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quadratic equation 2x^2-2x-4=0 watch

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    I need help with this
    quadratic equation 2x^2-2x-4=0
    Solve
    2x^2-2x-4=0
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    (Original post by Bybella)
    I need help with this
    quadratic equation 2x^2-2x-4=0
    Solve
    2x^2-2x-4=0

    All you have to do is punch that information into the quadratic formula which would be at the front of an exam paper? Are you doing Edexcel GCSE?

    If you need more help lemme know
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    2x²-2x-4=0
    Factor out a 2
    2(x²-x-2)
    You don't need quadratic equation for this one
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    (Original post by GUMI)
    2x²-2x-4=0
    Factor out a 2
    2(x²-x-2)
    You don't need quadratic equation for this one

    I mean you don't need to but you still could right?
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    (Original post by junayd1998)
    I mean you don't need to but you still could right?
    Yea given 2x²-2x-4=0
    a=2, b=2, c=4
    Just plug in the numbers you should get the same result but it's just stupid when you could factor
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    (Original post by Bybella)
    I need help with this
    quadratic equation 2x^2-2x-4=0
    Solve
    2x^2-2x-4=0
    (Original post by junayd1998)
    All you have to do is punch that information into the quadratic formula which would be at the front of an exam paper? Are you doing Edexcel GCSE?

    If you need more help lemme know
    hey buddy!
    (Original post by GUMI)
    2x²-2x-4=0
    Factor out a 2
    2(x²-x-2)
    2(x-2)(x+1)
    x=2 x=-1
    You don't need quadratic equation for this one
    going to latex this for the purpose of making it look nicer

     2x^2 -2x-4=0

    2\left(x^2 -x-2\right)=0

    2\left(x-2\right) \left(x+1\right)=0

    x=2\ x=-1
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    (Original post by GUMI)
    Yea given 2x²-2x-4=0
    a=2, b=2, c=4
    Just plug in the numbers you should get the same result but it's just stupid when you could factor

    I know but the reason I don't is because sometimes you are unable to factor so id rather just use the same method of punching in the digits.
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    (Original post by thefatone)
    hey buddy!

    going to latex this for the purpose of making it look nicer

     2x^2 -2x-4=0

    2\left(x^2 -x-2\right)=0

    2\left(x-2\right) \left(x+1\right)=0

    x=2\ x=-1
    Yeah guys im doing GCSE thank you
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    (Original post by thefatone)
    hey buddy!

    going to latex this for the purpose of making it look nicer

     2x^2 -2x-4=0

    2\left(x^2 -x-2\right)=0

    2\left(x-2\right) \left(x+1\right)=0

    x=2\ x=-1
    Also i have this question as well
    2x^2+4x= 16
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    (Original post by Bybella)
    Also i have this question as well
    2x^2+4x= 16
    ok so again move the 16 over to the left side by -16 to both sides
    take out 2 as a common factor then factorise
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    (Original post by thefatone)
    ok so again move the 16 over to the left side by -16 to both sides
    take out 2 as a common factor then factorise
    So does it become 2(x^2+4+x)=16
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    (Original post by Bybella)
    So does it become 2(x^2+4+x)=16
    no move 16 over first then take out common factor of 2
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    do i put an = 16 at the end of each thing i do like last time ?
    Also
    2(x+4)(x-2)
    (2x+8)(2x-4)????
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    (Original post by Bybella)
    do i put an = 16 at the end of each thing i do like last time ?
    Also
    2(x+4)(x-2)
    (2x+8)(2x-4)????
    what you've done is this
    2x²+4x=16
    -16 from both sides
    2x²+4x-16=0
    then this
    2(x+4)(x-2)

    (2x+8)(2x-4) <--- nothing wrong with this but it's not in its simplest form which mathematicians usually like to put things in
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    (Original post by thefatone)
    what you've done is this
    2x²+4x=16
    -16 from both sides
    2x²+4x-16=0
    then this
    2(x+4)(x-2)

    (2x+8)(2x-4) <--- nothing wrong with this but it's not in its simplest form which mathematicians usually like to put things in
    Then is it 4x^2+4?
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    (Original post by Bybella)
    Then is it 4x^2+4?
    oops my mistake are you solving for x?
    in which case the 2 shouldn't be in front of the x+4 bit i think
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    (Original post by Bybella)
    do i put an = 16 at the end of each thing i do like last time ?
    Also
    2(x+4)(x-2)
    (2x+8)(2x-4)????
    Just multiply the 2 by the first bracket, not both. So, you'd have (2x + 8)(x - 2), which then expands out to get the question.
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    (Original post by thefatone)
    oops my mistake are you solving for x?
    in which case the 2 shouldn't be in front of the x+4 bit i think
    What answer did you end up up getting for this ? I got X = 2 and X = -4
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    (Original post by Ishan_2000)
    Just multiply the 2 by the first bracket, not both. So, you'd have (2x + 8)(x - 2), which then expands out to get the question.

    What answer did you end up with? For x
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    (Original post by junayd1998)
    What answer did you end up with? For x
    So, we have:

    (2x+8)(x-2) = 0
    2x + 8 = 0 and
     x - 2 = 0

    I think you can solve those two equations for x now.
 
 
 
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