Datrys 2x^2-4x+3=2(x-p)^2+q

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

Datrys ar gyfer q

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

Camau Ateb

Defnyddio'r theorem binomaidd \left(a-b\right)^{2}=a^{2}-2ab+b^{2} i ehangu'r \left(x-p\right)^{2}.

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

Defnyddio’r briodwedd ddosbarthu i luosi 2 â x^{2}-2xp+p^{2}.

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

Cyfnewidiwch yr ochrau fel bod yr holl dermau newidiol ar yr ochr chwith.

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

Tynnu 2x^{2} o'r ddwy ochr.

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

Cyfuno 2x^{2} a -2x^{2} i gael 0.

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

Ychwanegu 4xp at y ddwy ochr.

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

Tynnu 2p^{2} o'r ddwy ochr.

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

Datrys ar gyfer p (complex solution)

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

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

Datrys ar gyfer p

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

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