- Rearrange it into the form
*ax*^{2}+*bx*+*c*= 0- zero must be on one side
- it is easier to use the side where
*a*is positive

- it is easier to use the side where

- zero must be on one side
- Factorise the quadratic and solve each bracket equal to zero
- If (
*x*+ 4)(*x*- 1) = 0, then either*x*+ 4 = 0 or*x*- 1 = 0- Because if
*A*×*B*= 0, then either*A*= 0 or*B*= 0

- Because if

- If (
- Factorise the quadratic and solve each bracket equal to zero
- To solve
- …solve “
**first bracket = 0**”:*x*– 3 = 0- add 3 to both sides: x = 3

- …and solve “
**second bracket = 0**”*x*+ 7 = 0- subtract 7 from both sides:
*x*= -7

- The
**two solutions**are*x*= 3 or*x*= -7

- The solutions have the opposite signs to the numbers in the brackets

- …solve “
- To solve
- …solve “first bracket = 0”
- 2
*x*– 3 = 0 - add 3 to both sides: 2
*x*= 3 - divide both sides by 2:
*x*=

- 2
- …solve “second bracket = 0”
- 3
*x*+ 5 = 0 - subtract 5 from both sides: 3
*x*= -5 - divide both sides by 3:
*x*=

- 3
- The two solutions are
*x*= or*x*=

- …solve “first bracket = 0”
- To solve
- it may help to think of
*x*as (*x*– 0) or (x) - …solve “first bracket = 0”
- (
*x*) = 0, so*x*= 0

- (
- …solve “second bracket = 0”
*x*– 4 = 0- add 4 to both sides:
*x*= 4

- The
**two solutions**are*x*= 0 or*x*= 4- It is a common
**mistake**to divide both sides by*x*at the beginning - you will**lose a solution**(the*x*= 0 solution)

- It is a common

- it may help to think of

- Use a calculator to check your final solutions!
- Calculators also help you to factorise (if you're struggling with that step)
- A calculator gives solutions to as
*x*= and*x*=- "Reverse" the method above to factorise!

- Warning: a calculator gives solutions to 12
*x*^{2}+ 2*x*– 4 = 0 as*x*= and*x*=- But 12
*x*^{2}+ 2*x*– 4 ≠ as these brackets expand to 6*x*^{2}+ ... not 12*x*^{2}+ ... - Multiply by 2 to correct this
- 12
*x*^{2}+ 2*x*– 4 =

- But 12

(a)

Solve

Set the first bracket equal to zero

*x* – 2 = 0

Add 2 to both sides

*x* = 2

Set the second bracket equal to zero

*x* + 5 = 0

Subtract 5 from both sides

*x* = -5

Write both solutions together using “or”

*x*** = 2 or x = -5**

(b)

Solve

Set the first bracket equal to zero

8*x* + 7 = 0

Subtract 7 from both sides

8*x* = -7

Divide both sides by 8

*x* =

Set the second bracket equal to zero

2*x* - 3 = 0

Add 3 to both sides

2*x* = 3

Divide both sides by 2

*x* =

Write both solutions together using “or”

*x*** = **** or x = **

(c)

Solve

Do not divide both sides by* x *(this will lose a solution at the end)

Set the first “bracket” equal to zero

(x) = 0

Solve this equation to find *x*

*x* = 0

Set the second bracket equal to zero

5*x* - 1 = 0

Add 1 to both sides

5*x* = 1

Divide both sides by 5

*x* =

Write both solutions together using “or”

*x*** = 0 or x = **

- To solve
*x*^{2}+*bx*+*c*= 0**replace**the first two terms,, with*x*^{2}+*bx***(**where*x*+*p*)^{2}-*p*^{2}*p*is half of*b*- this is called
**completing the square***x*^{2}+*bx*+*c*= 0 becomes- (
*x*+*p*)^{2}-*p*^{2}+*c*= 0 where*p*is half of*b*

