Pearl1323
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i am having problem with one question on my exercise ,
Q the roots of the equation ax^2+bx+c=0 differ by 1. prove that b^2-a^2-4ac=0
pls help!!!
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Kevin De Bruyne
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(Original post by Pearl1323)
i am having problem with one question on my exercise ,
Q the roots of the equation ax^2+bx+c=0 differ by 1. prove that b^2-a^2-4ac=0
pls help!!!
What have you tried so far?
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Pearl1323
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um... i try k(x-alpha)(x-beta) which i know is not quite right
i don't get what differ by 1 actually want me to do1!!
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tiny hobbit
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(Original post by Pearl1323)
i am having problem with one question on my exercise ,
Q the roots of the equation ax^2+bx+c=0 differ by 1. prove that b^2-a^2-4ac=0
pls help!!!
Call your roots alpha and alpha + 1
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Kevin De Bruyne
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(Original post by Pearl1323)
um... i try k(x-alpha)(x-beta) which i know is not quite right
i don't get what differ by 1 actually want me to do1!!
If you quote in the thread users will be able to see that you have replied

What are the two solutions to the quadratic equation in the question?
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oniisanitstoobig
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Look into the two roots from the quadratic equation & see what it means for the two roots to differ by 1.
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TeeEm
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(Original post by homeworkbazaar)
if you need some help for homework help. i think this is very good site homeworkbazaar.co.uk
nice plug.
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DFranklin
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(Original post by oniisanitstoobig)
Look into the two roots from the quadratic equation & see what it means for the two roots to differ by 1.
Probably not the expected approach, but undoubtedly the easiest here...
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Pearl1323
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so it should be k(x-alpha) (x-beta-1) like that ?
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DFranklin
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(Original post by Pearl1323)
so it should be k(x-alpha) (x-beta-1) like that ?
No. If you want to do it this way, then the general form is going to be

k(x-alpha)(x-beta).

But since you know the roots differ by 1, you can (without loss of generality) assume beta = alpha+1, and so rewrite as

k(x-alpha)(x-alpha-1).

However, for this problem the method suggested by oniisanitstoobig is going to be easier.
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Pearl1323
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what is the other method? ><
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TeeEm
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(Original post by Pearl1323)
so it should be k(x-alpha) (x-beta-1) like that ?
there are 2 ways of doing this

verification (very easy)
K(x-alpha)(x-alpha-1) = 0 (the k is not actually needed)
multiply out into a three term quadratic in x
plug the coefficients into the expression given and done

properly (a bit of algebra)
alpha + (alpha + 1) = -b/a
alpha(alpha + 1 ) = c/a
eliminate alpha between the 2 equations
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DFranklin
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(Original post by TeeEm)
there are 2 ways of doing this

verification (very easy)
K(x-alpha)(x-alpha-1) = 0 (the k is not actually needed)
multiply out into a three term quadratic in x
plug the coefficients into the expression given and done

properly (a bit of algebra)
alpha + (alpha + 1) = -b/a
alpha(alpha + 1 ) = c/a
eliminate alpha between the 2 equations
I've now posted this 3 times, but in this case, I think it's pretty clear the shortest answer is to just use the quadratic formula.
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TeeEm
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(Original post by DFranklin)
I've now posted this 3 times, but in this case, I think it's pretty clear the shortest answer is to just use the quadratic formula.
it is indeed, but conceptually for a student the hardest approach.
Once conceived indeed the workings are 2 lines..
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DFranklin
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(Original post by TeeEm)
it is indeed, but conceptually for a student the hardest approach.
Once conceived indeed the workings are 2 lines..
I dunno, I think it's the obvious approach (it's what I'd expect 90% of non-FM A-level students to do). It only becomes unobvious when you're in the middle of a topic where the answer to everything is to express x^2+bx+c in terms of alpha and beta.
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TeeEm
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(Original post by DFranklin)
I dunno, I think it's the obvious approach (it's what I'd expect 90% of non-FM A-level students to do). It only becomes unobvious when you're in the middle of a topic where the answer to everything is to express x^2+bx+c in terms of alpha and beta.
agreed.

(most students which have knowledge of relationships between roots and coefficients as soon as "... roots differ by 1 ...", autopilot takes over )
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Pearl1323
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thanks for the help ><
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Pearl1323
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#18
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so that solution is :
(2alpha+1)^2-(alpha^2+alpha)-4(alpha^2+alpha) ?
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