quadratic formula and simplification

Question

Consider the following equation:

$$x=\frac{(p+x(1-p))\cdot c}{1-(p+x(1-p))(1-c)}$$

Solution should be $$x=\frac{cp}{(1-c)(1-p)}$$

I tried to rewrite the first equation in such that:$$x^2[c(1-p )-(1-p)] + x[2cp - c + 1-p]-pc=0$$

Applying quadratic formula leads to something, since the term under the root cannot be simplified.

Is there another way to compute that?

Raffaele · Accepted Answer · 2020-11-10T10:42:19.913

1

$$x^2 (c+p-cp-1)-x (c+p-2 c p-1)-c p=0$$ Discriminant is $$D=(c+p-2 c p-1)^2-4cp (c+p-cp-1)=c^2+p^2+1+2 c p-2 c-2 p=(c+p-1)^2$$ so $$x=\frac{(c+p-2 c p-1)\pm(c+p-1)}{c+p-cp-1};\;c\ne 1\land p\ne 1$$ denominator $p-cp+c-1=p(1-c)-1(1-c)=(1-c)(p-1)$ $$x_1=\frac{c+p-2 c p-1-c-p+1}{2(1-c)(p-1)}=\frac{-cp}{(1-c)(p-1)}=\frac{cp}{(1-c)(1-p)}$$ and $$x_2=\frac{c+p-2 c p-1+c+p-1)}{2(1-c)(p-1)}=\frac{-2 (c-1) (p-1)}{2(1-c)(p-1)}=1$$ if $p=1\lor c=1$ then $x=1$

edited Nov 10 '20 at 10:42

answered Nov 09 '20 at 21:44

Raffaele

26,371

Thank you very much. I got my mistake:) – Tim Nov 09 '20 at 21:50
BTW do you know $p\ne 1, c \ne 1$? If not, you need to consider the cases where $p=1$ or $c=1$. – Neat Math Nov 09 '20 at 22:25
@NeatMath Thank you! i edited my answer including what I forgot – Raffaele Nov 09 '20 at 22:33
@Raffaele. I noticed that your result differs a litte bit. For $x_1$ you have (1-c)(p-1) in the denominator. I need (1-c)(1-p)? – Tim Nov 10 '20 at 10:25
Is there any $-$ sign missing? – Tim Nov 10 '20 at 10:34
I have edited my answer. There was a sign mistake actually. – Raffaele Nov 10 '20 at 10:43

quadratic formula and simplification

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