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

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

Solve for q

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

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-p\right)^{2}.

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

Use the distributive property to multiply 2 by x^{2}-2xp+p^{2}.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2x^{2} from both sides.

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

Combine 2x^{2} and -2x^{2} to get 0.

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

Add 4xp to both sides.

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

Subtract 2p^{2} from both sides.

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

Solve for p (complex solution)

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

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

Solve for p

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

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