- (

- rearrange this equation to
**make**(using ±√)*x*the subject

- For example, solve
*x*^{2}+ 10*x*+ 9 = 0 by completing the square*x*^{2}+ 10*x*becomes (*x*+ 5)^{2}- 5^{2}- so
*x*^{2}+ 10*x*+ 9 = 0 becomes (*x*+ 5)^{2}- 5^{2 }+ 9 = 0 - make
*x*the subject (using ±√)

- (
*x*+ 5)^{2}- 25^{ }+ 9 = 0 - (
*x*+ 5)^{2}= 16 *x*+ 5 = ±√16*x*= ±4 - 5*x*= -1 or*x*= -9

- (

- If the equation is
*ax*^{2}+*bx*+*c*= 0 with a number in front of*x*^{2}, then divide both sides by*a*first, before completing the square

- The quadratic formula actually comes from completing the square to solve
*ax*^{2}+*bx*+*c*= 0*a*,*b*and*c*are left as letters, to be as general as possible

- You can see hints of this when you solve quadratics
- For example, solving
*x*^{2}+ 10*x*+ 9 = 0- by completing the square, (
*x*+ 5)^{2}= 16 so*x*= ± 4 - 5 (from above) - by the quadratic formula, = ± 4 - 5 (the same structure)

- by completing the square, (

- For example, solving

- When making
*x*the subject to find the solutions at the end, don't expand the squared brackets back out again!- Remember to use ±√ to get
**two**solutions

- Remember to use ±√ to get

Solve by completing the square

Divide both sides by 2 to make the quadratic start with *x*^{2}

Halve the middle number, -4, to get -2

Replace the first two terms, *x*^{2} - 4*x*, with (*x* - 2)^{2} - (-2)^{2}

Simplify the numbers

Add 16 to both sides

Square root both sides

Include the ± sign to get two solutions

Add 2 to both sides

Work out each solution separately

*x* = 6 or *x* = -2

- A quadratic equation has the form:
*ax*^{2}+*bx*+*c*= 0 (as long as*a*≠ 0)- you need "= 0" on one side

- The
**quadratic formula**is a formula that gives both solutions:

- Read off the values of
*a*,*b*and*c*from the equation **Substitute**these into the formula- write this line of working in the exam
- Put
**brackets**around any**negative numbers**being substituted in

- To solve 2
*x*^{2}- 7*x*- 3 = 0 using the quadratic formula:*a*= 2,*b*= -7 and*c*= -3- Type this into a calculator
- once with + for ± and once with - for ±

- The
**solutions**are*x*= 3.886 and*x*= -0.386 (to 3 dp)**Rounding**is often asked for in the question- The calculator also gives these solutions in
**exact form**(surd form), if required *x*= and*x*=

- The part of the formula under the square root (
*b*^{2}– 4*ac*) is called the**discriminant** - The sign of this value tells you if there are 0, 1 or 2 solutions
- If
*b*^{2}– 4*ac*> 0 (positive)- then there are 2 different solutions

- If
*b*^{2}– 4*ac*= 0- then there is only 1 solution
- sometimes called "two repeated solutions"

- If
*b*^{2}– 4*ac*< 0 (negative)- then there are no solutions
- If your calculator gives you solutions with i terms in, these are "complex" and not what we are looking for

- Interestingly, if
*b*^{2}– 4*ac*is a perfect square number ( 1, 4, 9, 16, …) then the quadratic expression could have been factorised!

- If

- Yes to
**check**your final answers, but a method must still be shown as above

- Make sure the quadratic equation has "= 0" on the right-hand side, otherwise it needs rearranging first
- Always look for how the question wants you to leave your final answers
- for example, correct to 2 decimal places

Use the quadratic formula to find the solutions of the equation 3*x*^{2} - 2*x* - 4 = 0, giving your answers correct to 3 significant figures.

Write down the values of *a*, *b* and *c*

*a* = 3, *b* = -2, *c* = -4

Substitute these values into the quadratic formula,

Put brackets around any negative numbers

Input this into a calculator

Use + for ± to get the first solution

*x* = 1.53518...

Input this into a calculator a second time

Use - for ± to get the second solution

*x* = 0.86851...

Present both answers together (using the word "or" between them)

Round the answers correct to 3 significant figures (note how this affects the number of decimal places)

*x* = 1.54 or *x* = 0.869

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