Factorising Quadratics Flashcards
What formula do we use for the ac method
ax² + bx + c
When do we use this method?
When a is more than 1
If we have 4x^2 + 9x + 5, what values do a, b and c have?
a=4, b=9 and c=5
Example 1: Using: 2x^2+9x+4, factorise them using the ac method
a=2, b=9, c=4
We need two numbers that multiply to make a times c and add to make b
ac=8
b=9
1 and 8 times to make 8 and add to make b
I now rewrite the equation adding these numbers in: 2x^2+x+8x+4
I split it in half
First half: 2x^2+x
Factorise: x(2x+1)
Second half: 8x+4
Factorise: 4(2x+1)
Now I add the x and the 4 together and put them in brackets: (x+4)
And finally multiply this by (2x+1)
(x+4)(2x+1)
Example 2:
Factorise 3x^2-11x+6
- Find a, b and c: a=3, b=-11, c=6
- Find two numbers that multiply to give ac and add to give b. ac=18 and b=-11. These two numbers are -9 and -2. Remember two negatives multiply to give a positive.. and add to give a negative
- Make these numbers the two values of x, so instead of -11x it’s -9x and -2x
- Rewrite the equation: 3x^2-9x-2x+6
- Split it in half and factorise each side
First half:3x^2-9x
Factorise it: 3x(x-3)
Second half: -2x+6
Factorise it: -2(x-3) - Put the parts outside the brackets into brackets, so: (3x-2)
- Multiply both brackets together: (3x-2)(x-3)
- Check by expanding the brackets: 3x^2-9x-2x+6
- As the expanding brackets work then it has been done correctly
When you are solving a quadratic equation what are you finding?
The points where the graph crosses the x-axis. These are called roots or solutions. However, in an equation like this: (x+3)(x+3)=0, x would be -3 both times. This would mean there would only be one point at which the axis on the graph is crossed.
What are the steps to finding the roots/solutions when there is an equals, e.g. =0 at the end of the equation?
- Make sure the equation=0
- Factorise
- Set each bracket=0
- Solve each bracket
Example 3:
x^2-3x-10=0
- We need two numbers that multiplies to give -10 and adds to give -3. Note we don’t need to use the ac method anymore as there is no number before the x^2 bit (or coefficient). 2 and -5 multiply to give -10 and add to give -3
- Put it in brackets: (x+2)(x-5)=0
- Set each bracket to 0 and solve
First bracket: x+2=0, so x=-2
Second bracket: x-5=0, so x=5 - Answers: x=5 and x=-2
Example 4:
Find the roots of 2n^2+n=10
- Set it equal to 0 by minusing 10, so: 2n^2+n-10=0
- We need two numbers that multiplies to give -10 and adds to give 1 (multiplies to give ac and adds to give b). These numbers are 5 and -4
- Rewrite the equation with these numbers as the coefficient of n: 2n^2+5n-4n-10=0
- Split it in half and factorise
First half: 2n^2+5n
Factorise: x(2n+5)
Second half: -4n-10
Factorise: -2(2n+5) - Put the numbers outside the brackets into brackets to give: (n-2)
- Put the whole equation together again: (n-2)(2n+5)=0
- Set each bracket equal to 0
First bracket: n-2=0
Solve: n=2
Second bracket: 2n+5=0
Solve: 2n=-5, n=2.5 or -5/2 - Answers: n=2, n=-5/2
Example 5:
2x^2+7x=0
- We don’t use ac method as there is no c, so we go straight to factorisation: x(2x+7)=0
- Either x or 2x+7 must be equal to 0.
- If we set x as 0 then there is no extra work to be done and this is one of our answers
- Set the brackets as 0: 2x+7=0
- Solve: 2x=-7, so x=-3.5 or -7/2
- Answers: x=0, x=-7/2
Example 6:
Solve 4x^2-81=0
- Both 4 and 81 are square numbers, so we find the square roots of both numbers and put them in brackets (difference of 2 squares) giving: (2x+9)(2x-9)=0. Notice how there is a plus and an add in the brackets so we end up with a -81 rather than +81.
- Set each bracket equal to 0 and solve:
First bracket: 2x+9=0
Solve: 2x=-9. x=-9/2
Second bracket: 2x-9=0
Solve: 2x=9, x=9/2 - Answers: x=-9/2, x=9/2
What two reasons are there for why we cannot use the ac method for factorising all the time?
Because there is no c value, and the numbers we are looking for are not integers
What is the quadratic formula
x=-b add or minus the root of: b^2-4ac all over 2a
Example 7:
Solve x^2+5x+1=0. Give answer to 1dp
- As it says give answer to 1dp, we know we have to use the quadratic formula.
- Find values for a, b and c. a=1, b=5, c=1
- Insert the numbers into the equation: x=-5 plus or minus root 5^2-4x1+1 all over 2x1
- Insert this into a calculator, once using the plus and once the minus for the part before the root.
- Answers: x=-4.8 (to 1dp) and x=-0.2 (to 1dp)
Example 8:
Solve 12x^2-13x+4=13x^2-5x+1 to 1dp
- First sort out the x^2 by taking away 12x^2 from both sides of the equation: -13x+4=x^2-5x+1
- Now sort out the other x bit by adding 13x to both sides: 4=x^2+8x+1
- Now, so we can get the equals zero part, minus 4 from both sides: x^2+8x-3=0
- Find the values of a, b and c. a=1, b=8 and c=-3
- Insert them into the quadratic formula to give: -8 plus or minus root 8^2-4x1x-3 all over 2x1
- Put it into a calculator, first with a plus before the root and then a minus.
- Answers: x=0.4
x=-8.4
