Réitigh 2x^2-4x+3=2(x-p)^2+q

$$2 x ^ { 2 } - 4 x + 3 = 2 ( x - p ) ^ { 2 } + q$$

Réitigh do q.

$$q=4px-4x-2p^{2}+3$$

Céimeanna Réitigh

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

$$2x^{2}-4x+3=2\left(x^{2}-2xp+p^{2}\right)+q$$

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

$$2x^{2}-4x+3=2x^{2}-4xp+2p^{2}+q$$

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

$$2x^{2}-4xp+2p^{2}+q=2x^{2}-4x+3$$

Bain 2x^{2} ón dá thaobh.

$$-4xp+2p^{2}+q=2x^{2}-4x+3-2x^{2}$$

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

$$-4xp+2p^{2}+q=-4x+3$$

Cuir 4xp leis an dá thaobh.

$$2p^{2}+q=-4x+3+4xp$$

Bain 2p^{2} ón dá thaobh.

$$q=-4x+3+4xp-2p^{2}$$

Réitigh do p. (complex solution)

$$p=\frac{\sqrt{4x^{2}-8x-2q+6}}{2}+x$$

$$p=-\frac{\sqrt{4x^{2}-8x-2q+6}}{2}+x$$

Réitigh do p.

$$p=\frac{\sqrt{4x^{2}-8x-2q+6}}{2}+x$$

$$p=-\frac{\sqrt{4x^{2}-8x-2q+6}}{2}+x\text{, }q\leq 2x^{2}-4x+3$$