Anyone take A Level Math first year? Or did?

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JBroll

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x^2 + 4x + c = (x + a)^2 + b

x^2 +4x +c = x^2 + 2ax + a^2 + b

4x + c = 2ax + a^2 + b

Since the coefficients of the x^1 and the x^0 must be equal (as x varies), 4x=2ax and thus 4=2a or a=2, and c=a^2+b or c=4+b

Given also that the equation x^2 + 4x + c = 0 has unequal real roots:
iii. find the range of possible values of c.

Using the discriminant (b^2-4ac, the part of the quadratic formula that we take the square root of), we can determine what values of c give real roots.

If the discriminant is 0, there is one repeated real root, because the solutions given by the quadratic equation are of the form (-b +/- sqrt (discriminant))/2a and if the discriminant is zero the solutions are -b/2a. If the discriminant is real and greater than 0, there are two real roots. If the discriminant is negative, the roots are imaginary because the square root of a negative number is a complex number. You are then solving for b^2-4ac > 0, I'll leave that to you.

More when I can get back on...

Jeff
 

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LiesThatBind

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Ah damn, sorry jeff, but i dont get any of that :(

What happened to the x^2 on the left at the on the 3rd step? :S
 

JBroll

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First, I expanded the (x+a)^2, and on the next line I subtracted x^2 from both sides.

Jeff
 

LiesThatBind

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oh yeah i see the 2 x^2 now thanks but then i dont get the next bit..
Your in america right?
 

LiesThatBind

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Since the coefficients of the x^1 and the x^0 must be equal (as x varies), 4x=2ax and thus 4=2a or a=2, and c=a^2+b or c=4+b

This bit.

And damn im looking to get some private tuition, but i dont know who to get here.
 

JBroll

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Okay... since those two functions are equal, for any value of x, it only stands to reason that the coefficients of the x^1 terms must be equal. If that isn't clear, consider the equation at x=0.

4(0) + c = 2a(0) + a^2 + b

c=a^2+b

Now, substitute a^2+b for C in the original equation:

4x + c = 2ax + c

Subtracting c from both sides,

4x=2ax or a=2

Plugging a=r back into c=a^2+b we have

c=4+b or b=c-4

I usually tutor for free so you can just keep IMing or PMing me if you have to.

Jeff
 

LiesThatBind

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Okay... since those two functions are equal, for any value of x, it only stands to reason that the coefficients of the x^1 terms must be equal. If that isn't clear, consider the equation at x=0.

4(0) + c = 2a(0) + a^2 + b

c=a^2+b

Now, substitute a^2+b for C in the original equation:

4x + c = 2ax + c

Subtracting c from both sides,

4x=2ax or a=2

Plugging a=r back into c=a^2+b we have

c=4+b or b=c-4

I usually tutor for free so you can just keep IMing or PMing me if you have to.

Jeff

Oh thanks, im pretty much up on Statistics (over subject we done), but core 1 i am forgetting how to do stuff, and now we have started core 2 (trigonometry etc), im struggling a bit on that and it just make me so stressed :scream:
 

JBroll

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I fucking hate statistics... or at least the statistics the university makes me take, as it's under the College of Business and they just make it all formula-memorizing with no real substance. Trig is annoying, but if you know the law of cosines you're set for life. Just like in Algebra, where everything is some bastardization of the quadratic formula, trig is going to deal with laws of cosines twisted into funky shit...

Jeff
 

LiesThatBind

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I fucking hate statistics... or at least the statistics the university makes me take, as it's under the College of Business and they just make it all formula-memorizing with no real substance. Trig is annoying, but if you know the law of cosines you're set for life. Just like in Algebra, where everything is some bastardization of the quadratic formula, trig is going to deal with laws of cosines twisted into funky shit...

Jeff

Yeah im gettin the sine and cosine rule, very slowly, and statistics is the easier of the 2 just very long lol.
 

LiesThatBind

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Exactly! It's not forcing you to think about it in the morning, like derivatives or integrals :lol:. Fucking maths are gonna melt my brain.

Haha, yeah, you can't forget anything about squares, cos all it is is area, perimeter, all sides are the same and all angles are the same. Haha.

:bowdown:
 

JBroll

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lol... yes, but there is a little more to trig than that ;)

What I was saying was that all of it could be pulled from that.

What I was saying about the laws of sines and cosines being the same is that it takes very little work to derive one from the other.

Jeff
 
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