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

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
Reply 2
Avatar for GUMI
GUMI
19
2x²-2x-4=0
Factor out a 2
2(x²-x-2)
You don't need quadratic equation for this one
(edited 8 years ago)
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?
Reply 4
Avatar for GUMI
GUMI
19
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
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! :biggrin:
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

2x2−2x−4=0 2x^2 -2x-4=0

2(x2−x−2)=02\left(x^2 -x-2\right)=0

2(x−2)(x+1)=02\left(x-2\right) \left(x+1\right)=0

x=2 x=−1x=2\ x=-1
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.
Reply 7
Avatar for Bybella
Bybella
OP
8
Original post by thefatone
hey buddy! :biggrin:

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

2x2−2x−4=0 2x^2 -2x-4=0

2(x2−x−2)=02\left(x^2 -x-2\right)=0

2(x−2)(x+1)=02\left(x-2\right) \left(x+1\right)=0

x=2 x=−1x=2\ x=-1


Yeah guys im doing GCSE thank you
Reply 8
Avatar for Bybella
Bybella
OP
8
Original post by thefatone
hey buddy! :biggrin:

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

2x2−2x−4=0 2x^2 -2x-4=0

2(x2−x−2)=02\left(x^2 -x-2\right)=0

2(x−2)(x+1)=02\left(x-2\right) \left(x+1\right)=0

x=2 x=−1x=2\ x=-1


Also i have this question as well
2x^2+4x= 16
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
Reply 10
Avatar for Bybella
Bybella
OP
8
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
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
Reply 12
Avatar for Bybella
Bybella
OP
8
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)????
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
Reply 14
Avatar for Bybella
Bybella
OP
8
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?
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
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.
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
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
Original post by junayd1998
What answer did you end up with? For x


So, we have:

(2x+8)(x−2)=0(2x+8)(x-2) = 0
2x+8=02x + 8 = 0 and
x−2=0 x - 2 = 0

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

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