Solve x^2+3x+4=(x+a)^2+b

$$x ^ { 2 } + 3 x + 4 = ( x + a ) ^ { 2 } + b$$

Solve for b

$$b=-a\left(2x+a\right)+3x+4$$

Solution Steps

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

$$x^{2}+3x+4=x^{2}+2xa+a^{2}+b$$

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

$$x^{2}+2xa+a^{2}+b=x^{2}+3x+4$$

Subtract x^{2} from both sides.

$$2xa+a^{2}+b=x^{2}+3x+4-x^{2}$$

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

$$2xa+a^{2}+b=3x+4$$

Subtract 2xa from both sides.

$$a^{2}+b=3x+4-2xa$$

Subtract a^{2} from both sides.

$$b=3x+4-2xa-a^{2}$$

Solve for a (complex solution)

$$a=-\left(\sqrt{x^{2}+3x-b+4}+x\right)$$

$$a=\sqrt{x^{2}+3x-b+4}-x$$

Solve for a

$$a=-\left(\sqrt{x^{2}+3x-b+4}+x\right)$$

$$a=\sqrt{x^{2}+3x-b+4}-x\text{, }b\leq x^{2}+3x+4$$