)(x 5)
Put in +/- signs (2x-4)(x+5)=0
Put =0 on each bracket (2x-4)=0 & (x+5)=0
So therefore x= 2 or -5
2)2x²-13x-7=0
(2x+1)(x-7)=0
2x=-1
x=1/2 or 7
3) 3x²+13x+4=0
(3x+1)(x+4)=0
3x=-1
x= -1/3 or -4
4) 2x² +11x-6=0
(2x+1)(x-6)
2x=-1
x= -1/2 or 6
5.)5x²-17x-12=0
(5x+3)(x-4)=0
5x=-3
x= -3/5 or 4
6.) 5x²-7x-6=0
(5x+3)(X-2)=0
5x=-3
x= -3/5 or 2
7.) 2x²+7x+5=0
(2x+5)(x-1)
2x=-1
x=-1/2 or 5
8) 7x²+22x+3=0
(7x+1)(x+3)=0
7x =-1
x=-1/7 or x=-3
9.)2x²+7x+3=0
(2x+1)(x+3)
2x=-1
x= -1/2 or – 3
A Table of Results for Quadratic Equations where a does not equal 1
Equation A B C Roots
Rules Shown from The Table
- The roots multiplied by each other and then by a always equal c
- C divided by A equals the multiple of the roots
G.C.S.E Maths Coursework Extension
Cubics
The general equation for a cubic is ax3+bx2+cx+d
Although we cannot factorise a cubic yet I looked up on the internet how to make a cubic equation from its roots here is how I did it
Example
The roots of this equation are x= -2, +3&+1
1. First of all put the roots into three brackets with opposite +/- signs
(x+2)(x-3)(x-1)=0
2. Next factorise two of the brackets into one
(x2+2x-3x-6)(x-1)=0
3. Then factorise the two remaining brackets
(x2-x-6)(X-1)=0
4. Next tally up everything out of the brackets
X2-x2-6x-x2+x+6=0
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
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.
