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

I misread your original post inside the root.
You should get a (b^2+b-1)^2 inside the (numerator) root ...

Reply 9

mqb2766

Original post by chatterclaw73

Expand (the numerator) under the root sign ...

Reply 10

chatterclaw73

I tried this.
x=b^2+b+1 +/- sq rt((b^4+3b^3-2b^2-2b+1)/2(b+1)

Reply 11

mqb2766

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 ...

(edited 5 years ago)

Reply 12

chatterclaw73

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

Reply 13

mqb2766

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?

Reply 14

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.

Original post by mqb2766

Which part don't you understand?

Reply 15

mqb2766

Original post by chatterclaw73

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.

Reply 16

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.

Reply 17

mqb2766

Original post by chatterclaw73

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.

Reply 18

chatterclaw73

I am a self learner doing C1. I this a normal question?

Original post by mqb2766

Reply 19

mqb2766

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?