#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.
0
3 years ago
#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.
Its not a quadratic, there is no b^2 term?
1
#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?
0
3 years ago
#4
(Original post by chatterclaw73)
Sorry. the equation is (b+1)x^2-(b^2+b+1)x+b=0.
I would have thought the discriminant would be ok, what did you get?
0
#5
sq rt((b^2+b+1)^2-4(b+1)b)
Last edited by chatterclaw73; 3 years ago
0
3 years ago
#6
(Original post by chatterclaw73)
sq rt((b+1)^2-4(b+1)b)
Don't forget you're dividing by (b+1) outside the root, so take that inside ...
0
#7

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 ...
Last edited by chatterclaw73; 3 years ago
0
3 years ago
#8
(Original post by chatterclaw73)

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 ...
0
#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.

(Original post by mqb2766)
You should get a (b^2+b-1)^2 inside the (numerator) root ...
0
3 years ago
#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.
Expand (the numerator) under the root sign ...
0
#11
I tried this.
x=b^2+b+1 +/- sq rt((b^4+3b^3-2b^2-2b+1)/2(b+1)
0
3 years ago
#12
(Original post by chatterclaw73)
I tried this.
x=b^2+b+1 +/- sq rt((b^4+3b^3-2b^2-2b+1)/2(b+1)
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 ...
Last edited by mqb2766; 3 years ago
0
#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 ...
0
3 years ago
#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.
Which part don't you understand?
0
#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.
(Original post by mqb2766)
Which part don't you understand?
0
3 years ago
#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.
See post 12. Under the root sign you have
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.
0
#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.
0
3 years ago
#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.
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.
0
#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.
0
3 years ago
#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?
0
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