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chatterclaw73
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#1
(b+1)^2-(b^2+b+1)+b=0. I have tried using the formula, but the discriminant ends up really messy. Any help would be greatly appreciated.
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mqb2766
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#2
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#2
(Original post by chatterclaw73)
(b+1)^2-(b^2+b+1)+b=0. I have tried using the formula, but the discriminant ends up really messy. Any help would be greatly appreciated.
(b+1)^2-(b^2+b+1)+b=0. I have tried using the formula, but the discriminant ends up really messy. Any help would be greatly appreciated.
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chatterclaw73
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#3
Sorry. the equation is (b+1)x^2-(b^2+b+1)x+b=0.
(Original post by mqb2766)
Its not a quadratic, there is no b^2 term?
Its not a quadratic, there is no b^2 term?
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mqb2766
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#4
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#4
(Original post by chatterclaw73)
Sorry. the equation is (b+1)x^2-(b^2+b+1)x+b=0.
Sorry. the equation is (b+1)x^2-(b^2+b+1)x+b=0.
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chatterclaw73
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#5
sq rt((b^2+b+1)^2-4(b+1)b)
Last edited by chatterclaw73; 3 years ago
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mqb2766
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#6
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#6
(Original post by chatterclaw73)
sq rt((b+1)^2-4(b+1)b)
sq rt((b+1)^2-4(b+1)b)
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chatterclaw73
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#7
I have thought about your answer but I do not understand. Sorry!
I get sq rt((b^2+b+1)^2)/4(b+1)^2+b/b+1) for discriminant!!
I get sq rt((b^2+b+1)^2)/4(b+1)^2+b/b+1) for discriminant!!
(Original post by mqb2766)
Don't forget you're dividing by (b+1) outside the root, so take that inside ...
Don't forget you're dividing by (b+1) outside the root, so take that inside ...
Last edited by chatterclaw73; 3 years ago
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mqb2766
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#8
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#8
(Original post by chatterclaw73)
I have thought about your answer but I do not understand. Sorry!
I get sq rt((b^2+b+1)^2)/4(b+1)^2+b/b+1) for discriminant!!
I have thought about your answer but I do not understand. Sorry!
I get sq rt((b^2+b+1)^2)/4(b+1)^2+b/b+1) for discriminant!!
You should get a (b^2+b-1)^2 inside the (numerator) root ...
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chatterclaw73
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#9
I will post all my working and hopefully make things a little clearer.
Solve (b+1)x^2-(b^2+b+1)x+b=0
x=b^2+b+1 +/-sq rt((b^2+b+1)^2-4(b+1)b)/2(b+1). This is as far as I get and then the algebra gets messy. (well mine does!!) Thankyou for taking the time to assist.
Solve (b+1)x^2-(b^2+b+1)x+b=0
x=b^2+b+1 +/-sq rt((b^2+b+1)^2-4(b+1)b)/2(b+1). This is as far as I get and then the algebra gets messy. (well mine does!!) Thankyou for taking the time to assist.
(Original post by mqb2766)
I misread your original post inside the root.
You should get a (b^2+b-1)^2 inside the (numerator) root ...
I misread your original post inside the root.
You should get a (b^2+b-1)^2 inside the (numerator) root ...
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mqb2766
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#10
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#10
(Original post by chatterclaw73)
I will post all my working and hopefully make things a little clearer.
Solve (b+1)x^2-(b^2+b+1)x+b=0
x=b^2+b+1 +/-sq rt((b^2+b+1)^2-4(b+1)b)/2(b+1). This is as far as I get and then the algebra gets messy. (well mine does!!) Thankyou for taking the time to assist.
I will post all my working and hopefully make things a little clearer.
Solve (b+1)x^2-(b^2+b+1)x+b=0
x=b^2+b+1 +/-sq rt((b^2+b+1)^2-4(b+1)b)/2(b+1). This is as far as I get and then the algebra gets messy. (well mine does!!) Thankyou for taking the time to assist.
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chatterclaw73
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#11
mqb2766
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#12
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#12
b^4 + 2b^3 + 3b^2 + 2b + 1 - 4b(b+1)
the first set of terms is essentially a trinomial expansion (r + s + t)^2, which is obviously similar to the binomial expansion (pascal's triangle). Then including the last term
b^4 + 2b^3 - b^2 - 2b + 1
This is a (trinomial) expansion of (b^2 + b - 1)^2.
Obviously, the root is now easy ...
Last edited by mqb2766; 3 years ago
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chatterclaw73
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#13
Not sure you should be putting obviously in your reply. I still do not understand you and now i feel stupid.
(Original post by mqb2766)
Just looking under the root sign for the moment (couple of typos), you had
b^4 + 2b^3 + 3b^2 + 2b + 1 - 4b(b+1)
the first set of terms is essentially a trinomial expansion (r + s + t)^2, which is obviously similar to the binomial expansion (pascal's triangle). Then including the last term
b^4 + 2b^3 - b^2 - 2b + 1
This is a (trinomial) expansion of (b^2 + b - 1)^2.
