Réitigh left(1-pright)x^2+5=2x

$$\left( 1-p \right) { x }^{ 2 } +5 = 2x$$

Réitigh do p.

$$p=\frac{x^{2}-2x+5}{x^{2}}$$

$$x\neq 0$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun 1-p a mhéadú faoi x^{2}.

$$x^{2}-px^{2}+5=2x$$

Bain x^{2} ón dá thaobh.

$$-px^{2}+5=2x-x^{2}$$

Bain 5 ón dá thaobh.

$$-px^{2}=2x-x^{2}-5$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(-x^{2}\right)p=-x^{2}+2x-5$$

Roinn an dá thaobh faoi -x^{2}.

$$\frac{\left(-x^{2}\right)p}{-x^{2}}=\frac{-x^{2}+2x-5}{-x^{2}}$$

Má roinntear é faoi -x^{2} cuirtear an iolrúchán faoi -x^{2} ar ceal.

$$p=\frac{-x^{2}+2x-5}{-x^{2}}$$

Roinn 2x-x^{2}-5 faoi -x^{2}.

$$p=1+\frac{5-2x}{x^{2}}$$

Réitigh do x. (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{5p-4}+1}{1-p}\text{; }x=\frac{-\sqrt{5p-4}+1}{1-p}\text{, }&p\neq 1\\x=\frac{5}{2}\text{, }&p=1\end{matrix}\right.$$

Réitigh do x.

$$\left\{\begin{matrix}x=\frac{\sqrt{5p-4}+1}{1-p}\text{; }x=\frac{-\sqrt{5p-4}+1}{1-p}\text{, }&p\neq 1\text{ and }p\geq \frac{4}{5}\\x=\frac{5}{2}\text{, }&p=1\end{matrix}\right.$$