Réitigh 4x^2+12x=4(x+p)^2-q

$$4 x ^ { 2 } + 12 x = 4 ( x + p ) ^ { 2 } - q$$

Réitigh do q.

$$q=4\left(p\left(2x+p\right)-3x\right)$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(x+p\right)^{2} a leathnú.

$$4x^{2}+12x=4\left(x^{2}+2xp+p^{2}\right)-q$$

Úsáid an t-airí dáileach chun 4 a mhéadú faoi x^{2}+2xp+p^{2}.

$$4x^{2}+12x=4x^{2}+8xp+4p^{2}-q$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$4x^{2}+8xp+4p^{2}-q=4x^{2}+12x$$

Bain 4x^{2} ón dá thaobh.

$$8xp+4p^{2}-q=4x^{2}+12x-4x^{2}$$

Comhcheangail 4x^{2} agus -4x^{2} chun 0 a fháil.

$$8xp+4p^{2}-q=12x$$

Bain 8xp ón dá thaobh.

$$4p^{2}-q=12x-8xp$$

Bain 4p^{2} ón dá thaobh.

$$-q=12x-8xp-4p^{2}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$-q=-8px+12x-4p^{2}$$

Roinn an dá thaobh faoi -1.

$$\frac{-q}{-1}=\frac{-8px+12x-4p^{2}}{-1}$$

Má roinntear é faoi -1 cuirtear an iolrúchán faoi -1 ar ceal.

$$q=\frac{-8px+12x-4p^{2}}{-1}$$

Roinn 12x-8xp-4p^{2} faoi -1.

$$q=8px-12x+4p^{2}$$

Réitigh do p. (complex solution)

$$p=-\frac{\sqrt{4x^{2}+12x+q}}{2}-x$$

$$p=\frac{\sqrt{4x^{2}+12x+q}}{2}-x$$

Réitigh do p.

$$p=-\frac{\sqrt{4x^{2}+12x+q}}{2}-x$$

$$p=\frac{\sqrt{4x^{2}+12x+q}}{2}-x\text{, }q\geq -4x^{2}-12x$$