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...because I'm too dim to do my maths. I'm sure it's simple really but I have:
2x(squared) + 7x - 2 = 0
When you replace x with either -2 or 1/4, the equation is correct but I can't seem to prove this using the quadratic formula or by trying to solve the equation by using brackets. Are there any other ways of proving this?
Any help would be much appreciated.
2x(squared) + 7x - 2 = 0
When you replace x with either -2 or 1/4, the equation is correct but I can't seem to prove this using the quadratic formula or by trying to solve the equation by using brackets. Are there any other ways of proving this?
Any help would be much appreciated.
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munn wrote:
> Are you dimwits forgetting that -4 squared = 164?
>
> -2 does work.
>
> But only if 2x is in brackets
>
> 2xsquared = 2 * xsquared
>
> (2x)squared = 2*x all squared
Exactly... nice use of the edit button.
> Are you dimwits forgetting that -4 squared = 164?
>
> -2 does work.
>
> But only if 2x is in brackets
>
> 2xsquared = 2 * xsquared
>
> (2x)squared = 2*x all squared
Exactly... nice use of the edit button.
munn wrote:
> Are you dimwits forgetting that -4 squared = 164?
>
> -2 does work.
Actually it just equals 16 :p but the only way that could be relevant is if the question is if you're working out (2*-2)^2. Which you shouldn't be, unless you've written the question wrong...
> Are you dimwits forgetting that -4 squared = 164?
>
> -2 does work.
Actually it just equals 16 :p but the only way that could be relevant is if the question is if you're working out (2*-2)^2. Which you shouldn't be, unless you've written the question wrong...
Are you dimwits forgetting that -4 squared = 164?
-2 does work.
But only if 2x is in brackets
2xsquared = 2 * xsquared
(2x)squared = 2*x all squared
-2 does work.
But only if 2x is in brackets
2xsquared = 2 * xsquared
(2x)squared = 2*x all squared
JFH wrote:
> The answers were in the back of the book, I wasn't meant to know that
> x = -2 or 1/4. The problem is getting to those two answers from
>
> 2x(squared) + 7x - 2 = 0
>
> without it looking like I looked in the back. And for those of you
> don't believe they are the right answe I'll prove -2 now.
>
> 2*-2 = -4
>
> -4*-4 = 16
>
> 7*-2 = -14
>
> 16 - 14 - 2 = 0
well, I can't be bothered to work out what you did, but I'm pretty sure this proves -2 doesn't work:
f(x) = 2x^2 + 7x - 2
f(-2) = 2(-2)^2 + 7(-2) - 2
= 2(4) + 7(-2) - 2
= 8 - 14 - 2
= -8
(the f(x) just means that function, and f(-2) means subbing -2 into that function, so for it to prove that -2 is an answer, the equation should come out as 0)
Just stick it through the formula and see what you get out...
There's something that calculates it automatically here [URL]http://www.1728.com/quadratc.htm[/URL] so you don't have to write out about 200 lines (slight exaggeration) to find out you went wrong on the second line.
Ohh, I see, you've multiplied 2 by -2 before squaring the -2. I think unless it's all in a bracket like this; (2x)^2 then you should sqaure the x before doing any multiplying. At least... that's what we were taught.
I mean, the squared symbol applies only to the term it's put onto, if it's 2x^2, it applies only t the x. However, if it's (2x)^2 then the whole thing is hit by it.
> The answers were in the back of the book, I wasn't meant to know that
> x = -2 or 1/4. The problem is getting to those two answers from
>
> 2x(squared) + 7x - 2 = 0
>
> without it looking like I looked in the back. And for those of you
> don't believe they are the right answe I'll prove -2 now.
>
> 2*-2 = -4
>
> -4*-4 = 16
>
> 7*-2 = -14
>
> 16 - 14 - 2 = 0
well, I can't be bothered to work out what you did, but I'm pretty sure this proves -2 doesn't work:
f(x) = 2x^2 + 7x - 2
f(-2) = 2(-2)^2 + 7(-2) - 2
= 2(4) + 7(-2) - 2
= 8 - 14 - 2
= -8
(the f(x) just means that function, and f(-2) means subbing -2 into that function, so for it to prove that -2 is an answer, the equation should come out as 0)
Just stick it through the formula and see what you get out...
There's something that calculates it automatically here [URL]http://www.1728.com/quadratc.htm[/URL] so you don't have to write out about 200 lines (slight exaggeration) to find out you went wrong on the second line.
Ohh, I see, you've multiplied 2 by -2 before squaring the -2. I think unless it's all in a bracket like this; (2x)^2 then you should sqaure the x before doing any multiplying. At least... that's what we were taught.
I mean, the squared symbol applies only to the term it's put onto, if it's 2x^2, it applies only t the x. However, if it's (2x)^2 then the whole thing is hit by it.
The answers were in the back of the book, I wasn't meant to know that x = -2 or 1/4. The problem is getting to those two answers from
2x(squared) + 7x - 2 = 0
without it looking like I looked in the back. And for those of you don't believe they are the right answe I'll prove -2 now.
2*-2 = -4
-4*-4 = 16
7*-2 = -14
16 - 14 - 2 = 0
2x(squared) + 7x - 2 = 0
without it looking like I looked in the back. And for those of you don't believe they are the right answe I'll prove -2 now.
2*-2 = -4
-4*-4 = 16
7*-2 = -14
16 - 14 - 2 = 0
Paradox: wrote:
> Off the top of my head Gamezfreak and Chipseh are pretty good
> mathmeticians.
Hehe I wish!
I'm rubbish at Maths, I got a D at GCSE.
My college asked for a C minimum for me to do the Computing course, obviously I didn't meet the requirements, but I just went to the lesson anyway and my computing teacher had only found out about 15 lessons later.
And guess what I got in AS Computing...
...A.
> Off the top of my head Gamezfreak and Chipseh are pretty good
> mathmeticians.
Hehe I wish!
I'm rubbish at Maths, I got a D at GCSE.
My college asked for a C minimum for me to do the Computing course, obviously I didn't meet the requirements, but I just went to the lesson anyway and my computing teacher had only found out about 15 lessons later.
And guess what I got in AS Computing...
...A.
Yep.
But since you already have your two values, it's easiest to just sub them in and show that they get the equation to zero.
Of course, the problem here is that doesn't happen.
And apparently your two x values for that equation should be 0.26556 and -3.7656.
But since you already have your two values, it's easiest to just sub them in and show that they get the equation to zero.
Of course, the problem here is that doesn't happen.
And apparently your two x values for that equation should be 0.26556 and -3.7656.
If I remember correctly, completing the square is another proof method.
Worth a shot.
I also suck at maths though.
One of the clevery computery-minded people can probably help, they're mostly good at maths.
Off the top of my head Gamezfreak and Chipseh are pretty good mathmeticians.
I also suck at maths though.
One of the clevery computery-minded people can probably help, they're mostly good at maths.
Off the top of my head Gamezfreak and Chipseh are pretty good mathmeticians.
I could be missing something here, but from where I'm standing, neither -2 or 1/4 will satisfy that equation.
You're trying to prove x = 1/4 and x = -2 ?
You're trying to prove x = 1/4 and x = -2 ?
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