Obviously, the root is now easy ...
Just looking under the root sign for the moment (couple of typos), you had
b^4 + 2b^3 + 3b^2 + 2b + 1 - 4b(b+1)
the first set of terms is essentially a trinomial expansion (r + s + t)^2, which is obviously similar to the binomial expansion (pascal's triangle). Then including the last term
b^4 + 2b^3 - b^2 - 2b + 1
This is a (trinomial) expansion of (b^2 + b - 1)^2.
Obviously, the root is now easy ...
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mqb2766
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#14
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#14
(Original post by chatterclaw73)
Not sure you should be putting obviously in your reply. I still do not understand you and now i feel stupid.
Not sure you should be putting obviously in your reply. I still do not understand you and now i feel stupid.
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chatterclaw73
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#15
I will state the question again and show all my working in case I gave wrong info.
(B+1)x^2-(B^2+B+1)x+B=0. Solve for x.
(B^2+B+1)+/-Sq rt((B^2+B+1)^2-4(B+1)B)/2(B+1)
(B^2+B+1)+/- Sq rt(B^4+2B^3-B^2-2B+1)/2(B+1)
After this I do not understand how to simplify the discriminant further.
(B+1)x^2-(B^2+B+1)x+B=0. Solve for x.
(B^2+B+1)+/-Sq rt((B^2+B+1)^2-4(B+1)B)/2(B+1)
(B^2+B+1)+/- Sq rt(B^4+2B^3-B^2-2B+1)/2(B+1)
After this I do not understand how to simplify the discriminant further.
(Original post by mqb2766)
Which part don't you understand?
Which part don't you understand?
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mqb2766
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#16
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#16
(Original post by chatterclaw73)
I will state the question again and show all my working in case I gave wrong info.
(B+1)x^2-(B^2+B+1)x+B=0. Solve for x.
(B^2+B+1)+/-Sq rt((B^2+B+1)^2-4(B+1)B)/2(B+1)
(B^2+B+1)+/- Sq rt(B^4+2B^3-B^2-2B+1)/2(B+1)
After this I do not understand how to simplify the discriminant further.
I will state the question again and show all my working in case I gave wrong info.
(B+1)x^2-(B^2+B+1)x+B=0. Solve for x.
(B^2+B+1)+/-Sq rt((B^2+B+1)^2-4(B+1)B)/2(B+1)
(B^2+B+1)+/- Sq rt(B^4+2B^3-B^2-2B+1)/2(B+1)
After this I do not understand how to simplify the discriminant further.
B^4 + 2B^3 - B^2 - 2B + 1
which is
(B^2 + B - 1)^2
Take the square root, and put it with the other parts of the discriminant.
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chatterclaw73
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#17
Ok thanks for your patience and understanding. How would I know that the discriminant could be factorised in this way. Is it just experience? you are obviously way more advanced than I am! thanks again.
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mqb2766
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#18
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#18
(Original post by chatterclaw73)
Ok thanks for your patience and understanding. How would I know that the discriminant could be factorised in this way. Is it just experience? you are obviously way more advanced than I am! thanks again.
Ok thanks for your patience and understanding. How would I know that the discriminant could be factorised in this way. Is it just experience? you are obviously way more advanced than I am! thanks again.
You'd be expected to know the
1 2 1
Binomial expansion which pops up in Pascal's triangle. It's also worth noting
(r + s)^2 - 4rs = (r - s)^2
Which is a simplified version of what you have.
Here you have something very similar in terms of coefficient values. The trinomial
(r + s + t)^2 = r^2 + 2rs + 2rt + s^2 + 2st + t^2
This is very similar to the coefficients in your case and a bit of fiddling with the signs gets you there.
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chatterclaw73
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#19
I am a self learner doing C1. I this a normal question?
(Original post by mqb2766)
Again post 12.
You'd be expected to know the
1 2 1
Binomial expansion which pops up in Pascal's triangle. It's also worth noting
(r + s)^2 - 4rs = (r - s)^2
Which is a simplified version of what you have.
Here you have something very similar in terms of coefficient values. The trinomial
(r + s + t)^2 = r^2 + 2rs + 2rt + s^2 + 2st + t^2
This is very similar to the coefficients in your case and a bit of fiddling with the signs gets you there.
Again post 12.
You'd be expected to know the
1 2 1
Binomial expansion which pops up in Pascal's triangle. It's also worth noting
(r + s)^2 - 4rs = (r - s)^2
Which is a simplified version of what you have.
Here you have something very similar in terms of coefficient values. The trinomial
(r + s + t)^2 = r^2 + 2rs + 2rt + s^2 + 2st + t^2
This is very similar to the coefficients in your case and a bit of fiddling with the signs gets you there.
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mqb2766
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#20
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#20
I really don't know where you got the question from, but my gut feeling is that factorising quartics into a quadratic square would be at the challenging end.
(Original post by chatterclaw73)
I am a self learner doing C1. I this a normal question?
I am a self learner doing C1. I this a normal question?
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