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    Help With Solving Quadratic Equations shanemacken2000

    Last year I learnt the box method but have since forgotten it and my HL Maths teacher told me today when I asked for help that I should be able to do the brackets method for HL Maths and wouldn't refresh my memory. Can somebody help?

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      E.Devitt

      Look at the book or you're junior cert maths notes and restudy it

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      shanemacken2000

      I have no notes and it isn't in my maths book so that's why I am asking...

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      smcshane

      It's pretty difficult to write in this format but there are a few types of quadratics that you will see for example

      x^2 +10x +24 = 0

      x^2 -10x -24 = 0

      2x^2 +5x +3 = 0

      some with positive and negative numbers or some with a number in front of the X^2.

      for the three examples above

      Ex 1.

      x^2 +10x +24 = 0 Draw your two sets of brackets

      ( ) ( ) =0 You are dealing with the letter/variable x so put it in brackets

      (x ) (x ) =0

      Next look at the last number which is +24, what are the pairs of numbers that multiply to give you +24

      +1 and +24

      +2 and +12

      +3 and +8

      +4 and +6

      -1 and -24

      -2 and -12

      -3 and -8

      -4 and -6

      Once you have all the pairs of numbers, add or subtract depending on their signs to see which pair give you the number in front of your x term, which is +10 in this example. so from above

      +1 and +24 = 25

      +2 and +12 = 14

      +3 and +8 = 11

      +4 and +6 = 10

      -1 and -24 = -25

      -2 and -12 = -14

      -3 and -8 = -11

      -4 and -6 = -10

      You can see its the pair of +4 and +6. So you can fill them in to your brackets above, doesn't matter in which order

      (x ) (x ) = 0

      (x +4) (x +6) = 0

      let each bracket = 0 to find your two roots

      x +4 = 0 and x +6 =0 and solve

      x = -4 and x = -6

      Naturally this is written out in a long way but you dont have to go through all the pairs of numbers when you get used to it like i did above, you get used to spotting what numbers it will be.

      Ex2

      x^2 -10x -24 = 0 Draw your two sets of brackets

      ( ) ( ) =0 You are dealing with the letter/variable x so put it in brackets

      (x ) (x ) =0

      Next look at the last number which is -24, what are the pairs of numbers that multiply to give you -24, they will have to have different signs this time to get -24

      +1 and -24

      +2 and -12

      +3 and -8

      +4 and -6

      -1 and +24

      -2 and +12

      -3 and +8

      -4 and +6

      Once you have all the pairs of numbers, add or subtract depending on their signs to see which pair give you the number in front of your x term, which is +10 in this example. so from above

      +1 and -24 = -23

      +2 and -12 = -10

      +3 and -8 = -5

      +4 and -6 = -2

      -1 and +24 = 23

      -2 and +12 = 10

      -3 and +8 = 5

      -4 and +6 = 2

      You can see its the pair of +2 and -12. So you can fill them in to your brackets above, doesn't matter in which order

      (x ) (x ) = 0

      (x +2) (x -12) = 0

      let each bracket = 0 to find your two roots

      x +2 = 0 and x -12 =0 and solve

      x = -2 and x = +12

      Again i wrote out all the pairs of numbers but you can see once you now the right pair you dont have to keep writing the rest.

      Finally in example 3 you have to be careful because of the the number in front of x^2. I do this differently but you can do it the same as above but you have to be careful about the pairs you use. one of the numbers will have to be multiplied by 2 in this case before you put them together.

      i do it the following way

      Ex 3

      2x^2 +5x +3 = 0

      Multiply the number in front of x^2 by the constant. In this case 2 multiply by +3 = +6

      so write out the pairs of numbers that multiply to get +6 so they will have to have same signs

      +1 and +6

      +2 and +3

      -1 and -6

      -2 and -3

      find the pair that go together to give you the number in front of x term which is +5

      +1 and +6 = 7

      +2 and +3 = 5

      -1 and -6 = -7

      -2 and -3 = -5

      so the pair is +2 and + 3. This is where it differs slightly

      i rewrite the original question which is

      2x^2 +5x +3 = 0

      to

      2x^2 +2x +3x +3 = 0 (so basically i've split the 5x up to look like +2x +3x...means the same)

      split the whole thing in two

      2x^2 +2x +3x +3 = 0

      Factorise each

      2x( x+1) +3(x+1) = 0 (you will notice the bracket is the same in both)

      Put the bits outside bracket together and then use the other bracket. Your brackets will look like

      (2x + 3) ( x + 1) = 0

      Solve both = 0

      2x + 3 = 0 and x + 1 = 0

      2x = -3 and x = -1

      x = -3/2 and x = -1

      So you have your two answers.

      Finally all that being said, you can use the quadratic formula (the -b +/- formula) for all quadratics and it will work out, its just you said you wanted brackets method.

      Its tough to write it all out here so i hope this helps and the best of luck!

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      smcshane

      Sorry that comes out really really really long but just to clear up the example 3 when i said split it in two because the spaces didnt come out

      split the whole thing in two

      2x^2 +2x ..............and +3x +3

      then factorise

      2x(x+1) ................and 3(x+1) then take the outside terms for one bracket and use the other bracket. continue from here in previous post.

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      shanemacken2000

      That is brilliant thank you so much!

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